Probability And Statistics Solutions

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 3 Discrete Random Variables and Probability Distributions

Page 127 Problem 1 Answer

Given : Number of specimens of basaltic rock=10

Number of specimens of granite=10

The total specimens=20

The laboratory assistant selects randomly 15 out of these 20 specimens.

To find: We will find the pmf of the number of granite specimens selected for analysis.

We will suppose that the successes is represented as(S) and failures is represented as (F) in the population.

Also, the population and M successes and N−M failures.

Now, taking X=number of successes in a random sample size,n

The pmf will be p(x,nM,N)=(M/x)(N−M/n−x)/(N/n)

For all integers,x , max{0,n−M+m}≤x≤min{n,M}

Here, the success is a granite and failure is basaltic rock. The specimens are10

We have M=10,N=20,So,N−M=10

Suppose that we have 4 specimens of granite then it is sure that the number of specimens of basaltic rock will be more than 10 but since the total number of specimens of basaltic rock is 10.

We conclude that it is not possible for it to have more than 10 specimens.

max{0,n−N+M}≤x≤min{n,M};

max{0,15−20+10}≤x≤min{15,10}; 5≤x≤10

We know X has a hyper geometric distribution with parameters,

n=15,

N=20,

M=10

Hence, the pmf of X is – P(X=x) = p(x;15,10,20)

p(x;15,10,20)=(10/x)(20−10/15−x)/(20/15)

for x ϵ{5.6.7.8.9.10}

​We will calculate the pmf at given values as follows:

At  x = 5,pmf is 0.0163

At x=6, pmf is 0.1354

Atx=7, pmf is 0.3483

Atx=8, pmf is 0.3483

Atx=9, pmf is 0.1354

At x=10, pmf is 0.0163 .

The required pmf are given below:

x=5 pmf is 0.0163

​x=6 pmf is 0.1354

x=7 pmf is 0.3483​

x=8 pmf is 0.3483​

x=9 pmf is 0.1354​

x=10 pmf is 0.0163​

Page 127 Problem 2 Answer

Given : The number of specimens in granite =10

The number of specimens in basaltic rock =10

Total number of specimens are 20.

The laboratory assistant  randomly select 15 of the specimens for analysis.

We need to find the probability that all specimens of one of the two types of rock are selected for analysis.

We will find the pmf.

Hence, the probability that all specimens of one kind are taken from the sample.

We will denote this by Y-P(Y)=P(X=5)+P(X=10)

We obtain:

N=20,

n=15,

M=10

P(X=5 )=p(5;15,10,20)

p(5;15,10,20)=(10/5)(20−10/15−5)/(20/15)

P(X=5 )=p(5;15,10,20)

p(5;15,10,20)= 0.0163

P(X=10 )=p(10;15,10,20)

p(10;15,10,20)=(10/5)(20−10/15−5)/(20/15)

P(X=10 )=p(10;15,10,20)

p(10;15,10,20)=0.0163

​Using the obtained values, we get :

P(Y)=P(X=5)+P(X=10)

P(Y)=0.0163+0.0163

P(Y)=0.0326

We obtain the probability that all specimens of one of the two types of rock are selected for analysis as: 0.0326

Page 127 Problem 3 Answer

Given -N=20

M=10

n=15

​We need to determine probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value.

To get the answer, we first have to find its mean.

We obtain the mean of the random variable X

as follows:

E(X)= nM/N

E(X)= 15⋅10/20

E(X)=7.5​

Standard deviation = √Variance

So, to get the value of standard deviation, we need to find the variance.

We get:

V(X)= (N−n/N−1)⋅nM/N⋅(1−M/N)

V(X)= (20−15/20−1)⋅15⋅10/20⋅(1−10/20)

V(X)= (5/19)⋅7.5⋅(1−1/2)

V(X)= (5/19)⋅7.5⋅(1/2)

V(X)= 0.98684

​Hence standard deviation is 0.9934.

Now, we will determine the desired probability.

We note that:

E(X)−Standard deviation<X<E(X)+Standard Deviation

7.5 − 0.9934 <X<7.5+0.9934/6.5066<X<8.4934

​We obtain : pmf ofP(X=7)andP(X=8)-P(X=7)=p(7;15,10,20)

p(7;15,10,20)=(10/7)(20−10/15−7)/(20/15)

P(X=7)=p(7;15,10,20)

p(7;15,10,20)= 0.3483

P(X=8)=p(8;15,10,20)

p(8;15,10,20)=(10/8)(20−10/15−8)/(20/15)

P(X=8)=p(8;15,10,20)

p(8;15,10,20)= 0.3483                                      ​

P(6.5066)<X<P(8.4934) = P(7≤X≤8)

P(6.5066)<X<P(8.4934) = P(X=7)+ P(X=8)

P(6.5066)<X<P(8.4934) = 0.3483 + 0.3483

P(6.5066)<X<P(8.4934) = 0.6966

​The required probability is: P(6.5066)<X<P(8.4934) = 0.6966.

Page 127 Problem 4 Answer

Given that the total number of industrial firms =50.Out of 50 firms, the selected firms for inspection =10.Therefore, N = 50

n = 10

M = 15

We will find the pmf of the number of firms visited by the inspector that are in violation of at least one regulation if 15 of the firms are actually violating at least one regulation.

We have sample of 10 is more than 10 % of the population so, we cannot use the Binomial distribution.

We assume X be a random variable that denote the number of firms that violate at least one regulation from 10 randomly selected firms out of 50 of which 15 violate at least one regulation.

We obtain the pmf as follows:

P(X=x)=h(x;n,M,N)=(M/x)(N−M/n−x)/(N/n)

P(X=x)=h(x;10,15,50)=(15/x)(50−15/10−x)/(50/n)

P(X=x)=h(x;10,15,50)=(15/x)(35/10−x)/(50/n)

We obtain the required pmf as:

P(X=x)=(15/x)(35/10−x)/(50/n)

Page 127 Problem 5 Answer

Given -The total numbers of the firms in the area= 500.

Also,150 are in violation.We will approximate the probability mass function by a simple pmf.

We will use the pmf of binomial distribution.

As N is very large, so X is approximated to a Binomial random variable with probability p.

We get:

p = 150/500

p = 0.3

Here n = 10

So, the pmf of X is b(x;10,0.3).

Now,

b(x;n,p) = b(x;10,0.3)

b(x;10,0.3) =(10/x)(0.3)x(1−0.3)10−x .

​We obtain the pmf as :

b(x;10,0.3) =(10/x)(0.3)x(1−0.3)10−x .

Page 127 Problem 6 Answer

Given -The total number of industrial firms =.50

Out of 50 firms, the selected firms for inspection =10

N=50

n=10

M=15​

We defineX= the number among the 10 visited that are in violation.

We need to get the mean and variance both for the exact pmf and the approximating pmf.

For the given data, the probability of successful trials is calculated as follows:

p = 150/500

p = 150/500

p = 3/10

p = 0.3

Now, mean of a hypergeometric distribution is the number of draws multiplied by the number of successes, divided by the population size.

E(X) = nM/N

E(X) = 10×15/50

E(X) = 3 ​

Now, we calculate the variance of the hypergeometric distribution :

V(X) = nM/N× N−M/N×N−n/N−1

V(X) = 10×15/50× 50−15/50×50−10/50−1

V(X) = 3× 0.7×0.816

V(X) = 1.71428 (approx. 1.7143)

The approximated pmf is calculated as: We know that the mean of a binomial distribution is the product of the sample size n and the probability p.

E(X) = np

E(X) = 10⋅(0.3)

E(X) = 3 .

We will now obtain variance :

V(X) = npq

V(X) = np(1 − p)

V(X) = 10×(0.3)(1 − 0.3)

V(X) = 2.1 .

​For exact pmf : E(X) is 3

V(X) is 1.74128

For approximated pmf:E(X) is 3

V(X) is 2.1

Page 128 Problem 7 Answer

This exercise is based on Negative Binomial DistributionAs every family needs two female children, the total number of female children in three families,

r=2+2+2

=6

Now, the probability of male children born is defined as success and the probability of female children born is defined as failure

The probability of success and failure will be equal for each independent trial

P(female) = P( male)

=1/2

=0.5

We can get the required expectation as follows:

E(X)=r(1−p)/p

We will calculate the probability mass function of X as follows:

P(X)=nb(x;r,p)

Putting the values of r=6 And p=0.5

we obtain : P(X)=nb(x;6,0.5)

We know: E(X)=r(1−p)/p

Putting the values of r and p given in the question

we get,E(X)=6(1−0.5)/0.5

=6(0.5)/0.5=6

For each brother expected we want to have two female children.

The total number of female children for three families will be=3×2=6

The expected number of male children born to each other is the same as we calculated from mathematical expression of E(X)

The pmf of X= the number of male children born to brothers is n b(x;6,0.5)

The expected value for the number of male children born to each other calculated using expression and calculated directly is the same i.e.6

Page 128 Problem 8 Answer

Given – The drought length Y is the number of consecutive time intervals in which the water supply remains below a critical value Y0, preceded by and followed by periods in which the supply exceeds this critical value (a surplus).

The cited paper proposes a geometric distribution with forp=0.409  this random variable.

We will find the probability that a drought lasts exactly 3 intervals and at most 3 intervals.

We will use geometric distribution.

using the given data, we obtain the probability that a drought lasts exactly for 3 intervals as follows :

P(Y=3)=(1−0.409)³×0.409

= 0.844427

= 0.0844

​Now, we calculate the probability that a drought lasts at most3

intervals as follows:

=(1−0.409)0×0.409+(1−0.409)¹×0.409+(1−0.409)²×0.409+(1−0.409)³×0.409

=0.409+0.2417+0.1429+0.0844

=0.8780

​We obtain:

P(Y=3)=0.0844.

P(Y≤3)=0.8780.

Page 128 Problem 9 Answer

For the given data, we need to find the probability that the length of a drought exceeds its mean value by at least one standard deviation.

We note that the mean is given as μ=1−p/p

The standard deviation is given as σ=√1−p/p²

Firstly, we will calculate the mean of random variable Y as follows:

μ=1−p/p

=1−0.409/0.409

=1.445

Now, we will calculate the standard Deviation of the random variable Y as follows:

σ=√1−p/p²

=√1−0.409/(0.409)²

=1.88

​Finally, we calculate the required probability as follows:

P(Y≥μ+σ)=P(Y≥1.445+1.88)

=P(Y>3.325)

=P(Y>3)

=1−P(Y≤3)

=1−P(Y=0)−P(Y=1)−P(Y=2)+P(Y=3)

=1−{(1−0.409)0×0.409+(1−0.409)¹×0.409+(1−0.409)²×0.409+(1−0.409)³×0.409}

=1−{0.409+0.2417+0.1429+0.0844}

=1−0.878

=0.122

​We obtain the probability of the event that the length of a drought exceeds its mean value by at least one standard deviation as:0.122

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics

Page 24 Problem 1 Answer

Given,To make the Steam-and-Leaf Display:

We select the leading digits, then the digits that come after the leading digits are the leaves.

Then we list the parent values ​​that we have received in a column.

Then we comment on the leaf for all data values ​​associated with the respective parent value.

Then we write down the units of the stem and leaves.

The data given to us in tabular form is:

5.9	7.2
7.3	6.3
8.1	6.8
7	7.6
6.8	6.5
7	6.3
7.9	9
8.2	8.7
7.8	9.7
7.4	7.7
9.7	7.8
7.7	11.6
11.3	11.8
10.7

If we follow our explanation above, we first obtain the leading digits. These leading digits are:5,6,7,8,9,10,11

Listing the leading digits:

Stem
5
6
7
8
9
10
11

Recording the leaves for each and every observation and obtain the Stem-and-leaf display:

Stem	​ Leaf
5	9
6	33588
7	234677889
8	127
9	77
10	7
11	368
Stem: ones	Leaf: tenths

Since the representative strength value is the value that is in the middle, in this case, it would be  7.7.

The observations are also spread out and the values around our representative strength value aren’t symmetrically spread around the middle.

Therefore, the Stem-and-Leaf display is obtained as Stem

The representative strength value is 7.7 and the observations are spread out.

Page 24 Problem 2 Answer

To get the representative value of the force, we will first write down all the values ​​on a screen of the root leaf.

The representative force value is the value in the middle of the root leaf lattice. If the data is symmetrical about the value of the representative force, then the data is said to be undistorted.

The stem-and-leaf display of the values is given as:​

Stem	Leaf
5	9
6	33588
7	234677889
8	127
9	77
10	7
11	368
Stem: ones	Leaf: tenths

The representative strength value for the data is the value in the middle which is 7.7.

The observations are spread out and not symmetric about the representative strength value rather they are more positively skewed.

Hence,the data is not symmetrical and the values aren’t symmetrically distributed around the representative strength value.

It makes sense that the dataset is skewed positively.

Page 24 Problem 3 Answer

Given, In order to obtain the values that lie outside, we’ll first note down all the observations in a stem-and-leaf display.

If some of the values in the data are abnormally deviating, then they are said to be outliers.

The stem-and-leaf display of the values is given as:

Stem	​ Leaf
5	9
6	33588
7	234677889
8	127
9	77
10	7
11	368
Stem: ones	Leaf: tenths

We notice that 5.9 and 11.8 differ the most from the rest of the values but not abnormally.

Therefore we can’t say that they are outliers.

Hence,we notice that 5.9 and 11.8 differ the most from the rest of the values but not abnormally.

Therefore we can’t say that they are outliers there are no outliers in the dataset.

Page 24 Problem 4 Answer

Given,In order to calculate the proportion of strength observations in this sample exceed 10 MP a, we have to calculate the total number of observed values.

After that we will determine the number of observations that are greater than 10 MP a.

We consider taking the ratio of the both in which the number of observations which are greater than 10 MP a is the numerator and total number of observations is the denominator.

The total number of observations: n=27

Number of observations that are greater than10

MP a:10.7,11.3,11.6,11.8 x=4

​The proportion of strength observations in this sample exceed 10 MP a:X/n=4/27

⇒0.1481

Converting the proportion into percentages we get: 14.81%≈15%

Therefore, the proportion of strength observations in this sample which exceed 10 MPa: 15%

Page 24 Problem 5 Answer

Given, To make the Steam-and-Leaf Display:

Stem (tenths)

0.3

0.4

0.5

0.6

0.7

We choose the leading digits, following that, the digits coming after the leading digits are the leaves.

Then list the values of the stem were obtained in a column.

Note down the leaf for all data values pertaining with the particular stem value.

We then note down the units of the stem and leaves.

The leading digits (tenths) for the dataset are:0.3,0.4,0.5,0.6,0.7

Listing the leading digits:​

Stem	Leaf
0.3	156678
0.4	1122222345667880
0.5	14458
0.6	26678
0.7	5
Steam: tenths	Leaf: hundreths

Recording the leaves for each and every observation and obtain the Stem-and-leaf display:

Stem	​ Leaf
0.3	156678
0.4	1.12222E+15
0.5	14458
0.6	26678
0.7	5
Steam: tenths	Leaf: hundreths

We see that the dataset is positively skewed with a representative value of 0.445

Therefore,the dataset following batch of exam scores is positively skewed, uni-modal with a representative value of 0.445.

Page 25 Problem 6 Answer

Given, In a study of author productivity a large number of authors were classified according to the number of articles they had published during a certain period.

The results were presented in the given frequency distribution.

We need to construct a histogram corresponding to this frequency distribution.

Also write the most interesting feature of the shape of the distribution.

Now,Based on the given frequency distribution the histogram is plotted as by taking number of papers on horizontal line.

Height of the rectangle bar represents the frequency.

From observing the above plot,the histogram is completely skewed to the right.

Hence,the histogram is highly positive skewed.

Therefore,by using the frequency distribution the histogram plotted can be as follows:

Page 25 Problem 7 Answer

Given,In a study of author productivity a large number of authors were classified according to the number of articles they had published during a certain period.

The results were presented in the given frequency distribution.

We need to find the proportion of these authors published at least five papers.

Also find the proportion for at least ten papers and proportion for more than ten papers.

Now, To find the proportion of these authors published at least five papers:

P(X≥5)=1−P(X<5)

=1−{P(X=1)+P(X=2)+P(X=3)+P(X=4)}

=1−{784/1309+204/1309+127/1309+50/1309}

=1−1165/1309            ​

=144/1309

=0.11​

Similarly, for the proportion for at least ten papers:

P(X≥10)=39/1309

≈0.03

Then,the proportion for more than ten papers:

P(X>10)=32/1309

≈0.024

​Therefore,the proportion of these authors published at least five papers is obtained as 11%.

Similarly,for the proportion of these authors published for at least ten papers is obtained as 3%.

The proportion of these authors published for more than ten papers is obtained as 2.4%.

Page 25 Problem 8 Answer

Given, In a study of author productivity a large number of authors were classified according to the number of articles they had published during a certain period.

The results were presented in the given frequency distribution.

Suppose the five 15s, three 16s, and three 17s had been lumped into a single category displayed as≥15.

We need to verify that whether we can draw a histogram with the given data with explanation.

Now, Suppose the five 15s, three 16s, and three 17s had been lumped into a single category displayed as ≥15.

As we can observe that the class is≥15 has no upper bound.

Hence, we are unable to draw a histogram because of the class≥15 with no upper bound specified.

So,from the given data of class which is greater than 15 with no upper bound specified then we are unable to draw a histogram

Page 25 Problem 9 Answer

Given,In a study of author productivity a large number of authors were classified according to the number of articles they had published during a certain period.

The results were presented in the given frequency distribution.

Suppose that instead of the values15,16 and 17 being listed separately, they had been combined into a 15−17 category with frequency 11.

We need to explain whether we able to draw a histogram or not with proper explanation

Now,Combined class of15−17 category with frequency 11.

Width of the class is 2.

So,the area of 11 of the bar equals to product of height and width.

Hence,we can draw histogram with the given data.

Therefore,instead of the values 15,16 and 17 being listed separately, they had been combined into a 15−17 category with frequency 11 then we can draw

histogram with width of class 2 and are a11 of the bar equals to product of height and width.

Page 25 Problem 10 Answer

Given,Total number of sampled wafers are100.

Frequency distribution of these 100 sampled wafers i.e.,

We need to find the proportion of the sampled wafers that had at least one particle.

Then find the proportion of the sampled wafers that had at least five particles.

Let us determine the proportion of the sampled wafers that had at least one particle.

In the given frequency distribution,Total number of sampled wafers are100.

Now, there is only first sampled wafer i.e., “0”, which does not have any particle.

Thus, the number of sampled wafers that had at least one particle are(100−1).

ORthe number of sampled wafers that had at least one particle are: 2+3+12+11+15+18+10+12+4+5+3+1+2+1=99

Therefore, proportion of the sampled wafers that had at least one particle is given by:

p= number of sampled wafers that had at least one particle

Total number of sampled wafers

This implies, p=99%

Let us determine the proportion of the sampled wafers that had at least five particles.

In the given frequency distribution,Total number of sampled wafers are 100.

Number of sampled wafers that had less than 5 particles are:100−(11+12+3+2+1)=71

OR the number of sampled wafers that had at least five particles are: 15+18+10+12+4+5+3+1+2+1=71

Therefore, proportion of the sampled wafers that had at least five particles is given by:

p= number of sampled wafers that had at least five particles

Total number of sampled wafers

p=71/100

p=0.71

This implies, p=71%

Therefore,the proportion of the sampled wafers that had at least one particle is 0.99=99 %.

Proportion of the sampled wafers that had at least five particles is 0.71=71%

Page 25 Problem 11 Answer

Given, Total number of sampled wafers are 100.

Frequency distribution of these 100 sampled wafers i.e.,

we need to find the proportion of the sampled wafers having particles between five and ten, inclusive.

Then we have to find proportion of the sampled wafers having particles strictly between five and ten.

We need to find out the proportion of the sampled wafers having particles between five and ten, inclusive

That is, we need to find out the proportion of the sampled wafers having particles between five and ten, including 5 and 10.

It is given that total number of sampled wafers are 100.

Number of sampled wafers having particles between five and ten, including 5 and 10.

15+18+10+12+4+5=64 Using proportion formula, The proportion of the sampled wafers having particles between five and ten, inclusive is: p= number of sampled wafers having particles between 5 and 10 , inclusive total number of sampled wafers

p=64/100

p=0.64

This implies, p=64%

We need to find out the proportion of the sampled wafers having particles strictly between five and ten.

That is, we need to find out the proportion of the sampled wafers having particles between five and ten, not including 5 and 10.

It is given that total number of sampled wafers is 100.

Number of sampled wafers having particles strictly between five and ten are: 18+10+12+4=44

Using proportion formula,

The proportion of the sampled wafers having particles strictly between five and ten is:

p= number of sampled wafers having particles strictly between 5 and 10

total number of sampled wafers

p=44/100

p=0.44

This implies, 44% .

Hence,from the above the proportion of the sampled wafers having particles between five and ten, inclusive is 0.64=64%.and the proportion of the sampled wafers having particles strictly between five and ten is 0.44=44%.

Page 25 Problem 12 Answer

Given:,frequency distribution of 100 sampled wafers.

We need to draw a histogram taking relative frequency on the vertical axis and number of particles on horizontal axis.

Since, it is given that total sample size is 100.

Therefore, relative frequency is obtained by dividing each frequency with100.

We need to draw a histogram of this frequency distribution such that the height of the bars are equal to the relative frequency and width of all bars is equal.

Now, the required histogram with relative frequencies on y−axis is:

In this histogram, we observed that It is not perfectly symmetric, that means it is almost symmetrical.

It has only one hump, that means it is unimodal.

There are more data points on the left side, that means it is slightly positively skewed.

Therefore, we constructed a histogram using relative frequency on the vertical axis and described the shape of the histogram.

It is not perfectly symmetric, that means it is almost symmetrical.

It has only one hump, that means it is uni-modal.

There are more data points on the left side, that means it is slightly positively skewed.

Page 26 Problem 13 Answer

Given, An important characteristic of such an individual cell is its inter-division time representative IDT data:

To construct a histogram of the original data using class intervals 10−<20,20−<30,…

To construct a histogram for the transformed data using the interval1.1−<1.2,1.2−<1.3,…

To describe the effect of the transformation.

In a histogram, width of the bars should be equal and height should be equal to the frequency.

To draw the histogram for the original data, we consider the following table:

Therefore, the required histogram is:

  

To draw the histogram for the transformed data, we consider the following table:

Therefore, the required histogram is:

The frequency distribution histogram of the original data is positively skewed as there are more data points on the left side and the highest bar is on the left side.

The frequency distribution histogram of the transformed data is slightly symmetric as there are more data points in the center and the highest bar is in the center.

Thus, we can conclude that the transformation made the distribution of the given data set more symmetric.

So,we have constructed the histogram for original and transformed data.

The transformation made the distribution of the given data set more symmetric.

Page 27 Problem 14 Answer

The given data:

11 59 81 105 161

14 61 84 105 168

20 65 85 112 184

23 67 89 118 206

31 68 91 123 248

36 71 93 136 263

39 74 96 139 289

44 76 99 141 322

47 78 101 148 388

50 79 104 158 513​

We need to tell why can a frequency distribution not be based on the class intervals 0−50,50−100,100−150 and so on.

If suppose, we classify the distribution based on class intervals0−50,50−100 and so on, then just by looking, we would not tell that in which interval the end term in located like for example we can not tell that in which interval 50 is located, either 0−50 or 50−100.

Hence,we cannot classify the given data into class intervals due to overlapping issue.

Page 27 Problem 15 Answer

Given data,

11  59  81  105  161

14  61  84  105  168

20  65  85  112  184

23  67  89  118  206

31  68  91  123  248

36  71  93  136  263

39  74  96  139  289

44  76  99  141  322

47  78  101  148  388

50  79  104  158  513

We need to construct the frequency distribution and histogram.

We will construct the frequency distribution table as follows:

We have found the relative frequency by the frequency of particular interval divided by the total frequency.

The histogram is positively skewed.

It has representative value around 150. It is unimodal.

Hence,the required histogram of the data using class boundaries 0,50,100,…,is

Page 27 Problem 16 Answer

Given data is

11 59 81 105 161

14 61 84 105 168

20 65 85 112 184

23 67 89 118 206

31 68 91 123 248

36 71 93 136 263

39 74 96 139 289

44 76 99 141 322

47 78 101 148 388

50 79 104 158 513

​We need to construct a frequency distribution and histogram of the natural logarithms of the lifetime observations.

We will firstly determine the natural logarithm of each data point.

2.40 4.08 4.39 4.65 5.08

2.64 4.11 4.43 4.65 5.12

3.00 4.17 4.44 4.72 5.21

3.14 4.20 4.49 4.77 5.33

3.43 4.22 4.51 4.81 5.51

3.58 4.26 4.53 4.91 5.57

3.66 4.30 4.56 4.93 5.67

3.78 4.33 4.60 4.95 5.77

3.85  4.36 4.62 5.005.96

3.91 4.37 4.64 5.06 6.24​

Now we will construct the frequency distribution:

The approximate interval length = max value − min. value

√n=6.24−2.40/√50≈0.55

Since, it is not strict, we take0.5 and start from 6.25and using these we can calculate the number of intervals and their lengths.

The histogram of the given data will be

This histogram is more symmetrical than the histogram formed by the original values.

The representative value is nearly 4.5.It is unimodal.

The histogram for the log of the terms of the data is given by

This histogram is more symmetrical than the histogram formed by the original values.

The representative value is nearly 4.5.It is unimodal.

Page 27 Problem 17 Answer

According to the question, we need to find the proportion of the lifetime observations in this sample are less than 100 and proportion of the observations are at least 200 from the data:​11

59 81 105 161

14 61 84 105

168 20 65 85

112 184 23 67

89 118 206 31

68 91 123 248

36  71 93 136

263 39 74 96

139 289 44 158

76 99 141 322

47 78 101 148

388 50 79 104

​For the portion of the data less than 100 , we will sum up all the relative frequencies of the desired intervals.

We get 0.18+0.38=0.56 which is nearly56% .

Similarly, for the portion of the data which is at least 200,we will sum up all the relative frequencies of the desired intervals.

We get0.04+0.04+0.02+0.02+0.02=0.14 which is nearly 14%.

Hence,the portion of data which is less than 100 is 56%.

The portion of data which is at least200 is 14%.

Page 27 Problem 18 Answer

Given, Human measurements provide a rich area of application for statistical methods.

Based on the article “A Longitudinal Study of the  Development of Elementary School Children’s Private Speech” reported on a study of children talking to themselves.

It was thought that private speech would be related to IQ, because IQ is supposed to measure mental maturity, and it was known that private speech decreases as students progress through the primary grades.

The study included 33

students whose first-grade IQ scores are given here:

We need to explain the data and comment on any interesting features .

Now, By plotting the given data by using Box plot then:

From the above we can observe that the data is quite large and there is positive skew.

The smallest value is 82 is separated from the remaining big sample data and can be considered outlier.

So,by plotting the given data using box plot the data is quite large and there is positive skew.

The smallest value is 82 is separated from the bulk of data so it can be considered outlier.

Page 27 Problem 19 Answer

Given, Consider the following data on types of health complaint made by tree planters.

The data is consistent with percentages given in the article “Physiological Effects of Work Stress and Pesticide Exposure in Tree Planting by British Columbia Silviculture Workers,” Ergonomics.

Obtain frequencies and relative frequencies for the various  categories,

We need to draw a histogram.

Now, By tabulating the frequency and relative frequency of occurrence of health complaints from the given data:

Complaint	​ Frequency	​ Relative Frequency
B	7	0.1167
C	3	0.05
F	9	0.15
J	10	0.1667
M	4	0.0667
N	6	0.1
O	21	0.35

To obtain relative frequency divide by total number of observations 60.

​By constructing histogram using the data:

Complaint	​ Frequency	​ Relative Frequency
B	7	0.1167
C	3	0.05
F	9	0.15
J	10	0.1667
M	4	0.0667
N	6	0.1
O	21	0.35

So, by obtaining frequencies and relative frequencies for the various  categories as follows:

By constructing histogram using the data as follows:

Page 24 Problem 20 Answer

Given,

Failed component =126.

Incorrect component 210.

Insufficient solder =67.

Excess solder =54.

Missing component =131

From the above data, we need to construct a Pareto chart.

With the help of the given data, we will construct the Pareto diagram keeping following things in mind:

We will take equal width and the height will be equal to the frequency of the respective observation.We will rank the bars from highest to lowest.

Hence, we get Hence, the required Pareto diagram is represented on bar diagram as:

Page 27 Problem 21 Answer

Given, According to the question, we need to compute the cumulative frequency and relative cumulative frequency of the following data:​

5434	​4948	​4521	4570	4990	5702	5241
5112	5015	4659	4806	4637	5670	4381
4820	5043	4886	4599	5288	5299	4848
5378	5260	5055	5828	5218	4859	4780
5027	5008	4609	4772	5133	5095	4618
4848	5089	5518	5333	5164	5342	5069
4755	4925	5001	4803	4951	5679	5256
5207	5621	4918	5138	4786	4500	5461
5049	4974	4592	4173	5296	4965	5170
4740	5173	4568	5653	5078	4900	4968
5248	5245	4723	5275	5419	5205	4452
5227	5555	5388	5498	4681	5076	4774
4931	4493	5309	5582	4308	4823	4417
5364	5640	5069	5188	5764	5273	5042
5189	4986

Cumulative frequency of any observation can be calculated by adding up all the frequencies of previous observations.

Relative cumulative frequency of any observation=  Cumulative frequency of that observation /100

We know that frequency is defined as the number of times a value occurs in the given data.

The frequency, cumulative frequency, relative cumulative frequency of the given data are given below:

Hence,we have calculated the cumulative frequency and relative cumulative frequency of every value in the data with the help of definitions as follows:

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics

Page 34 Problem 1 Answer

Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the given observation.

15.0​  13.0  ​18.0​  14.5   ​12.0​  11.0​   8.9   ​8.0

We need to determine the values of the sample mean, sample median, and 12.5% trimmed mean, and compare these values.

Now,Considering the data:

15.0​  13.0  ​18.0​  14.5   ​12.0​  11.0​   8.9   ​8.0

We need to find the mean: xˉ​=1/n∑xi

=1/8{15+13+18+14.5+12+11+8.9+8.0}

=12.55

For sample median:

We need to arrange the data in ascending order as follows:

8.0  15.0​8.9  18.0​11.0​12.0​13.0​14.5

As number of observations n=8 is even then, Median be (8/2)^th and (8/2+1)^th

So, average of the fourth and fifth values of ordered values gives median:

​x~=12+13/2=25/2=12.5 hence,12.5% trimmed mean is determined by eliminating smallest 12.5 % and largest 12.5 % in the sample then using remaining numbers we have to obtain mean.

xtr(12.5)ˉ=8.9+11+12+13+14.5+15/6

=74.4/6

=12.4

From the obtained values, Sample median and trimmed mean are bit smaller than sample mean.

Hence,the values of the sample mean, sample median, and 12.5% trimmed mean obtained as:

Mean:xˉ=12.55

Median:xˉ=12.5

Trimmed mean:xˉ=12.4

Sample median and trimmed mean are bit smaller than sample mean.

Page 34 Problem 2 Answer

Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations:

15.0​  13.0​  18.0  14.5​  12.0​  11.0​  8.9​  8.0

​We need to find by how much could the smallest sample observation, currently 8.0, be increased without affecting the value of the sample median.

Now,From the given observations:

15.0​  13.0​  18.0  14.5​  12.0​  11.0​  8.9​  8.0

​By how much could the smallest sample observation, currently 8.0, be increased without affecting the value of the sample median:

The smallest observation from the data is 8.0

Without changing median then we have to add such that it should be less than 12.

Hence,we can conclude that by adding 4 to the smallest observation without changing the median.

Therefore,by adding 4 to the smallest observation without effecting the value of sample median.

Page 34 Problem 3 Answer

Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations

15.0  ​13.0​  18.0  ​14.5​  12.0​  11.0​   8.9  ​8.0

​Suppose we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi.

We need to find whether it is necessary to re-express each observation in ksi, or can the values calculated in part (a) be used directly.

Now,The sample mean,median and trimmed mean of the observations obtained in 35(a) as follows:

Mean: xˉ=12.55 Median: xˉ=12.5 Trimmed mean: xˉ=12.4​

To convert them form pounds per square inch to kilogram per square inch:

We need to divide the obtained values by 2.2 such that it gets converted to ksi.

New sample mean​

=12.55/2.2

=5.70​

New sample median​

=12.50/2.2

=5.68​

New Trimmed mean​

=12.4/2.2

=5.64​

Hence,it is verified that there is no need to recalculate the each value to convert them form psi to ksi.

So,as we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi then there is no need to recalculate the each value.

We can obtain them by directly dividing each obtained values with 2.2.

New sample mean=5.70 ksi

New sample median=5.68 ksi

Trimmed mean=5.64 ksi

Page 34 Problem 4 Answer

Given,Using the article “Snow Cover and Temperature Relationships in North America and Eurasia” the statistical techniques to relate the amount of snow cover on each continent to average continental temperature:

There are ten observations on October snow cover for Eurasia during a period.

We need to find report a representative, or typical, value of October snow cover for this period.

Given data- 6.5/7.9​12.0/21.9​14.9/12.5​10.0/14.5​10.7/9.2

​For the given data, 21.9 is an outlier, so trimmed mean would be good choice for the researcher.

For the trimmed mean remove the smallest and the largest values-

xtr(10)=12.0+14.9+10.0+10.7+7.9+12.5+14.5+9.2/8

=91.7/8

=11.4625​

Hence, based on the article “Snow Cover and Temperature Relationships in North America and Eurasia” using statistical techniques to relate the amount of snow cover on each continent to average continental temperature report as representative, or typical, value of October snow cover for the period is represented by 10% trimmed mean which is 11.4625.

Page 40 Problem 5 Answer

Given,Based on the study of Life distribution of micro-drills listed in increasing order on drill life time:

Using the Data on drill lifetime in certain brass alloy:

We need to find sample median,25% trimmed mean, 10% trimmed mean, and sample mean, and compare them.

The ordered values are-11,14,20,23,31,36,39,44,47,50,59,61,65,67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,112,118,123,136,139,141,148,158,161,168,184,206,248,263,289,322,388,513

The given data has even values so the sample median is-

Therefore, Median =x²⁵+x²⁶/2

=91+93/2

=184/2

=92

​For 25% trimmed mean, the trimmed value is,

Trimmed value =n× Percentage

=50×0.25

=12.5≈13​

So delete 13 Now, the remaining data is, observations on both sides-

67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,105,112,118,123,136,139.

Therefore, the mean is,

xTr−(0.25)

​=1/nTr(∑xTr)

=1/24(67+68+71+……+139)

=1/24(2274)

=94.75​

For 10% trimmed mean, the trimmed value is,

Trimmed value =n× percentage

=50×0.10

=5​

So delete 5 observations on both sides-

Now, the remaining data is,11,14,20,23,31,36,39,44,47,50,59,61,65,67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,112,118,123,136,139,141,148,158,161,168,184,206,248,263,289,322,388,513

Therefore, the mean is,

xTr−(0.10)

=1/nTr(∑xrr)

=1/40(36+39+44+47+…+248)

=1/40(4089)

=102.225​

The sample mean.

xˉ=1/n∑xi

=1/50(11+14+20+….+388+513)

=1/50(5963)

=119.26​

So, we can observe that the sample mean has more value than the 10%  trimmed mean and 10% trimmed mean has more value than the 25% trimmed mean.

Thus, we can conclude that the mean value is decreasing when trimming the data.

Hence,the sample median is92,25% trimmed mean is 94.75,10%  trimmed mean is 102.225 and sample mean is 119.26.

On comparing the obtained values, we can conclude that the mean value is decreasing when trimming the data.

Page 41 Problem 6 Answer

Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows:

A sample of n=10 automobiles for a crash test.

We need to find-the value of the sample proportion of successes.

SSFFSSSFSS

S means success and F damage.

So here we have 7 successes and 3 damages , total 10 tests.

The sample proportion of successes ​x/n

​=7/10

=0.7

​Hence,the value of the sample of automobiles selected, and each was subjected to a 5 -mph crash test and proportion of successes is 0.7

Page 41 Problem 7 Answer

Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows:

A sample of n=10 automobiles for a crash test.

We need to find xˉ and compare it with x/n.

Here we have to replace each S with 1 a  and each F with a 0 .

So it becomes 1,1,0,0,1,1,1,0,1,1

Herexˉ=7/10

=0.7

So it can be observed that xˉ is similar to the proportion of success xn.

Hence,after replacing each S with a 1 and each F with a 0 ,the mean:

xˉ=0.7 and it is equal to xn.

Page 41 Problem 8 Answer

Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows

A sample of n=25 automobiles for a crash test.

We need to find the number of these would have to be S′s to give xn=.80 .

We currently have actually 7 S′s and 3F′s-And we want xn=0.8n would be 25,

x=25∗0.8

=20​

So we need 20 S′s  and we have 7.

So 13S′s from the new 15 cars is required in order to satisfy the condition.

So, by including 15 more cars in the experiment 13S′s

would have to be from new 15 tests, so that xn=0.8

 

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics

Page 12 Problem 1 Answer

Given, Statement says that we need to”Give three different examples of concrete populations and three different examples of hypothetical populations.”

Example of concrete populations:

All student who take JAM exam in the year 2021.

All workers at your University.

Number of students participate in recent college quiz competition.

Examples of hypothetical populations:

All possible samples of a particular type of cancer tissue.

Average students at your University participated in scholarship exam in next academic year.

Page length of scientific papers published in 2015.

Hence,some of the example of concrete populations are as follows:

All student who attend JAM exam in the year 2021.

All workers at your University.Strength of students participated in recent college quiz competition.

Similarly,some examples of hypothetical populations are as follows:

All probable samples of a particular type of cancer tissue.

Average students at your University attended scholarship exam for the next academic year.

Page length of scientific papers published in 2015

Page 12 Problem 2 Answer

Given,We need to give an example of a probability question and also an example of an inferential statistics question for each one of your concrete and hypothetical populations.

Probability questions for concrete populations:

What is the probability of getting an Ace? What is the probability of getting at least 1 King? What is the probability of getting 2 heads when three coins are flipped once?

Probability questions for hypothetical populations: What is the probability that you threw a stone more than 100 meters? What is the probability of the drug being more toxic than 15(units)?What is the probability of student going to have average grade bigger than 4.9?

Inferential static question for concrete population: Are the basketball athletes of 2013 more accurate than those of 2003?

Inferential static question for hypothetical population: Can we conclude the average grade of all students average grades will be below 4 the given 5 sampled students at your University have an average grade of 2.3,4.2,4.3,3.9,4 ?

Probability questions for concrete populations:

What is the probability of getting an Ace? What is the probability of getting at least 1 King? What is the probability of getting 2 heads when three coins are flipped once?

Probability questions for hypothetical populations: What is the probability that you threw a stone more than 100 meters? What is the probability of the drug being more toxic than 15 (units)? What is the probability of student going to have average grade bigger than 4.9?

Inferential static question for concrete population:

Are the basketball athletes of 2013 more accurate than those of 2003?

Inferential static question for hypothetical population:

Can we conclude the average grade of all students average grades will be below 4 the given 5 sampled students at your University have an average grade of 2.3,4.2,4.3,3.9,4  ?

Page 12 Problem 3 Answer

Given, We need to find that if scores from the SI group a sample from an existing population.

And If we do find those, what are they?If not, what is the most resemble conceptual population?

No, the score from the SI group a sample from an existing population.

In the SI program of this sort all student taking large static course who participate in this sort.

Therefore,it is clear that the score from the SI group is a sample taken from an existing population.

From the above,SI program of this kind all student enrolling in large static course who participate in this sort.

Page 12 Problem 4 Answer

Given,By randomly dividing the students into two groups rather than leaving an option to each student to choose which group to join.

The advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join is:

It safeguards against various bias.

It helps in ensuring that those who are in SI groups are as similar as possible to the student in the control group.

Therefore,the advantages are observed as: It helps and safeguard against various bias.

It helps in ensuring that those who are in SI groups are as similar as possible to the student in the control group.

Page 12 Problem 5 Answer

Given, Statement in the article says that “Supplemental Instruction: An Effective Component of Student Affairs Programming” (J. of College Student Devel ., 1997: 577−586 ).

Now, The investigators put all students in the treatment group because there is no firm basis for assessing the effectiveness of SI.

The SI scores cannot be reasonably compared with anything.

Therefore,the investigators put all students in the treatment group because of the reason that there is no strong basis for assessing the effectiveness of SI.

These SI scores cannot be compared having proper reason with anything.

Page 12 Problem 6 Answer

Given: A certain city is naturally divided into ten districts.

A real estate appraiser selects a sample of single-family homes that could be used as a basis for developing an equation to predict the estimated value based on characteristics such as age, size, number of bathrooms, distance from the home. next school etc.

When all single family homes in the city is given then we can generate the simple random samples or the stratified random sample.

In stratified random sample we take the simple random sample from each of the district neighborhoods.

Values of all desired variables would be determined from each of the selected homes.

Now, we can conclude that their exists a finite, identifiable population of objects from which to sample.

So, this is the enumerative study.

We can conclude that there is a finite and identifiable population of objects to sample from, so this is the enumerative study.

Page 12 Problem 7 Answer

Given: The flow rate through a solenoid valve in an automobile emission control system is an important characteristic.

An experiment was conducted to investigate how the flow rate depends on three factors: armature length, spring force and coil depth.

Two different levels (low and high). ) of each factor, and a single flow observation was made for each combination of levels.

Total factors are 3.

Each factors receive one of two possible level.

Low level is represented by L.

High level is  represented by H.

The total observations are 8 are 8 are given as :

{LLL, LLH, LHL, LHH, HLL, HLH, HHL, HHH}

Hence,the total observations that resulting data set consisted are 8 are given as :

{ LLL, LLH,LHL, LHH,HLL,HLH,HHL,HHH}

Page 12 Problem 8 Answer

The flow rate through a solenoid valve in an automobile emission control system is an important characteristic.

An experiment was conducted to investigate how the flow rate depends on three factors: armature length, spring force and coil depth.

Two different levels (low and high). ) of each factor and a single flow observation was made for each combination of levels.

Enumerative study: This study is identifiable not ambiguous. In this study,  the elements of the population are well-defined and unchanging. In this study, a sampling frame which is a list of sample points to be collected is either available to an investigator or else can be constructed.

Analytic study: It is not an enumerative study. It is a process that does not even exit at the time of the study.

The study does not have a well-defined sampling frame and the impact of this study is highly localized and short term.

The given study is analytic study.

Because we are investigating the amount of flow through solenoid valve.

Hence, this can be improved in the future.

Therefore,an experiment was carried out to study how flow rate depended on three factors: armature length, spring load, and bobbin depth.

The given study is analytic study.

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability

Page 54 Problem 1 Answer

Given:1,2,3 and 4 universities are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4.

Then the two winners will play for the championship, and the two losers will also play. To list all the possible outcomes in S.

We shall first consider for university 1 and see the possible outcomes and then similarly we will have the possible outcomes for all the other universities.

Four universities 1,2,3 and 4 are participating in a holiday basketball tournament.

Let’s assume that the  university 1 wins the holiday basketball tournament which means that 1 beats 2 in the first game.

Now the second game university 3 can beat university 4 and university 4 can beat university 4

Therefore, all outcomes where university 1  wins the holiday basketball tournament will be

{1324,1342,1423,1432}  in the similar way we obtain the outcomes when university 2 wins, university 3 wins and university 4 wins the holiday basketball tournament.

Therefore we obtain the sample space as:

S={1324,1342,1423,1432,2314,2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}

The list of all the possible outcomes is S={1324,1342,1423,1432,2314,2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}.

Page 54 Problem 2 Answer

Given:1,2,3,4 universities are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4.

Then the two winners will play for the championship, and the two losers will also play.

Given:A denotes the event that 1wins the tournament.To list all the outcomes in A.

We will check for all the possibilities where the university 1will win the tournament.

Four universities 1,2,3 and 4 are participating in a holiday basketball tournament.

Let’s assume that the university 1 wins the holiday basketball tournament which means that 11 beats 2 in the first game.

Now the second game university 3 can beat university 4 and university 4 can beat university 4 Therefore, all outcomes where university 1 wins the holiday basketball tournament will be

A={1324,1342,1423,1432}

Outcomes of event A is A={1324,1342,1423,1432}

Page 54 Problem 3 Answer

Given: 1,2,3 and 4 universities are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4.

Then the two winners will play for the championship, and the two losers will also play.

Given: B denotes the event that 2 gets into the championship game.

To list all the outcomes in B We will check for all the possibilities where the university 2 gets into the championship game.

Now the university 2 gets into the championship game means that the university 2 wins first round game but they can still lose the championship game.

Therefore, we get: B={2314,2341,2413,2431,3214,3241,4213,4231}.

The list of  outcomes in B is: B={2314,2341,2413,2431,3214,3241,4213,4231}.

Page 54 Problem 4 Answer

A denotes the event that 1 wins the tournament.

B denotes the even that 2 gets into the championship game.

Obtained sample space: S={1324,1342,1423,1432,2314,2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}

Obtained outcomes of A and B:A={1324,1342,1423,1432}

B={2314,2341,2413,2431,3214,3241,4213,4231}​

To find A∪B and in A∩B.To find the outcomes of A′.

We will use the definitions to find the required values.

We haveA={1324,1342,1423,1432}

B={2314,2341,2413,2431,3214,3241,4213,4231}​

Therefore the list of outcomes which in A,B or both are:

A∪B={1324,1342,1423,1432,2314,2341,2413,2431,3214,3241,4213,4231}

The list of outcomes which is contained in both A and B are : A∩B=∅

Since there is no outcome common to both.

We have S={1324,1342,1423,1432,2314,2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}

And A={1324,1342,1423,1432}

Hence the list of outcomes which in S and not in A is: A′

={2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}

We obtain:A∪B={1324,1342,1423,1432,2314,2341,2413,2431,3214,3241,4213,4231}

A∩B=∅A′={2341,2413,2431,3124,3142,3214,3241,4123,4132,4213,4231}

​Page 54 Problem 5 Answer

Given figure of the system:

Given: The sub-system will function  if at least one of the two individual components functions.

To find the outcomes contained in event A that exactly two out of the three components function.

We will check for all the possibilities for the given condition.

Exactly two out of the three components are functioning means that we should have exactly two letters S in the events and one latter F, therefore, all outcomes contained in the event A is A={SSF,SFS,FSS}

The outcomes with two functioning and one nonfunctioning component is:   A={SSF,SFS,FSS}

Page 54 Problem 6 Answer

Given figure of the system:

Given: The sub-system will function  if at least one of the two individual components functions.

To find outcomes are contained in the event B that at least two of the components function.

We will check for all the possibilities for the given condition.

At least two of the components function means that it could be possible that all three function or exactly two are functioning that is the event B

B={SSS,SSF,SFS,FSS}

The list of outcomes are contained in the event B that at least two of the components function is:

B={SSS,SSF,SFS,FSS}.

Page 54 Problem 7 Answer

Given figure of the system:

Given: The sub-system will function  if at least one of the two individual components functions.

To find outcomes that are contained in the event C that the system functions.

We will check for all the possibilities for the given condition.

Now we have that the system functions if the component 1 functions (meaning that S has to be in the first place) and one of the other two has to work (meaning that at least one S

should be in second and third place).

Therefore we get: C={SSS,SFS,SSF}

The outcomes where the system functions are C={SSS,SFS,SSF}

Page 54 Problem 8 Answer

Obtained values of event A,B and A are

A={SSF,SFS,FSS}

B={SSS,SSF,SFS,FSS}

C={SSS,SFS,SSF}

​To list the outcomes: C′,A∪C,A∩C,B∪C, and B∩C

We will use the definitions to find the required values

We know that C′ is the complement of C which imply that system is not functioning.

We need to list all the outcomes which are not functioning forC′

Hence we get C′={FSS,FSF,FFS,FFF,SFF}

In order to calculate A∪C we need to list all outcomes which are in A or C. So, we getA∪C={SSS,SSF,SFS,FSS}.

In order to calculate A∩C we need to list all outcomes which are in A and C. So, we get A∩C={SSF,SFS}.

In order to calculate B∪C we need to list all outcomes which are in B or C. So, we get B∩C={SSS,SSF,SFS}.

In order to calculate B∩C we need to list all outcomes which are in B and C. So, we get B∩C={SSS,SSF,SFS}.

We obtain:

C′={FSS,FSF,FFS,FFF,SFF}

​A∪C={SSS,SSF,SFS,FSS}

A∩C={SSF,SFS}

B∩C={SSS,SSF,SFS}

B∩C={SSS,SSF,SFS}

​Page 55 Problem 9 Answer

Given: A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3,4 and 5) are second printings.

A student examines these books in random order, stopping only when a second printing has been selected.

To list all the possible outcomes in S We will write the sample space in tabular form.

The sample space for choosing the books is as follows:

When creating the table, you can always start with all the same letters and go down, this is a classic method for creating the table.

The outcomes in S is given as:

Page 55 Problem 10 Answer

Given: A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3,4 and 5) are second printings.

A student examines these books in random order, stopping only when a second printing has been selected.

To list the outcomes of A which denote the event that exactly one book must be examined.

We will make use of the sample space.

The sample space table is given as:

We need to select the book from second printings in order to stop the selection.

By observing the sample space from the table , we get A=1,2,3 or A=3,4,5.

The outcomes in event A can be any one of 1,2,3 or 3,4,5.

Page 55 Problem 11 Answer

Given:  A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3,4 and 5) are second printings.

A student examines these books in random order, stopping only when a second printing has been selected.

To list the outcomes of B which denote the event that book 5 is the one selected.

We will make use of the sample space.

The sample space table is given as:

With the help of the sample space observed in the table, we can see if the book5 is selected then the last selection number must be 5 as the student will stop selecting when he/she selects the copy5

Hence, the outcomes are 3,6,9,12,15 or 5,15,25,125,215

The outcomes of event B can be 3,6,9,12,15 or 5,15,25,125,215

Page 55 Problem 12 Answer

Given: A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3,4 and 5) are second printings.

A student examines these books in random order, stopping only when a second printing has been selected.

To list the outcomes of C which denote the event that book 1 is not examined.

We will make use of the sample space.

The sample space table is given as:

With the help of the sample space observed in the table, we can easily check for all the outcomes which do not contain 1.

Hence, the outcomes can be1,2,3,7,8,9 or 3,4,5,23,24,25

The outcomes of event C can be 1,2,3,7,8,9 or 3,4,5,23,24,25

Page 55 Problem 13 Answer

Given: An engineering construction firm is currently working on power plants at three different sites.

LetA{i}  denotes the event that the plant at site i is completed by the contract date.

Given event: At least one plant is completed by the contract date.

To use the operations of union, intersection, and complementation to describe each of the following events in terms of A{1},A{2}, and A{3}

To draw a Venn diagram, and shade the region corresponding to the given event.

In order to show that at least one plant from A1,A2 and A3 and is completed by the contract date, we will define an event which is denoted as A1∪A2∪A3.

The Venn Diagram is given below:

The event that at least one plant is completed by the contract date is given by: A1∪A2∪A3

The required Venn diagram is:

Page 55 Problem 14 Answer

Given: An engineering construction firm is currently working on power plants at three different sites.

A{i}  denotes the event that the plant at site i is completed by the contract date.

Given event: All plants are completed by the contract date.

To use the operations of union, intersection, and complementation to describe each of the following events in terms of A{1},A{2}, and A{3}

To draw a Venn diagram, and shade the region corresponding to the given event.

In order to get all plants completed by the contract data, we require A1∩A2∩A3 .

The Venn diagram for this is given below:

The event that all plants are completed by the contract date is given by A1∩A2∩A3

The required Venn Diagram is:

Page 55 Problem 15 Answer

Given: An engineering construction firm is currently working on power plants at three different sites.

A{i} denotes the event that the plant at site i is completed by the contract date.

Given event: Only the plant at site 1 is completed by the contract date.

To use the operations of union, intersection, and complementation to describe each of the following events in terms of A{1},A{2}, and A{3}

To draw a Venn diagram, and shade the region corresponding to given event.

In order to get only the plant at site 1 to be completed by the contract date, we will define the event as A1A2′∩A3′

Its Venn diagram is as follows:

The event that only the plant at site 1 is completed by the contract date is given by: A1∩A2′∩A3′

The required Venn diagram is:

Page 55 Problem 16 Answer

Given: An engineering construction firm is currently working on power plants at three different sites.

A{i} denotes the event that the plant at site i is completed by the contract date.

Given event: Exactly one plant is completed by the contract date.Use the operations of union, intersection, and complementation to describe each of the following events in terms of A{1},A{2}, and A{3} .

To draw a Venn diagram, and shade the region corresponding to given event.

We have to get exactly one plant to be completed by the contract date we define the event which is represented as(A1∩A2′∩A3′)∪(A1′∩A2∩A3′)∪(A1′∩A2′∩A3).

From part c) we can say that if only plant at site 1 is completed by that contract data then we get A1∩A2′∩A3′

Here, the plant site is not defined rather we just need that exactly one plant is completed by the contract date.

It doesn’t matter that the plant is of which site. We can add all three possibilities A1

∩A2′∩A3′ (site 1)A1′∩A2∩A3′,(site 2),A1′∩A2′∩A3  (site 3).

Hence, event is represented as (A1∩A2′∩A3′)∪(A1′∩A2∩A3′)∪(A1′∩A2′∩A3)

Its Venn diagram is represented below:

The event that exactly one plant is completed by the contract date is given by: (A1

∩A2′∩A3′)∪(A1′∩A2∩A3′)∪(A1′∩A2′∩A3)

The required Venn diagram is:

Page 55 Problem 17 Answer

Given: An engineering construction firm is currently working on power plants at three different sites.

A{i}  denotes the event that the plant at site i is completed by the contract date.

Given event:  Either the plant at site 1 or both of the other two plants are completed by the contract date.

To use the operations of union, intersection, and complementation to describe each of the following events in terms of A{1},A{2}, and A{3}, To draw a Venn diagram, and shade the region corresponding to given event.

In order to show that either the plant at site 1 or both of the other two plants are completed by the contract date, we will define an event which is represented as A1∪(A2∩A3).

This is because “or” means union and “and” means intersection.

Hence the Venn diagram is:

The event that either the plant at site 1or both of the other two plants are completed by the contract date is given by: A1∪(A2∩A3)

The required Venn diagram is:

Page 55 Problem 18 Answer

Given relationship for two events A and B: (A∪B)′=A′∩B′

To use a Venn diagram to verify the relationship.We will draw Venn diagram for both the events and show that it is equal.

We first construct(A∪B)′ as Venn diagram. It is the complement of A union B. the  diagram below, the shaded area represents the left side(A∪B)′

Now, we will construct  A′ and B′ as follows:

.

A′is represented by horizontal stripes.

A′is represented by vertical stripes.

The Venn diagram will look like:

Hence we can see that both the Venn diagrams are equal.

We have verified the relationship (A∪B)′=A′∩B′

by using Venn diagram.

Page 55 Problem 19 Answer

Given relationship for two events :(A∩B)′=A′∪B′

To use a Venn diagram to verify the relationship.We will draw Venn diagram for both the events and show that it is equal.

The shaded region in the given diagram represents(A∩B)′.

And A′∪B′ is represented by the area that has horizontal and vertical lines.

Hence we can see that both the Venn diagrams are equal.

We have verified the relationship (A∩B)′=A′∪B′ using Venn diagrams.

Page 55 Problem 20 Answer

Given example 2.10: A={Chevrolet, Pontiac, Buick }

B={Ford, Mercedes }

C={Lincoln, Chrysler }

To identity three events that are mutually exclusive.We will have to find the events whose intersection is a null set.

If A∩B=ϕ where A and B are two events then we can say that they are mutually exclusive.

Now, the events from example 2.10 are:

A=  {Chevrolet, Pontiac, Buick}

B={Ford, Mercedes}

C={Lincoln, Chrysler}

We can see that all the name card are car brand and their intersection is empty.

Hence, they are mutually exclusive.

Then, A∩B=∅and A∩C=∅and B∩C=∅

The mutually exclusive events are:

A={Chevrolet, Pontiac, Buick}

B={Ford, Mercedes}

C= {Lincoln, Chrysler}

Page 55 Problem 21 Answer

Given in example 2.10:

A={Chevrolet, Pontiac, Buick }

B={Ford, Mercedes }

C={Lincoln, Chrysler }

Given: No outcome common to all the three events.To determine if the three events will be mutually exclusive.We will give a counter example.

If there is no outcome common to all three of the events A,B ,and C  then these three events will not necessarily mutually exclusive.

For that lets take a counter example that is:

A1={Chevrolet, Buick}

B1={Buick, Mercedes}

C1={Mercedes, Lincoln}

As we can see that in A1 and B1  have Buick in common and in B1 and C1 have Mercedes in common which implies that are not mutually exclusive and there is no common car in all three events.

No. The three events are not necessarily mutually exclusive where there is no outcome common to all three of the events A,B, and C.

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics

Page 46 Problem 1 Answer

Given,Based on the paper “Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI”gave the following data on aortic root diameter (cm) and gender for a sample of patients having various degrees of aortic stenosis:

Data on aortic root diameter (cm) and gender for a sample of patients

We need to compare and contrast the diameter observations for the two genders.

Initially we need to find various parameters and then make a boxplot in order to compare the data given.

The sample means. xM

xF=3.1+3.4+3.4+3.4+3.5+3.6+3.7+3.7+3.8+3.8+3.9+4.0+4.0/13

=3.6385

=2.6+3.0+3.0+3.1+3.1+3.2+3.2+3.5+3.8+4.3/10

=3.28​

We can observe that the sample mean is greater for males than for females.

Standard deviation for both samples.

sM=√1/13−1[(3.1−3.6385)²+(3.4−3.6385)²+⋅(3.4−3.6385)²+(3.4−3.6385)²+(3.5−3.6385)²+(3.6−3.6385)²+(3.7−3.6385)²+(3.7−3.6385)²+(3.8−3.6385)²+]

(3.8−3.6385)²+(3.9−3.6385)²+(4.0−3.6385)²+(4.0−3.63)+85)²]

​​=0.2694​

sF=√1/10−1[(2.6−3.28)²+(3.0−3.28)²+(3.0−3.28)²+(3.1−3.28)²+(3.1−3.28)²+(3.2−3.28)²+(3.2−3.28)²+(3.5−3.28)²+(3.8−3.28)²+(4.3−3.28)²]=0.4780

​We can observe that the sample standard deviation is greater for females than for males.

Arranging each sample in the increasing order-

M:{3.1,3.4,3.4,3.4,3.5,3.6,3.7,3.7,3.8,3.8,3.9,4.0,4.0}

F:{2.6,3.0,3.0,3.1,3.1,3.2,3.2,3.5,3.8,4.3}​

For a box-plot we need-The median is given by the  element  for males and by the average of the for females.

xM/N=3.7

xF=3.1+3.2/2

=3.15 −  lower fourth M

=3.4 ,lower fourth F

=3.0 − upperfourth M

=3.8 ,upperfourth F

=3.5 − ​fs/M

Fs/F= upper fourth M− lower fourth M

=3.8−3.4

=0.4

= upper fourth F− lower fourth F

=3.5−3.0

=0.5

​M:smallest xi=3.1and largest xi=4.0

F:smallest xi=2.6 and largest xi=4.3

​The boxplot can be constructed as-

The values for the M-sample are greater than female.

The boxplot and standard deviations suggest that variability is smaller for the M-sample.

An outlier, , in the female sample exists. There is positive overall skew in the F-sample and negative skew in the M-sample.

Hence,based on the given data the mean, standard deviation, median, Q1,Q3 and fs

for M-sample are 3.6385,0.2694,3.7,3.4,3.8 and 0.4 respectively.

The mean, standard deviation, median, Q1 , Q3 and fs  for F-sample are 3.28,0.4780,3.15,3.0,3.5 and 0.5  respectively.

Thus the boxplot is-

The typical values for the M-sample are greater greater than female.

The boxplot and standard deviations suggest that variability is smaller for the M-sample. An outlier, 4.3, in the female sample exists.

There is positive overall skew in the F-sample and negative skew in the M-sample.

Page 46 Problem 2 Answer

Given,Aortic stenosis refers to a narrowing of the aortic valve in the heart.

The paper “Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI” gave the following data on aortic root diameter (cm) and gender for a sample of patients having various degrees of aortic stenosis:

Data on aortic root diameter (cm) and gender for a sample of patients.

We need to find 10% trimmed mean for each of the two samples, and compare to other measures of center.

First, calculate both the trimmed means one by one.

The 10% trimmed mean for the M-sample.

So, (.1)(13)=1.3 elements need to be eliminated from each end of the sample, i.e. eliminate 1

and then elements at each end.

pU/pF=100(1/13)

=7.692%

=100(2/13)

=15.385%​

So now trimmed mean for xtr(7.692)M and xtr(15.385)M/

xˉ​=3.4+3.4+3.5+3.6+3.7+3.7+3.8+3.8+3.9+4.0/11

=3.6545

=3.4+3.4+3.5+3.6+3.7+3.7+3.8+3.8+3.9/9

=3.6444

​Interpolate between  xm(7.692)M and xtr(15.385) M to find xtr(10) M.

xtr(10)​Mˉ=0.7⋅xtr(7.692)M−+0.3⋅xtr(15.385)M−

=(0.7)(3.6545)+(0.3)(3.6444)

=3.6515

We can observe that it is greater than the actual mean of the M-sample but smaller than the median.

For, xrr(10)F eliminate (.1)(10)=1 value from each end of the ordered  -sample.

xˉ=3.0+3.0+3.1+3.1+3.2+3.2+3.5+3.8/8

=3.2375​

We can observe that it is smaller than the actual mean of the F-sample but greater than the median.

Therefore,based on the paper “Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI” gave the following data on aortic root diameter (cm) and gender for a sample of patients having various degrees of aortic stenosis we get the following:

The 10% trimmed mean for M- sample and F-sample are 3.6515 and 3.2375 respectively.

We can observe that it is greater than for both M- sample and F-sample the trimmed mean is smaller than the actual mean but greater than the median.

Page 47 Problem 3 Answer

Given, Based on the article “Establishing mechanical property allowables for metals” we get:

From the given information, we get that:

The sample mean, the sample median, and the trimmed mean have almost the same values.

This shows that the data is very symmetrical and there is no presence of outliers.

If in the worst-case scenario there exists outliers then they are balanced out.

This means that the bigger value would be at the same distance from the mean as the smaller value would be.

All of the values are between122.2 and 147.7.

Variability of the data is very small.

Therefore, based on the article “Establishing mechanical property allowables for metals” the following observations: we can say that:

The data has a symmetric distribution.

There are balanced outliers present.

The range of the data is 25.5.

Page 47 Problem 4 Answer

Given, Based on the article “Establishing mechanical property allowables for metals” as follows.

From the given information, we get that:

Since we are already given the five-number summary, we need to determine whether outliers are present or not.

We distribute our outliers into two categories, extreme and mild outliers.

We need to construct boxplot based on the given data.

In this step, we will determine the fourth spread of our data:

fs=Q3−Q1

⇒138.25−132.95

⇒5.3​

In this step, we will determine the limits of mild and extreme outliers along with specific values of outliers if present.:

The criteria for any data to be an outlier is that it must be further than 1.5f{s} than either Q{1} Or Q{3}.

An outlier is said to be an extreme outlier if it is further than 3f{s} than either Q{1} or Q{3} otherwise, it is said to be a mild outlier.

Since fs=5.3 : ​1.5fs

=1.5×5.3

⇒7.95

3fs=3×5.3

⇒15.9

Then the limits for mild outliers are,

Q3+1.5fs

=138.25+7.95

⇒146.2

Q1−1.5fs

=132.95−7.95

⇒125​

The limits for extreme outliers are,

Q3+3fs

=138.25+15.9

⇒154.15

Q1−3fs

=132.95−15.9

⇒117.05

Therefore we can say that mild outliers lie between [117.05,125] and [146.2,154.15]

We can also say that extreme outliers are values that are smaller than 117.05 and greater than 154.15 .

Hence there are mild outliers present in our data. These are:

117.05<122.2,124.2,124.3<125 and 147<147.7,147.7<154.15

Therefore, our lower data is 122.2,124.2,124.3,125.6,126.3 and our high data is

144.1,144.5,144.5,147.7,147.7

The box plot for the exercise prompt is:

Therefore,based on the article “Establishing mechanical property allowables for metals” the following are obtained:

the final boxplot for the given statistical inference is given as:

From the box plot we can see that we have 3 outliers: 122.2,124.2, and 147.7.

The median of the box-plot is 135.4.

Page 47 Problem 5 Answer

Given, Based on the paper “Decreased Benzodiazepine Receptor Binding in Prefrontal Cortex in Combat-Related Post traumatic Stress Disorder” described the first study of benzodiazepine receptor binding in individuals suffering from PTSD the following data are given:

Variable	​N	​ Mean	SE Mean	TrMean	StDev	Minimum	Q1	Median	Q3
PSTD	13	32.92	2.75	33.82	9.93	10	26.5	37	39
Healthy	13	52.23	4.12	53.09	14.86	23	40.5	51	66.5

 

Variable	​ Maximum
PSTD	46
Healthy	72

We need to use various methods from this chapter to describe and summarize the data.

Construction of box plot using MINITAB

Hence,based on the paper “Decreased Benzodiazepine Receptor Binding in Prefrontal Cortex in Combat-Related Post traumatic Stress Disorder” data

the above box plot shows that PSTD has larger variance than Healthy data and also in PSTD the data distribution in left skewed whereas right skewed in Healthy data.

Page 47 Problem 6 Answer

Given,Based on the article “Can We Really Walk Straight?”reported on an experiment in which each of 20 healthy men  was asked to walk as straight as possible to a target 60 m away at normal speed the data as follows:

Use the methods developed in this chapter to summarize the data; include an interpretation or discussion wherever appropriate.

Count the number of observations which is repeated most often to get the mode.

Now,Frequency of observation is shown below :

Hence,based on the article “Can We Really Walk Straight?”reported on an experiment in which each of 20 healthy men  was asked to walk as straight as possible to a target 60 m away at normal speed the following are obtained as:

The required mode is 0.93.

Page 47 Problem 7 Answer

Given,Based on the article “Can We Really Walk Straight?”reported on an experiment in which each of 20

healthy men  was asked to walk as straight as possible to a target 60

m away at normal speed the data as follows:

We are given our dataset and we are asked to find the mode.

Since the mode is the most frequent value, we will parse through our dataset and find the data which occurs the most.

Then, resulting data is the mode for the cadence dataset.

The dataset given is, ​.95​.85​.92​.95​.93​.86​1.00​.92​.85​.81​.78​.93​.93

​1.05​.93​1.06​1.06​.96​.81​.96

We observe that the data 0.93 occurs the most with a frequency of 4.

Therefore, the mode for the data is 0.93.

Hence,based on the article “Can We Really Walk Straight?”reported on an experiment in which each of 20 healthy men  was asked to walk as straight as

possible to a target 60m away at normal speed the data as follows:

The mode for the cadence data is 0.93.

Page 47 Problem 8 Answer

Given, We are given the dataset of cadence and we are asked to define the modal category for a categorical sample.

In simple terms, the category which would contain the most observations would be known as the modal category.

Therefore,for a categorical sample, we would define the modal category as follows:

The category which would contain the most observations would be known as the modal category.

Page 47 Problem 9 Answer

Given, Specimens of three different types of rope wire were selected, and the fatigue limit (MPa) was determined for each specimen, resulting in the accompanying data as follows:

The five number summary of three different types of ropes are

We need to construct a comparative boxplot, and comment on similarities and differences.

The box plot of three different type of ropes is :

350	​358.00	​371.00	​384.00	​392.00
350	363	371	380	392
350	364	370	379	392

Since the distributions have same minimum and maximum value i.e.350 and 392 respectively, all the three have symmetric distribution.

But the first and fourth quadrant values are different.

Hence, we can conclude that Type1 is more spread out than the other two, therefore the data is more dispersed in Type1 with largest inter quartile range.

Page 47 Problem 10 Answer

Given,Specimens of three different types of rope wire were selected, and the fatigue limit (MPa) was determined for each specimen, resulting in the accompanying data as follows:

In order to plot the dot-plot for all the three different types of ropes, we will find the count for the data values in the data set.

We need to construct a comparative dot plot (a dot plot for each sample with a common scale) and comment on similarities and differences.

For type1 sample:

Therefore, the dot-plot for the following datasets are:

Hence,based on the given data the dot-plot for the dataset can be as as follows:

Page 47 Problem 11 Answer

Given, Specimens of three different types of rope wire were selected, and the fatigue limit (MPa) was determined for each specimen, resulting in the accompanying data:

In order to create a box-plot we need five numbers that will summarize the data.

They are:

1) smallest xi

2)lower fourth,

3)Median,

4)upper fourth,

5)largest xi

Therefore, the boxplot does not give an informative assessment of similarities and differences.

In this case, the dot plot explained the dataset in a more enhanced way.

Therefore the boxplot is not able to give an informative assessment of similarities and differences.

Page 48 Problem 12 Answer

Given,Based on the authors of the article “Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines” (Corrosion, 2009: 332–342) provided the data on which their investigation was based as follows:

To make the Steam-and-Leaf Display:

Stem	Leaves
0L	414141414343434000
0H	587979818181919000
1L	2.04E+24
1H	4040
2L	5959606891969690
2H	1021314649
3L	5774
3H	101830
4L	5858
4H	15
5L	75
5H	33
6L	6570
6H	13
7L	HI: 10.41,13.44
7H
8L

We need to construct a stem-and-leaf display in which the two largest values are shown in a last row labeled HI.

Select the leading digits, following that, the digits coming after the leading digits are the leaves.

Then list the values of the stem we obtained in a column.

Note down the leaf for all data values associated with the particular stem value.

Finally, note down the units of the stem and leaves.

The leading digits (ones) for the data in the exercise are:

0,1,2,3,4,5,6,7,8

Listing the leading digits we get:

Stem

0

1

2

3

4

5

6

7

8​

Let us denote:L (lower): the tenth digits which are less than 5.H (higher): the tenth digits which are more than or equal to 5.

We then have our stem to be:

Stem

0L

0H

1L

1H

2L

2H

3L

3H

4L

4H

5L

5H

6L

6H

7L

7H

8L

​Recording the leaves for each and every observation and obtain the Stem-and-leaf display:

Stem	​ Leaves
0L	414141414343434000
0H	587979818181919000
1L	2.04E+24
	4040
1H	5959606891969690
2L	1021314649
2H	5774
3L	101830
3H	5858
4L	15
4H	75
5L	33
5H
6L
6H
7L
7H	6570
8L	13  HI: 10.41,13.44

Hence ,based on the authors of the article “Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines” provided the data on which their investigation was based obtained as follows:

The stem-and-leaf display is given as:​​

Stem	​ Leaves
0L	414141414343434000
0H	587979818181919000
1L	2.04E+24
	4040
1H	5959606891969690
2L	1021314649
2H	5774
3L	101830
3H	5858
4L	15
4H	75
5L	33
5H
6L
6H
7L
7H	6570
8L	13  HI: 10.41,13.44

Page 48 Problem 13 Answer

Given,The authors of the article “Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines”provided the data on which their investigation was based as  follows:

As the given classes have different widths,

we should construct a density histogram make a table having class, frequency, relative frequency, density.

The formula for relative frequency for each class can be defined as the ratio of the frequency to the total number of observations.

As the given classes are having different widths so, we have to construct a density histogram.

For each class the density is given by the formula: rectangle height = relative frequency of the class/class width

Table is described below:-

After getting the desired values, the graph is constructed as

Hence,based on the authors of the article “Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines” (Corrosion, 2009: 332–342) provided the data on which their investigation was based:

A histogram graph is shown below

Page 48 Problem 14 Answer

Given, Based on the authors of the article “Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines” (Corrosion, 2009: 332–342) provided the data on which their investigation as follows:

By observing the comparative boxplot for different types of soils:

For all types of soil, values are quite similar:

The largest variability holders are in C and CL types.

For SYCL type of soil, there is mild positive overall skew.

significant positive overall skew is shown by the first three types i.e.(C,CL,SCL).

While negative skew id presented by the middle of the data.

For the first three types of soils include extreme one The values for all types of soil are almost same as they are represented in the graph.

Some show positive overall skew while some show mild positive overall skew.

Hence, Important features are described as below.

For all types of soil, values are quite similarThe largest variability holders are in C and CL types.

For SYCL type of soil, there is mild positive overall skew.significant positive overall skew is shown by the first three types i.e.(C,CL,SCL).

While negative skew id presented by the middle 50% of the data.

For the first three types of soils include extreme one

Page 48 Problem 15 Answer

Given,The article “Planning of City Bus Routes”gives the following information on lengths (km) for one particular system:

We need to draw a histogram for the above given data.

Hence,according to length and frequencies, histogram is drawn shown below:

Page 48 Problem 16 Answer

Given, The article “Planning of City Bus Routes”gives the following information on lengths (km) for one particular system: number which is less then 20:-6,23,30,35,32,48,42

Given number which is at least 30:- 27,11,2

N= some of all frequency that is 391​ Less than 20=6+23+30+35+32+48+42/N

=6+23+30+35+32+48+42/391

=216/391

=0.5524

Therefore, the proportion of route lengths are less than 20 is 0.5524.

atleast 30=27+11+2/N

=40/391

=0.1023​

Therefore, the proportion of routes have lengths of at least 30 is0.1023.

Therefore, the proportion of route lengths are less than 20 is 0.5524 .

Therefore, the proportion of routes have lengths of at least 30 is 0.1023 .

Page 48 Problem 17 Answer

Given,Based on the article “Planning of City Bus Routes” gives the following information on lengths (km) for one particular system:

The cumulative frequency table is,

Class	0−<0.5	0.14754	0.4918	0.2623
Frequency	9		4	2
Relative	0.14754	0.29508	0.06557	0.03279
Frequency		7	0.01311	0.00656
Density	0.29508	0.11475	​1.5−2
Frequency	7	0.05738	8
Relative Frequency	0.11475	​1−<1.5	0.13115
	0.11475	15
Density	​0.5−1	0.2459
	9

Here , N=391

90th  position is 90N

100=351.9

The cumulative frequency just greater than 351.9 is 378 . The corresponding class is 30−35 . The lower limit of the class is 30 . The corresponding frequency is 27.

Use the following formula to compute the 90th percentile,P∞=I+h/f (90N/100−C)

Here, l is the lower limit of the median class = 30,

h is the height of the median class= 5.

C is the prior cumulative of frequency of the median class = 351

On substituting the values in the above formula:

P90=30+5/27(90(391)/100−351)

=30+5/27(351.9−351)

=30.1667​

Therefore, the 90th  percentile is 30

Hence,based on the article “Planning of City Bus Routes” (J. of the Institution of Engineers, 1995: 211–215) gives the following information on lengths (km) for one particular system obtained as follows:

Therefore, the 90th percentile is 30.

Page 48 Problem 18 Answer

Given,  Lengths of bus routes for any particular transit system will typically vary from one route to another.

Based on the article “Planning of City Bus Routes”gives the following information on lengths (km) for one particular system:

l=18,

h=2,

f=42,

C=174

We need to find the median route length.

Compute the median route length.

N/2​=(391)/2

=195.5​

The cumulative frequency (c.f) just greater than 195.5 is 216. Median =l+h/f(N/2−C) on substituting, Median ​

=18+2/42(195.5−174)

=18+(0.0476)(21.5)

=18+1.0238

=19.0238

​Therefore, the median route length is 19.0238

Hence,based on the article “Planning of City Bus Routes”gives the following information on lengths (km) for one particular system: the median route length is 19.0238.

Page 49 Problem 19 Answer

Given,Consider the following time series in which x{t}=temperature in Fahrenheit of effluent at a sewage treatment plant on day t: 47,54,53,50,46,46,47,50,51,50,46,52,50,50

We need to plot xtˉagainst t on a two-dimensional coordinate system (a time-series plot) and does there  appear any pattern.

Now, Consider the times series of effluent at a sewage treatment plant on day t:

The data set has  a range of values from 46 to 54.

By plotting the data points on temperature time series plot we get:

By observing the time series plot the data appears to be cyclical pattern similar to the sine wave.

So,by plotting the temperature time series data points we get :

By observing the time series plot the data appears to be cyclical pattern similar to the sine wave.

Page 49 Problem 20 Answer

Given, Smoothing constant: forα=.1

Initially set X1ˉ=x1

forα=.5

We need to calculate the xtˉ

t	​x
1	47
2	47.7
3	48.23
4	48.41
5	48.17
6	47.95
7	47.85
8	48.07
9	48.36
10	48.53
11	48.27
12	48.65
13	48.78
14	48.9

for different values of smoothing constant given above.

Then we need to comment that for which value of α the xtˉseries is smoother.

x2ˉ​=α⋅x2+(1−α)⋅x1ˉ=(0.1)(54)+(1−0.1)(47)

=47.7 and thren use the same method to compute, x3ˉ

x3ˉ​=α⋅x3+(1−α)⋅x2ˉ=(0.1)(53)+(1−0.1)(47.7)=48.23

Now repeat this process to get a table for all values in the time series dataset, with α=0.1

Repeat the exponential smoothing technique with α=0.5 to obtain the another table of data,.

t	​x
1	47
2	50.5
3	51.75
4	50.88
5	48.44
6	47.22
7	47.11
8	48.55
9	49.78
10	49.89
11	47.94
12	49.97
13	49.99
14	49.99

The process is exactly the same as before, only with a different value of α Exponential Smoothing with α=0.5

The first series withα=0.1 seems to be smoother than the second series with α=0.5.

The first series is smoother because there is less variability between values .

Additionally, the first series switches between increasing and decreasing less frequency.

When the first series does switch between increasing and decreasing, the magnitude the switch is much less than that of the switches in the second series.

Therefore, after examining both series, α=0.1 yields a smoother series

Page 49 Problem 21 Answer

Given,Substituting Xt−1−=αXt−1 +(1−α)Xt−2−on the right-hand side of the expression forxˉ,then substitute Xt−2−in terms of  Xt−2 and Xt−3−,

We need to find the number of values xt,xt−1,…,x1 does xˉ  depend and what happens to the coefficient xt−k  on  as k  increases.

Perform the substitution as directed in the problem statement.

xtˉ​=α⋅xt+(1−α)⋅xt−1−=α⋅xt+(1−α)⋅[α⋅xt−1+(1−α)⋅xt−2−]

=α⋅xt+α⋅(1−α)⋅xt−1+(1−α)2⋅xt−2−

Repeat the process again and substitute x in terms of x and x . t−2−t−2 t−3−

xt = α ⋅ xt + α ⋅ (1 − α) ⋅ xt−1 + (1 − α) ⋅ x/2t−2−

= α ⋅ xt + α ⋅ (1 − α) ⋅ xt−1 + (1 − α) ⋅ α ⋅ x + (1 − α) ⋅ x2[ t−2 t−3−]

= α ⋅ xt + α ⋅ (1 − α) ⋅ xt−1 + α ⋅ (1 − α) ⋅ x + (1 − α) ⋅ x2t−2/3t−3−

= (1 − α) ⋅ x + α ⋅ (1 − α) ⋅ x3/t−3− ∑k=0 2 ki−k

​repeat this process, and recalling that xˉ = x1, using the following formula From

xˉt = (1 − α) ⋅ + α ⋅ (1 − α) ⋅ x t−1 xˉ1 ∑k=0 t−2 k t−k

= (1 − α) ⋅ x + α ⋅ (1 − α) ⋅ x t−1 1 ∑k=0 t−2 k t−k

the formula, depends on all of the previous values This is true because each value appears in the formula for so each value has an impact on the value of .

Thus depends on all of the previous values

xˉt

xˉt xt, xt−1,…, x1

Since is between zero and one, is between zero and one, so raising it to higher and higher powers causes the value to get smaller.

This is precisely what happens to the coefficient of as the value of k increases.

In the limit, this coefficient goes to zero as k goes to infinity.

Thus, as k increases the coefficient of decreases

So,xtˉ depends on all of the previous values xt,xt−1,…,x1 which aret−1 terms.

As k increases the coefficient of xt−k decreases.

Page 49 Problem 22 Answer

Given,The sample data X1,X2,…,Xn represents a time series.

X1ˉ=x1

If t is large, we need to find how sensitive is Xtˉ to the initialization X1ˉ=x1 and explain.

Since, a is between zero and one, 1−α is between zero and one, so raising it to higher and higher powers causes the value to get smaller.

This is precisely what happens to the coefficient of xt−k as the value of k increases.

In the limit, this coefficient goes to zero as k goes to infinity.

Therefore,it is proved that xtˉ  is very insensitive to the initialization x1ˉ=x1.

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics

Page 43 Problem 1 Answer

Given ,Based on the study of the relationship between age and various visual  functions (such as acuity and depth perception) reported the following observations on the area of scleral lamina from human optic nerve heads (“Morphometry of Nerve  Fiber Bundle Pores in the Optic Nerve Head of the Human”).

The following data is provided:

2.75,2.62,2.74,3.85,2.34,2.74,3.93,4.21,3.88,4.33,3.46,4.52,2.43,3.65,2.78,3.56,3.01

We have to calculate i∑xi and i∑x²_i

To calculate i∑xi we find

2.75+2.62+2.74+3.85+2.34+2.74+3.93+4.21+3.88+4.33+3.46+4.52+2.43+3.65+2.78+3.56+3.01

2.75+2.62+2.74+3.85+2.34+2.74+3.93+4.21+3.88+4.33+3.46+4.52+2.43+3.65+2.78+3.56+3.01=56.8

​Hence i∑xi=56.8

To calculate i∑x²i

we find 2.752+2.622+2.742+3.852+2.342+2.742+3.932+4.212+3.882+4.332+3.462+4.522+2.432+3.652+2.782+3.562+3.012

2.752+2.622+2.742+3.852+2.342+2.742+3.932+4.212+3.882+4.332+3.462+4.522+2.432+3.652+2.782+3.562+3.012=197.804

​Hence i∑xi²=197.804

So,based on the study of the relationship between age and various visual  functions (such as acuity and depth perception) reported the following observations on the area of scleral lamina from human optic nerve heads then the values as follows:

∑ixi=56.8 and ∑ixi²=197.804

​Page 43 Problem 2 Answer

Given, Based on the study of the relationship between age and various visual functions (such as acuity and depth perception) reported the following observations on the area of scleral lamina   from human optic nerve heads (“Morphometry of Nerve  Fiber Bundle Pores in the Optic Nerve Head of the Human”.

The given data is: 2.75, 2.62, 2.74, 3.85, 2.34, 2.74, 3.93,4.21, 3.88, 4.33, 3.46, 4.52, 2.43, 3.65, 2.78, 3.56,3.01.

we need to find the sample variance using the formula.

First we have to calculate the standard deviation using sample variance.

To find s² we calculate s²=  i∑xi²−(∑ixi)²/n

Since i∑xi=56.8 therefore,(∑x​i)²/n

=(56.8)/2

17(∑x​i)²/n

=189.779​

Also since  i∑xi²=197.804

We have s{2}=197.804−189.779

The sample variance  s{2}=8.025

To find standard deviation  we s see

s=(8.025)1/2

s=2.8329

The standard deviation s=2.8329

Hence, based on study of the relationship between age and various visual  functions (such as acuity and depth perception) reported the following observations on the area of scleral lamina from human optic nerve heads determined as follows:

sample variance s{2}=8.025 and the standard deviation s=2.8329 .

Page 44 Problem 3 Answer

Given, Based on the article “A Thin-Film Oxygen Uptake Test for the  Evaluation of Automotive Crankcase Lubricants” the data as follows:

Set of data is ​87​103​130​160​180​195​132​145​211​105​145​153​152​138​87​99​93​119​129​.

We need to find the sample variance and standard deviation.

To find sample variance  we calculate s²=i∑xi²−(∑ixi)²/n

Since ∑ixi=87+103+130+160+180+195+132+145+211+105+145+153+152+138+87+99+93+119+125+129

∑ixi=2563​

Therefore,

(∑x​i)²/n=(2563)²/19

(∑x​i)²/n=345735.2105

​Also​∑ixi²=872

+1032+1302+1602+1802+1952+1322+1452+2112+1052+1452+1532+1522+1382+872+992+932+1192+1252+1292

∑ixi²=368501​

So we have  s{2}=368501−345735.2105

s{2}=22765.7895

The sample variance

s{2}=22765.7895

To find the standard deviations use s=√s²

The standard deviation s=150.8834

Hence, based on the article “A Thin-Film Oxygen Uptake Test for the  Evaluation of Automotive Crankcase Lubricants” is determined as follows:

sample variance is s{2} =22765.7895 and

standard deviation is s=150.8834.

Page 44 Problem 4 Answer

Given,

The given data set is: 87​ 103 ​130 ​160​ 180 ​195 ​132 ​145​ 211​ 105 ​145 ​153​ 152​ 138​ 87​9 9​93 119​ 129

We need to find the sample variance and sample standard deviation when observations were re-expressed in hours without actually performing the reexpression.

Since in the re expression minutes are changed into hours the observations xi  are now  xi/60

Hence the new sample variance s1

2=i∑(xi/60)²−(∑ixi/60)²/n

∑i(x​I/60)²−(∑ixi/60)²/n

=∑​ixi²−​{∑x_i²/n

3600

s1²=s2/3600

The new sample variance s{1}{2}=6.3238.

To find the new standard deviation  s1 use s1=√s1²

s{1}=2.514722

The new standard deviation s{1}=2.514722

Hence,based on the article “A Thin-Film Oxygen Uptake Test for the  Evaluation of Automotive Crankcase Lubricants”from the following data we get:

new sample variance s{1}{2}=6.3238 and new standard deviation s1=2.514722.

Page 44 Problem 5 Answer

Given, Based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles”included the following data on grip strength (N) for a sample of 42 individuals:

The given data set is :

16​1 8​1 8​2 6​3 3​41​ 54​ 56​6 6​68​ 87​9 1​9 59 8​10 6​10 9​11 1​11 8​12 7​12 7​13 5​14 5​14 7​14 9​15 1​168

172 ​183​ 189​ 190 ​200 ​210​ 220 ​229​ 230​ 233​ 238​ 244 ​259 ​294 ​329 403

​we need to construct a stem-and-leaf display based on repeating each stem value twice.

To construct a stem-and-leaf display we take the hundreds digit as the stem and tens and ones digits as the leaf.

We repeat each stem value twice by adding L to the stem with leaves from 00 to 49  and H to the stem with leaves from to.

Stem	Leaves
OL	16 18 18 26 33 41
OH	54 56 66 68 87 91 95 98
1L	06 09 11 18 27 27 35 45 47 49
1H	51 68 72 83 89 90
2L	00 10 20 29 30 33 38 44
2H	59 54
3L	29
3H
4L	3

Comment on interesting feature :

Most data are grouped in the 1L leaf. The largest sample value,403  is very far from the bulk of the data. It seems to be an outlier.

Hence,based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles”

most data is grouped in 1L leaf and the largest sample value is 403 which seems to be outlier.

Page 44 Problem 6 Answer

Given, Based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles”included the following data on grip strength (N) for a sample of 42 individuals:

The given dataset is:

16​1 8​18 ​26​3 3​41​ 54​5 6​66 ​68​8 7​91​ 95 98​ 106​ 109​ 111​ 118​ 127 ​127 ​135​ 145 ​147 ​149​ 151 ​168​

172​ 183​ 189 ​190​ 200​ 210 ​220​ 229​ 230​ 233​ 238 ​244​ 259 294​ 329​ 403

We need to find the values of the fourths and the fourth spread.

First segregate the data into two classes larger half and smaller half

The smaller half of the data set is

16,18,18,26,33,41,54,56,66,68,87,91,95,98,106,109,111,118,127,127,135

​There are 21 observations hence the median is the 11th observation hence

The larger half of the data set is

145,147,149,151,168,172,183,189,190,200,210,220,229,230,233,238,244,259,294,329,403

​There are 21 observations hence the median is the11th

observation hence upper fourth =210 ​Fourth spread fs

given by fs= upper fourth − lower fourth

​fs=210−87

fs=123

Hence the Fourth spread fs=123

​Therefore,based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” included the

following data on grip strength (N) for a sample of 42 individuals the values of the fourths and the fourth spread obtained as follows:

The Fourth spread fs=123

​Page 44 Problem 7 Answer

Given, Based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles”included the following data on grip strength (N) for a sample of 42 individuals:

The given dataset is :

16​1 8​18​ 26​3 3​41 ​54​5 6​66 ​68​8 7​91 ​95  98 106 ​109 111 ​118​ 127 ​127 ​135 ​145​ 147​ 149​ 151​ 168  172 ​183​ 189 ​190 ​200​ 210​ 220 ​229​ 230​ 233 ​238 ​244 ​259 294​ 329​4 03

​we need to construct a box plot.

The box plot  for the data

There is slight positive skew of the data.

The variability is rather big and there is one outlier on the upper end, 403.

Hence, based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles”included the following data on grip strength (N) for a sample of 42 individuals:

The boxplot is a better representation of the data is as follows:

Comment : There is slight positive skew of the data and the variability is rather big and there is one outlier on the upper end, 403.

Page 44 Problem 8 Answer

Given, Based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” included the following data on grip strength (N) for a sample of 42 individuals:

The given dataset is :

​16​ 18 ​18​ 26 ​33​ 41​ 54​5 6​66 ​68​8 7​91 ​95 ​98​ 106​ 109 ​111 ​118 ​127 ​127 ​135​ 145 ​147 ​149

​151​ 168  172 ​183 ​189 ​190 ​200 210​ 220​ 229​ 230​ 233​ 238​ 244 ​259 294​ 329​ 403​

We need to find if there is any outlier or extreme outlier.

Now, lower fourth =87 upper fourth =210

​​Since Fourth spread fs given by ​fs= upper fourth − lower fourth

​fs=210−87

fs=123

Hence the Fourth spread fs=123​

An element is an outliner if it is less than 87−1.5fs or greater than 210+1.5fs

Since there are no negative data so an element is an outliner if it is greater than 394.5

The element 403 is an outlier.

An element is an extreme outliner if it is less than 87−3fs or greater than 210+3f

​Since there are no negative data so an element is an extreme outliner if it is greater than 579.

There are no elements greater than 579 hence there are no extreme outlier.

So,based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” included the following

data on grip strength (N) for a sample of 42 individuals the element 403 is an outlier there are no extreme outlier.

Page 44 Problem 9 Answer

Given, Based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” included the following data on grip strength (N) for a sample of 42 individuals:

The given dataset is:

16​1 8​18 ​26​3 3​41​ 54​5 6​66​ 68​8 7​91 ​95​9 8​10 6​10 9​11 1​11 8​12 7​12 7​13 5​14 5​14 7​14 9​15 1​16 8​17 2​18 3​18 9​19 0​200 ​210 ​220​ 229 ​230​ 233 238​ 244 ​259 ​294​ 329​ 403

​We need to find by how much could the observation 403, currently the largest, be decreased without affecting fs.

lower fourth =87 upper fourth=210

Since Fourth spread fs= upper fourth − lower fourth

fs=210−87

fs=123

Since the fourth spread f{s}=123

So to effect the fourth spread either the upper fourth should change or the lower fourth

should change since the largest element 403 is closer to upper fourth so to change the upper fourth

The observation should become less than the upper fourth

so for 403 to become less than 210 it should be reduced at least by 194.

Hence the largest element 403 can be decreased by 193 without changing the fourth spread.

Therefore,based on the article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” included the following data on grip strength (N) for a sample of 42 individuals then the largest element 403 can be decreased by 193 without changing the fourth spread.

 

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability

Page 71 Problem 1 Answer

Given- 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.

To find- ways to serve 3bottles of zinfandel and serving order is important.We will use permutation formula P(n,r)=n!

(n−r)! to solve the problem.

If 3 bottles of zinfandel are to be served, so the number of ways for selecting 3 out of 8 are P(8,3).

P(8,3)=8!

(8−3)!=8!

5!=8×7×6×5!

5!=8×7×6=336

​There are 336 ways to serve 3 bottles of zinfandel and serving order is important.

Page 71 Problem 2 Answer

Given-8 bottles of zinfandel,10  of merlot, and 12 of cabernet all from different wineries. To find- ways to serve randomly selected 6 bottles of wine from the 30.

Since the order is not important we will use combination formula C(n,r)=n!

r!(n−r)! to solve the problem.

The number of ways of to serve 6 bottles of randomly selected wine out of 30 servings-

​C(30,6)​=30!

6!(30−6)!=30!

6!×24!=593,775

​The number of ways of to serve 6 bottles of randomly selected wine out of 30 servings are 593,775.

Page 71 Problem 3 Answer

Given-8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.

To find- ways to obtain two bottles of each variety when 6bottles are randomly selected.

We will use combination formula C(n,r)=n!

r!(n−r)! to solve the problem because we need two of each variety thus order will matter.

For 6 bottles we need 2 from 8 bottles of zinfandel, 2 from 10 bottles of merlot, 2 from 12 bottles of cabernet, so total number of ways -8C2×10

C2×12

C2=8!

2!(8−2)!×10!

2!(10−2)!×12!

2!(12−2)!=8!

2!×6!×10!

2!×8!×12!

2!×10!=12!

2!×2!×2!×6!

=83,160

​There are 83,160 ways to obtain two bottles of each variety when 6 bottles are randomly selected.

Page 71 Problem 4 Answer

Given- 8 bottles of zinfandel, 10 of merlot, and 12  of cabernet all from different wineries.

Number of ways of selecting two bottles of each variety being chosen when 6 bottles are randomly selected obtained in c): 83160.

To find- probability that two bottles of each variety being chosen when 6 bottles are randomly selected.

We will find the probability of the event which is given by: Number of favourable outcomes.

Total number of outcomes

As we know that there are 83,160  ways of selecting two bottles of each variety being chosen when  6 bottles are randomly selected.

So, the probability that two bottles of each variety being chosen when 6 bottles are randomly selected.

P( Two bottles of each variety )=83,160/30

C6=0.14​

The probability that two bottles of each variety being chosen when 6 bottles are randomly selected is 0.14.

Page 71 Problem 5 Answer

Given-8 bottles of zinfandel, 10 of merlot, and 12 of cabernet all from different wineries.

To find-  the probability that all of the bottles are of same variety when 6 bottles are randomly selected.

We will use combination to solve the problem of all of them being the same variety and then use probability to find the answer.

If 6 bottles are randomly selected from 30 bottles, then the probability that all of them are of the same variety is,

P( all of them are the same variety )=8

C6+10

C6+12

C6/30

C6=28+210+924/593,775

=0.002

​The probability that all of the bottles are of same variety when 6 bottles are randomly selected is0.002 .

Page 71 Problem 6 Answer

Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 6

workers on the graveyard shift. A quality control consultant is to select 6 of the workers in-depth interview.

To find- number of selections that 6 of the workers to be selected for an in-depth interviews will be from the day shift and the probability that all 6 selected workers will be from the day shift.

We will firstly find the combination to select the workers using C(n,r)=n!

r!(n−r)! the find the probability.

The 6 workers are selected from 20 day shift workers. So, the number of ways are(20/6) by using combination formula-

C(n,r)=n!

r!(n−r)!

=(20/6)

=20!

6!×14!

=38760 ways

​The probability that all the 6 members are selected from day shift is-

P(all 6 selected workers will  be from the day shift )

​=(20/6)/(45/6)

=38,760/8,145,060

=0.0048

​The number of selections that 6 of the workers to be selected for an in-depth interviews will be from the day shift are 38760 .

The probability that all 6 selected workers will be from the day shift is 0.0048.

Page 71 Problem 7 Answer

Given:  Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.

A quality control consultant is to select 6of the workers in-depth interview.

To find the probability that all 6selected workers will be from the same shift.

We will first use the formula n/Cr=n!

r!(n−r)! to choose from different shifts and then take the probability.

Calculate the probability of all the chosen people to be in day shift,

D=20

C6/20

C6=20!

6!(20−6)!

20/C6=38760

​Calculate the probability of all the chosen people to be in night shift,

N=15/C6

15/C6=15!

6!(15−6)!

15/C6=5005

​Calculate of probability of all the chosen people to be in the graveyard shift,

G=10/C6

10/C6=10!

6!(10−6)!

10/C6=210

​Calculate the total number of the probability of picking 6 people out of 45,

T=45/C6

45/C6=45!

6!(45−6)!

45/C6=814506

​Now substitute the values in the formula,

P=38760+5005+210/814506

P=0.00539

​The probability that all 6 selected workers will be from the same shift is:0.00539.

Page 71 Problem 8 Answer

Given- A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and  10 workers on the graveyard shift.

A quality control consultant is to select 6 of the workers in-depth interview.

To find the probability that at least two different shifts will be represented among the selected workers The probability that all 6 selected workers will be from the same shift is obtained as: 0.00539 Hence by using this we will find the required probability.

The probability of all 6 chosen can be from one shift was calculated from the previous question,

P=0.00539

Hence the probability that at least two different shifts will be represented among the selected workers

=1−Probability that all 6 selected workers will be from the same shift.

=1−0.00539

=0.9946​

The total probability of at least two different shifts represented among the selected workers is 0.9946.

Page 71 Problem 9 Answer

Given- A production facility 20 employs workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.

A quality control consultant is to select 6 of the workers in-depth interview.

To find the probability that at least one of the shifts will be unrepresented in the sample of workers.

We will first find the number of ways of selection for different shifts not represented using nCr=n!

r!(n−r)! and then get the probability.

Finding the number of ways in which day shift is not represented,

A1=15+10

C6/25

C6=25!

6!(25−6)!

25/C6=177100

​Finding the number of ways in which  swing shift is not represented,

A2=10+20/C6

30/C6=30!

6!(30−6)!

30/C6=593775

​Finding the probability of graveyard shift not represented,

A3=20+15

C6/35

C6=35!

6!(35−6)!

35/C6=1623160

​Hence we get:P(A)=25

C6/45

C6=0.022

P(A2)=30

C6/45

C6=0.073

P(A3)=35

C6,45

C6=0.199

​Similarly we get:

Now add all the possibilities and subtract individual probabilities to find the solution,

P3=A1+A2+A3−(P(A1∩A2)+P(A2∩A3)+P(A1∩A3))+P(A1∩A2∩A3)

Since P(A1∩A2)+P(A2∩A3)+P(A1∩A3)is nothing but the probability of all the swifts together, which was calculated earlier, which is 312.

The value of P(A1∩A2∩A3)=0

Now substitute the value of all the values calculated in the formula,

Hence the required probability will be:

P3=P(A1)+P(A2)+P(A3)−(P(A1∩A2)+P(A2∩A3)+P(A1∩A3))+P(A1∩A2∩A3)

=0.022+0.073+0.199+0.00003+0.0006+0.005+0

=0.289

​The probability that at least one of the shifts will be unrepresented in the sample of workers is 0.289.

Page 72 Problem 10 Answer

Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs.

Given: three bulbs are randomly selected.To find the probability that exactly two of the selected bulbs are rated 75-W.

First, we will find the number of outcomes that is in how many ways two bulbs of 75−W can be removed by using combination Cr,n=(n/r) and then find the probability.

Number of 40 W bulbs-4

Number of 60 W bulbs-5

Number of 75 W bulbs-6

Hence 2 bulbs shall be taken from a group of 6 items and remaining one can be taken from 4+5=9

Hence the number of ways to do so is:

C2,6×C1,9=(6/2)×(9/1)

⇒C2,6×C1,9

=135​

The total outcomes are:

C3,15=(15/3)

C3,15=455

Hence if we consider A to be the event where exactly two bulbs of 75−W are selected then:

P(A)= number of favorable outcomes in A number of outcomes in the sample space

P(A)=135/455

P(A)=0.297

​The probability that exactly two of the selected bulbs are rated 75-W is 0.297.

Page 72 Problem 11 Answer

Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs.

Given: Three bulbs are randomly selected.to find the probability that all three bulbs selected have the same rating.

We will use the formula for disjoint events: P(A∪B∪C)=P(A)+P(B)+P(C)

It is possible that the 3 selected light bulbs are all from either of the groups 40 W, 60 W and 75 W.

That is, denote events B1,B2,B3 as all 3 light bulbs are from 40W,60W , and 75W, respectively.

The event B is the initial one – all 3 randomly selected are from the same group.

Hence, as the 3 events B1,B2,B3 are disjoint  we can apply:

P(A∪B∪C)=P(A)+P(B)+P(C)

⇒P(B)=P(B1∪B2∪B3)

P(B1∪B2∪B3)=P(B1)+P(B2)+P(B3)=C3,4/C3,15+C3,5/C3,15+C3,6/C3,15

=4/455+10/455+20/455

=34/455

=0.075.

​The probability that all three of the selected bulbs have the same rating is 0.075.

Page 72 Problem 12 Answer

Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75

-W bulbs Given: Three bulbs are randomly selected.

To find the probability that one bulb of each type is selected.

We will solve this by getting the favorable outcomes and using (n/1)=n.

We denote with C event that the 3 randomly selected light bulbs have representative in each of the three groups.

Then the required probability is given by:

P(C)= number of favorable outcomes in C number of outcomes in the sample space

P(C)=(4/1)×(5/1​)×(​6/1​)(​15/3​)

P(C)=4×5×6/455

P(C)=0.2637

​The probability that one bulb of each type is selected is 0.2637.

Page 72 Problem 13 Answer

Given: A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs Given: Bulbs are to be selected one by one until a 75-W bulb is found.

To find the probability that it is necessary to examine at least six bulbs.

We will use complement rule P(A)=1−P(Ac) to find the required probability.

Let D be the event that is is necessary to examine at least six bulbs

The probability of the required event may be calculated using the facts that the light bulbs are selected one-by-one, and that the selected light bulb before sixth is from group of 75 W (complement of D , in 1./2./3./4 or 5. turn 75W is selected) Therefore we have:

P{ At least six bulbs selected  to get 75W Bulb }=1−P{75W bulb selected  within first five bulbs }

P(D)=1−P(Dc)

=1−6/15−9/15×6/14−9⋅8/15⋅14×6/13−9⋅8⋅7

15⋅14⋅13×6

12−9⋅8⋅7⋅6

15⋅14⋅13⋅12×6/11

=1−0.958

=0.042

​The probability that it is necessary to examine at least six bulbs is 0.042.

Page 72 Problem 14 Answer

Given: An ATM personal identification number (PIN) consists of four digits, each a0,1,2,…8, or 9, in succession.To find the number of different possible PINs if there are no restrictions on the choice of digits.

We will use the product rule for k−tuples.

Finding the number of ways the PIN can be arranged,When there is no restrictions each digit can be selected in 10 ways.

So possible numbers of pin are: P=10×10×10×10

P=10000​

The total number of different possible PINs can be generated, if there are no restrictions on the choice of digits is 10000.

Page 72 Problem 15 Answer

Given:  An ATM personal identification number (PIN) consists of four digits, each a0,1,2,…8 , or 9 , in succession.

Given:  According to a representative at the author’s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited:

(1) all four digits identical

(2) sequences of consecutive ascending or descending digits, such as 6543

(3) any sequence starting with 19

To find the probability that the sequence of pin will be legitimate.

Calculating the probability of all the conditions,. The identical numbers are totally 10.

There are 7 combinations of ascending and 7 descending combinations. A total of 14 .

The first two numbers are 1and 9, then the probability is 10=100.

Now remove all the invalid probabilities from the total,

P=10000−10−14−100/10000

P=0.9876

The probability that the random pick will be a legitimate PIN is 0.9876

Page 72 Problem 16 Answer

Given:  An ATM personal identification number (PIN) consists of four digits, each0,1,2,…8, or 9, in succession.

Given: Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively.

He has three tries before the card is retained by the ATM (but does not realize that).

So he randomly selects the 2^nd and 3^rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try .

To find the probability that the individual gains access to the account.

First, we will find the favorable outcomes by using the product rule and then find the probability.

Since there is only two slots left and 10 digits, the number of favorable outcomes will be

S=10⋅10

S=100

​There was three attempt for trying a combination,

P=3/100

P=0.03 .

The probability that the individual gains access to the account is 0.03

Page 72 Problem 17 Answer

Given: An ATM personal identification number (PIN) consists of four digits, each0,1,2,…8, or 9, in succession.

Obtained probability in part c) 0.03

To find the probability that the individual gains access to the account if the first and last digits are 1 and 1, respectively.

We will get the favorable outcomes and then find the required probability.

Since the numbers are identical, there is a possibility of violation in the password.

It can not be1111 and can not contain continues numbers (1901,1902,1991).

Therefore, the total number of violation is 10+1=11

The total probability with three possible combination is,

P=3/100−11

P=3/89

P=0.0337

The probability for the individual to gain access to the account is 0.0337.

Page 72 Problem 18 Answer

Given:  A starting lineup in basketball consists of two guards, two forwards, and a center.

Given:  A certain college team has on its roster three centers, four guards, four forwards, and one individual (X) who can play either guard or forward.

To find the number of different lineups that can be created.

We shall use the combination formula nCr=n!

r!(n−r)! for the possibilities for which X can be picked or not.

There are three possibilities of player(X),

When (X) is not picked,

P=(2 out of 4 guards )+(2 out of 4 frowards )+(1 out of 3 center )

P=(4/C2)⋅(4/C2)⋅(3/C1)

P=6⋅6⋅3

P=108

When (X) is picked as guard,

P=(1 out of 4 guards )+(2 out of 4 frowards )+(1 out of 3 center )

P=(4/C1)⋅(4/C2)⋅(3/C1)

P=4⋅6⋅3

P=72

The same number is when the player is picked as forward.

P=(2 out of 4 guards )+(1 out of 4 forwards )+(1 out of 3 center )

P=(4/C2)⋅(4/C1)⋅(3/C1)

P=4⋅6⋅3

P=72

​Now add all the found variables,

T=108+72+72

T=252​

The number of starting lineups that can be created is 252.

Page 72 Problem 19 Answer

Given: A starting lineup in basketball consists of two guards, two forwards, and a center.

Given:  the roster has 5 guards, 5 forwards, 3 centers, and 2 “swing players” (X and Y) who can play either guard or forward.

To find the probability that the team consists of a legitimate starting lineup.

We will find the combination with different position of the individual player X (old team member) or (given team member) and them sum up all.

When the starting lineup is made without X or Y: C2,5⋅C2,5⋅C1,3

=(​5/2​)⋅(​5/2​)⋅(​3/1​)

=5!/2(5−2)!⋅5!/2!(5−2)!⋅3!/1!(3−1)!

=10⋅10⋅3

=300

X is guard and there is no Y:​C1,5⋅C2,5⋅C1,3

=(​5/1​)⋅(​5/2​)⋅(​3/1​)

=5!/1((5−1)!⋅5!/2(5−2)!⋅3!/1(3−1)!

=5⋅10⋅3

=150

Xis guard and there is no Y:C2,5⋅C1,5⋅C1,3

=(​5/2​)⋅(​5/1​)⋅(​3/1​)

=5!/2(5−2)!⋅5!/1!(5−1)!⋅3!/1!(3−1)!

=10⋅5⋅3

=150

Y is guard and there is no X:

​C1,5⋅C2,5⋅C1,3

=(​5/1​)⋅(​5/2​)⋅(​3/1​)

=5!/1((5−1)!⋅5!/2!(5−2)!⋅3!/1(3−1)!

=5⋅10⋅3

=150

Y is forward and there is no X:C2,5⋅C1,5⋅C1,3

=(​5/2​)⋅(​5/1​)⋅(​3/1​)

=5!/2!(5−2)!⋅5!/1!(5−1)!⋅3!/1(3−1)!

=10⋅5⋅3

=150

X is guard and Y is also guard:

1⋅C2,5⋅C1,3

=1⋅(​5/2​)⋅(​3/1​)

=1⋅5!/2!(5−2)!⋅3!/1!(3−1)!

=1⋅10⋅3

=30

X is guard and Y  is forward:

C1,5⋅C1,5⋅C1,3

=(​5/1​)⋅(​5/1​)⋅(​3/1​)

=5!/1(5−1)!⋅5!/1(5−1)!⋅3!/1(3−1)!

=5⋅5⋅3

=75

​Xis forward and Y is guard: C1,5⋅C1,5⋅C1,3

=(​5/1​)⋅(​5/1​)⋅(​3/1​)

=5!/1!(5−1)!⋅5!/1!(5−1)!⋅3!/1(3−1)!

=5⋅5⋅3

=75

X is forward and Y is forward:

C2,5⋅1⋅C1,3

=(​5/2​)⋅1⋅(​3/1​)

=5!/2!(5−2)!⋅1⋅3!/1!(3−1)!

=10⋅1⋅3

=30

​We will now sum up all the favorable possibilities.

300+4⋅150+2⋅30+2⋅75=1110

Total number of ways to choose 5 players from 15 members:

​(​15/5​)

​=15!

5!(15−5)!

=3003

The required probability of five constituting the legitimate starting lineup is:

P(A)= Number of favorable outcomes in A

Number of outcomes in the sample space

=1110/3003

=0.3696

​The probability of five constituting the legitimate starting lineup is 0.3696.

Page 72 Problem 20 Answer

Given: In five-card poker, a straight consists of five cards with adjacent denominations. You are dealt with a five-card hand.

To find the probability that it will be a straight with high card 10.

To find the probability that it will be straight.

To find the probability that it will be a straight flush.

We will use the combination formula Ck,n=n!

k!(n−k)! to find the favorable outcomes for each case and then find the probability.

Total ways to select 5 cards out of 52are

C5,52 =2,598,960

​(​52/5​)/52!

= 5!(52−5)!

We have a 10 high card, straight consists of six, seven, eight, nine, ten.

We will have 4 different types of cards of each type.

So, favorable outcomes will be 4⋅4⋅4⋅4⋅4=45

=1024

So, probability that out of 5 dealt cards we get a straight with high card10:

Number of favorable outcomes in A

Number of outcomes in the sample space =1024/2,598,960 = 0.000394

We have the straights with high card,5 high card,6 high card,7 high card,8 high card,9 high card,10 high card,J high card,Q high card K and high card A.So, there are10 cards.

We have calculated the desired outcomes to be 1024

Hence, we get P({ straight })=10⋅1024/2,598,960=0.00394.

​The number of ways to select straight flush is 10.4=40 as there are 10 different straights and we have 4 different straight flush for each straight.

We know that total ways of selecting 5 cards out of 52 is 2,598,960.

Therefore, the probability that the card is a straight flush is P({ straight flush })=40/2,598,960=0.00001539​

The probability that out of 5 dealt cards, it is straight with high card 10 is 0.000394.

The probability that the selected c.

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability

Page 62 Problem 1 Answer

Given: A mutual fund company offers its customers a variety of funds: a money-market fund, three different bond funds, two stock funds and a balanced fund.

Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows:

Money-market 20%

High-risk stock 18%

Short bond 15%

Moderate-risk stock 25%

Intermediate bond 10%

Balanced 7%

Long bond 5%

A customer who owns shares in just one fund is randomly selected.

To find the probability that the selected individual owns shares in the balanced fund.

We will make use of the percentages of customers in different funds given to us.

Let A be the event:

A=” Selected individual owns shares in the balanced fund”

Hence we have:

P(A)= The sum of all given percentage is =20%+18%+15%+25%+10%+7%+5%=100%

The probability of balanced fund is already given that is =7%=0.07

The probability that the selected individual owns shares in the balanced fund is 0.07.

Page 62 Problem 2 Answer

Given: A mutual fund company offers its customers a variety of funds: a money-market fund, three different bond funds two stock funds and a balanced fund.

Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows:

Money-market  20%

High-risk stock 18%

Short bond 15%

Moderate-risk stock 25%

Intermediate bond 10%

Balanced 7%

Long bond 5%

A customer who owns shares in just one fund is randomly selected.

To find the probability that the individual owns shares in a bond fund We will make use of the percentages of customers in different funds given to us.

Let B be the event:

B= ”Individual owns shares in a bond fund” and let

B1=”Individual owns shares in short bond funds”

B2=”Individual owns shares in intermediate bond funds”

B3=” Individual owns shares in long-term bond funds”

Hence we get

Probability of bond fund = short bond + Intermediate bond + long bond

⇒P(B)=P(B1)+P(B2)+P(B3)

=15%+10%+5%

=30%

=0.30

​The probability that the individual owns shares in a bond fund is 0.30.

Page 62 Problem 3 Answer

Given: A mutual fund company offers its customers a variety of funds: a money-market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund.

Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows:

Money-market 20%

High-risk stock 18%

Short bond 15%

Moderate-risk stock 25%

Intermediate bond 10%

Balanced 7%

Long bond 5%

A customer who owns shares in just one fund is randomly selected.

To find the probability that the selected individual does not own shares in a stock fund.

We will make use of the Complement proposition P(A)+P(A′)=1.

Let C be the event that:

C=”Selected individual does not own shares in a stock fund”

Probability of stock fund  =high risk stock + moderate risk stock

Hence P(C)=18%+25%

=43%

=0.43

Now, using complement rule: ​P(A)+P(A′)=1

⇒P(A′)=1−P(A)

​P( not stock fund )=1−P( stock fund )

P(C′)=1−P(C)

=1−0.43

​P(C′)=0.57

​The probability that the selected individual does not own shares in a stock fund is 0.57.

Page 62 Problem 4 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given: Ai={ awarded project i}

for i=1,2,3and

P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01

​To express in words A1∪A2 and compute its probability.We will use P(A∪B)=P(A)+P(B)−P(A∩B) to compute the probability.

To describe A1∪A2 in words as:

A1∪A2= {awarded project one or project two or both the projects}.

We know that P(A∪B)=P(A)+P(B)−P(A∩B)

Hence the probability of A1∪A2 is:

P(A1∪A2)=P(A1)+P(A2)−P(A1∩A2)

=0.22+0.25−0.11

=0.36

​A1∪A2 in words is described as {awarded project one or project two or both the projects}.

Probability A1∪A2 is 0.36

Page 62 Problem 5 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given: Ai={ awarded project i} , for

i=1,2,3 and ​P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01

​To express A1′∩A2′ in words and compute its probability.

We will use ​P(A′∩B′)=P((A∪B)′)=1−P(A∪B) to compute the probability.

Describing A1′∩A2′ in words as:

A1′∩A2′={awarded project neither 1or 2}

By De Morgan’s law

(A1∪A2)′=A1′∩A2′ is given.

Now the probability of A1′∩A2′ is:

P(A1′∩A2′)=P((A1∪A2)′)=1−P(A1∪A2)

=1−0.36

=0.64​

A1′∩A2′  in words is described as {awarded project neither 1 or 2}

Probability of  A1′∩A2′ is 0.64

Page 62 Problem 6 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given: Ai={ awarded project i} for i=1,2,3 and suppose that

P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01

​To express A1∪A2∪A3 in words and compute its probability.

We will use P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C) to compute the probability.

Describing A1∪A2∪A3 in words as:

A1∪A2∪A3 {awarded project 1 or project 2 or project 3}.

Now the probability of A1∪A2∪A3 is:

P(A1∪A2∪A3)=P(A1)+P(A2)+P(A3)−P(A1∩A2)−P(A1∩A3)−P(A2∩A3)+P(A1∩A2∩A3)

=0.22+0.25+0.28+0.11−0.05−0.07+0.01

=0.53​

A1∪A2∪A3 in words is described as {awarded project 1 or project 2 or project3 } Probability of  A1∪A2∪A3 is 0.53.

Page 62 Problem 7 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given:Ai={ awarded project i} , for i=1,2,3 and suppose that

P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01​

To expressA1′∩A2′∩A3′3 in words and compute its probability.We will use​

P(A′∩B′∩C′)=P(A∪B∪C)′

=1−P(A∪B∪C) to compute the probability.

Describing A1′∩A2′∩A3′  in words as:

A1′∩A2′∩A3′={none of the three projects was awarded}.

We know that ​

P(A′∩B′∩C′)=P(A∪B∪C)′

=1−P(A∪B∪C)​

Hence probability of A1′∩A2′∩A3′ is:

P(A1′∩A2′∩A3′)=P(A1∪A2∪A3)′

=1−P(A1∪A2∪A3)

=1−0.53

=0.47​

Describing A1′∩A2′∩A3′  in words as:

A1′∩A2′∩A3′= {none of the three projects was awarded}.

We know that ​

P(A′∩B′∩C′)=P(A∪B∪C)′

=1−P(A∪B∪C)​

Hence probability of A1′∩A2′∩A3′ is:

P(A1′∩A2′∩A3′)=P(A1∪A2∪A3)′

=1−P(A1∪A2∪A3)

=1−0.53

=0.47

​Page 62 Problem 8 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given: Ai={ awarded project i}, for i=1,2,3 and suppose that

P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01

To expressA1′∩A2′∩A3 in words and compute its probability.We will use P(A′∩B′∩C)=P(C)−P(A∩C)−P(B∩C)+P(A∩B∩C) to compute the probability.

Describing A1′∩A2′∩A3 in words as A1′∩A2′∩A3= {awarded project 3 and neither awarded project 1 and project 2}.

We know that: P(A′∩B′∩C)=P(C)−P(A∩C)−P(B∩C)+P(A∩B∩C)

Hence probability of A1′∩A2′∩A3 is: P(A1′∩A2′∩A3)=P(A3)−P(A1∩A3)−P(A2∩A3)+P(A1∩A2∩A3)

=0.28−0.05−0.07+0.01

=0.17​

A1′∩A2′∩A3 in words described as {awarded project 3 and neither awarded project 1 and project 2 }.

Probability of  A1′∩A2′∩A3 is 0.17

Page 62 Problem 9 Answer

Given: A computer consulting firm presently has bids out on three projects.

Given: Let Ai={ awarded project i} , for i=1,2,3 and suppose that

P(A1)=.22

P(A2)=.25

P(A3)=.28

P(A1∩A2)=.11

P(A1∩A3)=.05

P(A2∩A3)=.07

P(A1∩A2∩A3)=.01

​To express (A1′∩A2′)∪A3 in words and compute its probability.We will use P((A′∩B′)∪C)=P(A′∩B′∩C′)+P(C) to compute the probability.

Describing (A1′∩A2′)∪A3 in words as:

(A1′∩A2′)∪A3={awarded neither project 1 and project 2 or awarded project 3 }

We know that P((A′∩B′)∪C)=P(A′∩B′∩C′)+P(C)

Hence probability of (A1′∩A2′)∪A3 is:

P((A1′∩A2′)∪A3)=P(A1′∩A2′∩A3′)+P(A3)

=0.47+0.28

=0.75

​(A1′∩A2′)∪A3  in words described as {awarded neither project 1 and project 2 or awarded project 3 }

Probability of (A1′∩A2′)∪A3  is 0.75 .

Page 62 Problem 10 Answer

Given: 55% of all adults regularly consume coffee, 45% regularly consume carbonated soda, and 70% regularly consume at least one of these two products.

To find the probability that a randomly selected adult regularly consumes both coffee and soda.We will make use of the property: P(A∪B)=P(A)+P(B)−P(A∩B)

Let A =Event that the elderly consumes coffee.

B=Event that adult regularly consume carbonated soda.

C=Event that adult consumes coffee, carbonated soda or both.

Hence:​

P(A)=0.55

P(B)=0.45

P(C)=0.7

​Now to find the probability that a randomly selected adult regularly consumes both coffee and soda means we will have to fins: P(A∩B)

We know that: P(A∪B)=P(A)+P(B)−P(A∩B)

⇒P(A∩B)=P(A)+P(B)−P(A∪B)

​We can see that the event C=A∪B

Hence P(A∩B)=P(A)+P(B)−P(C)

⇒P(A∩B)=0.55+0.45−0.7

⇒P(A∩B)=0.3

​The probability that a randomly selected adult regularly consumes both coffee and soda is 0.3.

Page 62 Problem 11 Answer

Given: Suppose that 55% of all adults regularly consume coffee,  regularly consume carbonated soda, and 70% regularly consume at least one of these two products.

To find the probability that a randomly selected adult doesn’t regularly consume at least one of these two products.

We will use the property P((A∪B)′)=1−P(A∪B)

Let, A= Event that adult regularly consumes coffee

B=Event that adult regularly consumes carbonated soda

C=Event that adult regularly consumes coffee, soda or both

Now, the probability of these events is given as:

P(A)=0.55

P(B)=0.45

P(C)=0.7​

The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is the probability that the selected adult consumes only coffee or consumes only carbonated soda or does not consume both products that is (A∪B)′

Now, the probability that a randomly selected adult doesn’t regularly consume at least one of these two products is: P((A∪B)′)=1−P(A∪B)

=1−0.7

=0.3​

The probability that a randomly selected adult doesn’t regularly consume at least one of these two products is 0.3

Page 63 Problem 12 Answer

Given: A denotes the event that the next request for assistance from a statistical software consultant relates to the SPSS package and B denotes the event that the next request for assistance from a statistical software consultant relates to the SAS package.

Given:P(A)=0.30

P(B)=0.50​

To check why it is not that case that P(A)+P(B)=1

There are three conditions for P(A)+P(B)=1even if one is not satisfied then the result cannot hold.

Now A and B should be complementary but It does not hold here, as these days some new programs like Python , Mat Lab are used and different people have different opinions hence SPSS and SAS are not the only events that can occur.

Therefore we can say that they are not not complementary events

Hence, it does not hold all the conditions, therefore P(A)+P(B)≠1.

We described why it is not a case that P(A)+P(B)=1.

Page 63 Problem 13 Answer

Given: A denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package and:B denote the event that the next request for assistance from a statistical software consultant relates to the SAS package.

Given P(A)=0.30

P(B)=0.50​

To find:P(A′).We will use: P(A)+P(A′)=1

Since for any event A we have P(A)+P(A′)=1

This implies,

P(A′)=1−P(A)

P(A′)=1−0.3

P(A′)=0.7

​We calculated that P(A′)=0.7

Page 63 Problem 14 Answer

Given: A denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package and B denote the event that the next request for assistance from a statistical software consultant relates to the SAS package.

Given: P(A)=0.3

P(B)=0.5​

To find: P(A∪B).

We will use : P(A∪B)=P(A)+P(B)−P(A∩B)

Since A and B cannot occur at the same time, they are mutually exclusive events .

This implies that, A∩B=∅

Also, P(A∪B)=P(A)+P(B)−P(A∩B)

Then, we have P(A∪B)=P(A)+P(B)−P(∅)

Since,

P(∅)=0

P(A)=0.3

P(B)=0.5

Thus,P(A∪B)=P(A)+P(B)

P(A∪B)=0.3+0.5

P(A∪B)=0.8

​We calculated that. P(A∪B)=0.8

Page 63 Problem 15 Answer

Given: A denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package and B denote the event that the next request for assistance from a statistical software consultant relates to the SAS package,Given: P(A)=0.3

P(B)=0.5​

From part c) we obtain: P(A∪B)=0.8

To find:P(A′∩B′).We will use: De Morgan’s law and the complement property.

Using De Morgan’s law,A′∩B′=(A∪B)′

we get,P(A′∩B′)=P((A∪B)′)

Also, since P(A)+P(A′)=1

P((A∪B)′)=1−P(A∪B)

Thus, we get, P(A′∩B′)=1−P(A∪B)

Since, P(A∪B)=0.8

Therefore,

P(A′∩B′)=1−0.8

P(A′∩B′)=0.2

​We calculated that P(A′∩B′)=0.2.

Page 63 Problem 16 Answer

A box contains six 40 W bulbs , five 60W and four 75 W bulbs . Bulbs are selected at random .

To find the probability that at least two bulbs must be selected to obtain one that is rated 75W.

We will use the definition of probability and complement rule.

Consider, an event E such that

E= at least two bulbs must be selected to obtain a first 75W bulb

Then, the complementary event of E will be E′ the first bill is 75W bulb

Also, since P(E)+P(E′)=1

This implies, P(E)=1−P(E′)

Thus, P(at least two bills must be selected to obtain a first 75W bulb)

=1−P ( the first bill is 75Wbulb ).

It is given that there are:

Using probability formula, P(E′)= number of 75 W bulbs  total number of bulbs

Total bills in a wallet are 5+4+6=15 bills.

Number of  75 W bulb =4

Thus, P(E′)=4/15

Therefore, P(E)=1−4/15

P(E)=11/15

P( at least two bills must be selected to obtain a first 75 W bulb)=11/15=0.7333

​The probability that at least two bulbs must be selected to obtain a first 75W bulb is0.7333.

Page 63 Problem 17 Answer

Given: Total number of joints in one batch =10,000

Number of defective joints judged by Inspector A=724

Number of defective joints judged by Inspector B=751

Number of defective joints judged by at least one of the inspectors =1159.

To find the probability that the selected joint was judged to be defective by neither of the two inspectors.

Consider X= Event that selected defective joints were judged by Inspector A

Y= Event that selected defective joints were judged by Inspector B

Hence we get:

X∪Y={ selected defective joints were judged by inspectorA or B or both }

(X∪Y)′={ selected defective joints were judged neither by inspector A or B} to find P((X∪Y)′).

Since, P(E)+P(E′)=1

Then, P((X∪Y)′)=1−P(X∪Y)

Using, probability formula, we have P(X∪Y)= number of defective joints judged by at least one of the inspectors  total number of joints

P(X∪Y)=1159/10000

Therefore,P((X∪Y)′)=1−1159/10000

P((X∪Y)′)=8841/10000

P((X∪Y)′)=0.8841​

The probability that the selected joint was judged to be defective by neither of the two inspectors is 0.8841

Page 63 Problem 18 Answer

Given: Total number of joints in one batch =10,000

Number of defective joints judged by Inspector A=724

Number of defective joints judged by Inspector B=751

Number of defective joints judged by at least one of the inspectors=1159

To find: the probability that the selected joint was judged to be defective by the inspector B but not by inspector A.

consider X=Event that selected defective joints were judged by InspectorA

Y=Event that selected defective joints were judged by Inspector B

Hence X∪Y={selected defective joints were judged by inspectorA or B or both }

X∩Y={selected defective joints were judged by inspector A and B}

X′={selected defective joints were not judged by Inspector A}

(X′∩Y)={ selected defective joints were not judged by Inspector A and were judged by Inspector B}.i

We need to find the probability of X′∩Y.

Venn diagram for(X′∩Y) i.e., the selected joint was judged to be defective by inspector B but not by inspector A is:

From the Venn Diagram above, we can observe that,

P(X′∩Y)=P(Y)−P(X∩Y)

Also, sinceP(X∪Y)=P(X)+P(Y)−P(X∩Y)

This implies, P(X∩Y)=P(X)+P(Y)−P(X∪Y)

Using probability formula, P(X)= number of defective joints judged by inspector A

Total number of joints

P(X)=724/10000

P(X)=0.0724

​P(Y)= number of defective joints judged by inspector B

Total number of joints

P(Y)=751/10000

P(Y)=0.0751

P(X∪Y)= number of defective joints judged by inspector A or B or both

Total number of joints

P(X∪Y)=1159/10000

P(X∪Y)=0.1159

P(X∩Y)=0.0724+0.0751−0.1159

P(X∩Y)=0.0316

​Thus, P(X′∩Y)=0.0751−0.0316

P(X′∩Y)=0.0435

​The probability that the selected joint was judged to be defective by inspector B but not by inspector A=0.0435

Page 63 Problem 19 Answer

Given: The route used by a certain motorist in commuting to work contains two intersections with traffic signals.

Given probability to stop at the first signal: 0.4

Given probability to stop at the second signal: 0.5

Given probability to stop at at least one of the two signals: 0.6

To find the probability that he must stop at both the signals.We will use: P(A∩B)=P(A)+P(B)−P(A∪B)

Let us define event A to be when the motorist stops at the first signal with probability P(A)=0.4.

Let us also define event B to be when the motorist stops at the second signal with probability P(B)=0.5.

The probability of the event that he must stop at at least one of the two signals isP(A∩B)=0.7

By using P(A∩B)=P(A)+P(B)−P(A∪B) we get:

P(A∩B)=P(A)+P(B)−P(A∪B)

⇒P(A∩B)=0.4+0.5−0.7

⇒P(A∩B)=0.2

​Hence the probability that he must stop at both of the signals is P(A∩B)=0.2.

Page 63 Problem 20 Answer

Given : The route used by a certain motorist in commuting to work contains two intersections with traffic signals.

Given probability to stop at the first signal(event A): 0.4

Given probability to stop at the second signal(event B):0.5

Given probability to stop at at least one of the two signals: 0.6

To find the probability that he must stop at the first signal but not at the second signal.We will use: P(A∩Bc)=P(A)−P(A∩B)

We are given,

P(A)=0.4

P(B)=0.5

P(A∪B)=0.7

We are asked to calculate P(A∩Bc).

Using the formula for the intersection of the event A and the complement of the event B:P(A∩Bc)=P(A)−P(A∩B)∣

Substituting the values from Step1, we get:

P(A∩Bc)=P(A)−P(A∩B)

⇒0.4−0.2

⇒0.2

​Therefore the probability that he stops at the first signal but not at the second signal is P(A∩Bc)=0.2 .

Hence the probability that he stops at the first signal but not at the second signal is P(A∩Bc)=0.2 .

Page 63 Problem 21 Answer

Given: The route used by a certain motorist in commuting to work contains two intersections with traffic signals.

Given probability to stop at the first signal (event A): 0.4

Given probability to stop at the second signal(event B): 0.5

Given probability to stop at at least one of the two signals:

To find the probability that he stops at exactly one signal.

We are given,

P(A)=0.4

P(B)=0.5

P(A∪B)=0.7​

We are asked to calculate

P({ at exactly one signal })=P(A∪B)−P(A∩B)

⇒P=[P(A)−P(A∩B)]+[P(B)−P(A∩B)]

Substituting values from Step1, we get:

P=[P(A)−P(A∩B)]+∣P(B)−P(A∩B)]

⇒[0.4−0.2]+[0.5−0.2]

⇒[0.2]+[0.3]

⇒0.2+0.3

⇒0.5

​Therefore the probability that he will stop at exactly one signal is P=0.5.

Hence the probability that he will stop at exactly one signal is P=0.5.

Page 64 Problem 22 Answer

Given: The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C).

Given: Request for A: 40%

Request for B:55%

Request for C:70%

Request for A or B:63%

Request for A or C:77%

Request for B or C:85%

To determine the probability that next purchaser will request at least one of the three options.

We will use P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

The probability that the next purchaser will request at least one of the three options can be written as:

P(A∪B∪C)

We can also write this as: P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

where P(A)=0.4

P(B)=0.55

P(C)=0.70

P(A∪B)=0.63

P(A∪C)=0.77

P(B∪C)=0.85

​In this step, we will calculate P(A∩B),P(A∩C),P(B∩C), and P(A∩B∩C)

P(A∩B)=P(A)+P(B)−P(A∪B)∣

⇒0.4+0.55−0.63∣

⇒0.32 I

P(A∩C)=P(A)+P(C)−P(A∪C)∣

⇒0.4+0.7−0.77∣

⇒0.33 I

P(B∩C)=P(B)+P(C)−P(B∪C)∣

⇒0.55+0.7−0.8∣

⇒0.45 I

P(A∩B∩C)=P(A∪B∪C)−P(A)−P(B)−P(C)+P(A∩B)+P(A∩C)+P(B∩C)∣

⇒0.85−0.4−0.55−0.7+0.32+0.33+0.45∣

⇒0.3​

The Venn Diagram representation is:

Using the formula for intersection and substituting values, we get:

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)∣

⇒0.4+0.55+0.7−0.32−0.33−0.45+0.33

⇒0.85

Therefore the probability that the next purchaser will request at least one of the three options is 0.85.

Hence the probability that the next purchaser will request at least one of the three options is 0.85

Page 64 Problem 23 Answer

Given: The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C). Given:

Request for A:40%

Request for B:55%

Request for C:70%

Request for A or B:63%

Request for A or C:77%

Request for B or C:85%

​To determine the probability that the next purchaser will select none of the three options.

We will use: P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

The probability that the next purchaser will request at least one of the three options can be written as:

P(A∪B∪C)

We can also write this as:

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

⇒0.4+0.55+0.7−0.32−0.33−0.45+0.33

⇒0.85​

The probability of the next purchaser selecting none of the three options can be written as:

P((A∪B∪C)c)

We can also write this as:

P((A∪B∪C)c)=1−P(A∪B∪C)

⇒1−0.85

⇒0.15​

The Venn Diagram for the following exercise prompt would look like this:

Hence the probability of the next purchaser selecting none of the three options is 0.15

Page 64 Problem 24 Answer

Given: The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C). Given:

Request for A:40%

Request for B:55%

Request for C:70%

Request for A or B:63%

Request for A or C:77%

Request for B or C:85%

​To find the probability that next purchaser will request only an automatic transmission and not either of the other two options.

We will draw the Venn diagram and then solve.

The probability that the next purchaser will request only an automatic transmission option and not either of the other two can be written as P(A′∩B′∩C)

The probability that the next purchaser will request only an automatic transmission option and not either of the other two can be written as P(A′∩B′∩C)

We can substitute the respective values:

P(A′∩B′∩C)=P(C)−P(A∩C)−P(B∩C)+P(A∩B∩C)

⇒0.7−0.33−0.45+0.3

⇒0.22

​Hence the probability that the next purchaser will request only an automatic transmission option and not either of the other two options is 0.22

Page 64 Problem 25 Answer

Given: The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C).

Given: Request for A:40%

Request for B:55%

Request for C:70%

Request for A or B:63%

Request for A or C:77%

Request for B or C:85%

​To find the probability that the next purchaser will choose exactly of the three options.

We will draw a Venn diagram and then solve.

The complete Venn Diagram for the exercise prompt is:

Since we want to find the probability that the next purchaser will select exactly one of these three options, we will shade our Venn Diagram accordingly.

After shading, the resulting Venn Diagram is:

Taking the summation of the shaded area we get:

P(exactly one of the three )

⇒P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩B)+3P(A∩B∩C)∣

⇒0.05+0.08+0.22∣

⇒0.35​

Therefore the probability that the next purchaser will select exactly one of these three options is 0.35

Page 64 Problem 26 Answer

Given: An academic department with five faculty members— Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve on a personnel review committee.

Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper and then randomly selecting two.

To find the probability that both Anderson and Box will be selected.

We will find the total number of events that is possible in this case.

After which we will take the total number of events of “both Anderson and Box being selected”.

Let this be denoted by P(A).We will then find P(A)

In this step, we will find the total number of sample spaces possible. The sample space in this case is:

{A,B},{A,Cox},{A,Cr},{A,F},{B,Cox}

{B,Cr},{B,F},{Cox,Cr},{Cox,F},{Cr,F}​

Therefore the sample space has 10 events.

Let A= Event that both Anderson and Box are selected.

Then the events can be:{A,B}

Therefore, we can write P(A) to be: P({A,B})=1/10

Hence the probability that both Anderson and Box will be selected is 1/10.

Page 64 Problem 27 Answer

Given: An academic department with five faculty members— Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve on a personnel review committee.

Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper and then randomly selecting two.

To find the probability that at least one of the two members whose name begins with C is selectedWe will find the total number of events that is possible in this case.

After which we will take the total number of events of ” at least one of the two members whose name begins with C being selected”.

Let this be denoted by P(C).We will the find P(C)

In this step, we will find the total number of sample spaces possible. The sample space in this case is:

{A,B},{A,Cox},{A,Cr},{A,F},{B,Cox},

{B,Cr},{B,F},{Cox,Cr},{Cox,F},{Cr,F}

​Therefore the sample space has 10 events.

Let P(C)= Event that at least one of the two members whose name begins with C is selected.

We are told that at least one of the two members whose name begins with C is selected. Then the disjoint events can be: {A,Cox},{A,Cr},{B,Cox},{B,Cr},{Cox,Cr},{Cox,F},{Cr,F}

Therefore, we can write P(C) to be:

P(C)=P({A,Cox}+{A,Cr}+{B,Cox}+{B,Cr}+{Cox,Cr}+{Cox,F}+{Cr,F})

⇒7/10,​

Hence the probability that at least one of the two members whose name begins with C is selected is 7/10 .

Page 64 Problem 28 Answer

Given: An academic department with five faculty members— Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve on a personnel review committee.

Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper and then randomly selecting two.

Given: Five faculties have taught for 3,6,7,10, and 14 years, respectively.

To find the probability that the two chosen representatives have a total of at least15 years’ teaching experience there.

We will find the total number of events that is possible in this case.

After which we will take the total number of events of “at least 15 years’ teaching experience”.

Let this be denoted by P(B). We will then find P(B)

In this step, we will find the total number of sample spaces possible. The sample space in this case is:

{3,6},{3,7},{3,10},{3,14},{6,7},{6,10},{6,14},{7,10},{7,14},{10,14}∣

Therefore the sample space has 10 events.

We are told that at least 15  years of teaching experience is required. Then the disjoint events can be:

{3,14},{6,10},{6,14},{7,10},{7,14},{10,14}

Let B= Event that representative have at least 15 years of teaching experience

Therefore, we can write P(B) to be:

P(B)=P({3,14}+{6,10}+{6,14}+{7,10}+{7,14}+{10,14})

⇒6/10

Hence the probability  that the two chosen representatives have a total of at least 15 years’ teaching experience is 6/10.

Page 64 Problem 29 Answer

Given : A family consisting of three members, A, B , and C goes to a medical clinic that always has a doctor at each stations 1,2, and 3.

During a certain week each member of the family visits a clinic and is assigned a random station.

Given: Any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other individuals have been assigned.

To find the probability that all three family members are assigned to the same station.We will find the total number of sample space that is possible in this case.

After which we will take the total number of events of “all the three family members are assigned to the same station”.

Let this be denoted by P(A). We will then find P(A)

In this step, we will find the total number of sample spaces possible.

Since there are 3 stations and 3 family members. Then the sample space will consist of:

n=33

n=27

Therefore the sample space is 27 events.

Let A= Event that all three family members go to the same station.

Then the events can be: (1,1,1),(2,2,2),(3,3,3)

We can write P(A) to be:

P(A)=P((1,1,1)+(2,2,2)+(3,3,3))

=3/27​

⇒P(A)=1/9

Hence the probability that all three family members are assigned to the same station is 1/9.

Page 64 Problem 30 Answer

Given: A family consisting of three members, A,B , and C goes to a medical clinic that always has a doctor at each stations 1,2 and 3.

During a certain week each member of the family visits a clinic and is assigned a random station.

Given: Any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other individuals have been assigned.

To find the probability that at most two family members are assigned to the same station.

We will find the total number of events that is possible in this case.

After which we will take the total number of events of “at most two family members are assigned to the same station”.

Let this be denoted by P(B). We will find the total number of events that is possible in this case.

After which we will take the total number of events of “at most two family members are assigned to the same station”.

Let this be denoted by P(B). We will then find p (B)

In this step, we will find the total number of sample spaces possible.

Since there are 3 stations and 3 family members. Then the sample space will consist of:

n=33

n=27

Therefore the sample space is 27 events.

Let B be the event that at most two family members go to the same station.

Then the events which do not follow can be:

B{c}=(1,1,1),(2,2,2),(3,3,3)

We can write P(B) as:

P(B)=1−P(Bc)

=1−1/9

=8/9

Hence the probability that at most two family members are assigned to the same station is 8/9⋅

Page 64 Problem 31 Answer

Given: A family consisting of three members, A,B , and C goes to a medical clinic that always has a doctor at each stations 1,2 and 3.

During a certain week each member of the family visits a clinic and is assigned a random station.

Given: Any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other individuals have been assigned.

To find the probability that every family member is assigned to a different station.

We will find the total number of events that is possible in this case.

After which we will take the total number of events of “every family member is assigned to a different station”.

Let this be denoted by P(C). We will then find P(C)

In this step, we will find the total number of sample spaces possible.

Since there are 3 stations and 3 family members. Then the sample space will consist of:

n=33

n=27

Therefore the sample space is 27 events.

Let C be the event that every family member is assigned a different station.

Then the events can be:

(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)∣

We can write P© to be P/(C)

=P({(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)}⟩

=6/27​

⇒P(C)=2/9

Hence the probability that every family member is assigned to a different station is 2/9.

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability

Page 80 Problem 1 Answer

Given: The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups.

Given  probability table is:

	Blood Group	O	A	B	AB
Ethnic Group	1	0.082	0.106	0.008	0.004
	2	0.135	0.141	0.018	0.006
	3	0.215	0.2	0.065	0.02

Given events:A={ type A selected },B={ type B selected }, and C={ ethnic group 3 selected }

​To calculate P(A),P(C), and P(A∩C)

We will make use of the given table in order to find the probabilities.

The probability of selecting type A is the sum of column A of joint probability table.

P(A)=0.106+0.141+0.2

=0.447

The probability of selecting group 3 is the sum of row 3 of the given joint probability table.

The probability of A∩C is row 3 and column A

P(A∩C)=0.2

We obtain:

P(A)=0.447

P(C)=0.5

P(A∩C)=0.2

​Page 80 Problem 2 Answer

Given: The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups.

Given: Joint probability table:

	Blood Group	O	A	B	AB
Ethinc Group	1	0.082	0.106	0.008	0.004
	2	0.135	0.141	0.018	0.006
	3	0.215	0.2	0.065	0.02

Given events:

A={ type A selected }

B ={ type B selected }

and C={ ethnic group 3 selected }

​To calculate: P(A∣C) and P(C∣A)

To explain in context what each of these probabilities represents.

We will calculate the required conditional probabilities by using the formula: P(A∣B)=P(A∩B)

P(B) where P(B)>0.

From previous part P(A∩C)=0.2

By using the  definition of conditional probability

P(A∣C)=P(A∩C)/P(C)

=0.2/0.5

=0.4

This represents that if we know that an individual comes from ethnic group 3, the probability that the individual has blood type A is 0.4.

and P(C∣A)= P(C∩A)/P(A)

0.2/0.447

= 0.447​

This represents that if we know that an individual has blood type A the probability that the individual comes from ethnic group 3 is 0.447.

We obtain:

P(A∣C)=0.4

P(C∣A)=0.447​

P(A∣C) represents that if we know that an individual comes from ethnic group 3 , the probability that the individual has blood type A is 0.4.

P(C∣A) represents that if we know that an individual has blood type A the probability that the individual comes from ethnic group3 is 0.447.

Page 80 Problem 3 Answer

Given: The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups.

We are given the table:

	Blood Group	O	A	B	AB
Ethnic Group	1	82	0.106	0.008	0.004
	2	0.135	0.141	0.018	0.006
	3	0.215	0.2	0.065	0.02

Given events: ​

A={ type A selected }

B={ type B selected }  and C={ ethnic group 3 selected }

​To find the probability a given individual is from ethnic group 1 if he or she does not have type B blood.

We will first find the probability of not have a B blood then use conditional probability.

We need to find P(D∣B′).

P(D∩B′)=0.082+0.106+0.004

=0.192

We know P(B′)=1−P(B)

=1−(0.008+0.018+0.065)

=0.909

By using the definition of conditional probability, we get

P(D∣B′)=P(D∩B′)

P(B′)= 0.192/0.909

=0.211

​The probability that the given individual is from ethnic group 1 given that he or she does not have type B blood is 0.211.

Page 80 Problem 4 Answer

Given:  An individual is randomly selected from the population of all adult males living in the United States Given the following events:

A={ Individual person is 6 feet tall }

B={ Individual person is a professional basketball player}

​To check among P(A∣B) and P(B∣A), which is larger.We will check using the conditional probability formula: P(A∣B)=P(A∩B)

P(B),P(B)>0

We are not given the exact data of how many people are taller than 6 feet.

By logic, we can say that number of people who are taller than 6 feet are more than the number of professional basketball players.

So, N(A)>N(B)

N(A)>N(B)

We know P(A)= Number of favourable outcomes of A/ Number of outcomes in sample space

=N(A)/N>N(B)/N

=P(B)

​Therefore,⇒P(A)>P(B)

⇒1

P(A)<1/P(B)

Obviously, P(A∩B)>0

So,  P(A∩B)

P(A)<P(A∩B)

P(B) which is equivalent to   P(B∣A)<P(A∣B)

For events

A={ Individual person is 6 feet tall }

B={ Individual person is a professional basketball player}

​we obtain P(B∣A)<P(A∣B) because N(A)>N(B)

Page 80 Problem 5 Answer

Given: A certain system can experience three different types of defects. Given: that Ai={ event of the type i ;=1,2,3}

Given probabilities:

P(A1)=0.12

P(A2)=0.07

P(A3)=0.05

P(A1∪A2)=0.13

P(A1∪A3)=0.14

P(A2∪A3)=0.1

P(A1∩A2∩A3)=0.01

​To find the probability that the system has type 2 defect, given that it has type 1 defect.

We will use the Conditional probability formula P(A∣B)=P(A∩B)

P(B),P(B)>0

We need to find P(A2∣A1).

By using conditional probability and data given in the question, we get

P(A2∩A1) =0.06

=P(A1)+P(A2)−P(A1∪A2)

= 0.12+0.07−0.13

​= P(A2∣A1)=P(A​2∩A​1)/P(A​1)

=0.06/0.12

=0.5​

The probability  that the system have a defect of type 2 given that the system has type 1 defect is 0.5

Page 80 Problem 6 Answer

Given: A certain system can experience three different types of defects.

Given: that Ai ={ event of the type i ;=1,2,3}

Given probabilities:

P(A1)=0.12

P(A2)=0.07

P(A3)=0.05

P(A1∪A2)=0.13

P(A1∪A3)=0.14

P(A2∪A3)=0.1

P(A1∩A2∩A3)=0.01​

To find the probability the system has all three types of defects given that it has type 1

defect. We will use the conditional prob

We need to find P(A1∩A2∩A3∣A3).

Using the definition of conditional probability and the data given in the question, we get

P(A1∩A2∩A3∣A1)=P[(A​1∩A​2∩A​3)∩A​3]/P(A​1)

=P(A​1∩A​2∩A​3)/P(A​1)

=0.01/0.12

=0.0833

​The probability that system has all three types of defects given that it has the type 1 defect is 0.0833.

Page 80 Problem 7 Answer

Given:  A certain system can experience three different types of defects.

Given: that Ai ={ event of the type i ;=1,2,3}

Given probabilities:

P(A1)=0.12

P(A2)=0.07

P(A3)=0.05

P(A1∪A2)=0.13

P(A1∪A3)=0.14

P(A2∪A3)=0.1

P(A1∩A2∩A3)=0.01

​To find the probability that the system has exactly one type of defect given that it has at least one type of defect.

We will use the Venn diagrams to obtain the probabilities of the individual events and then use the conditional probability.

We will obtain the following probabilities in order to solve further.

P(A1∩A2)=P(A1)+P(A2)−P(A1∪A2)

=0.12+0.07−0.13

=0.06

P(A1∩A3)=P(A1)+P(A3)−P(A1∪A3)

=0.12+0.05−0.14

=0

P(A2∩A3)=P(A2)+P(A3)−P(A2∪A3)

=0.07+0.05−0.1

=0.02​

We observe that A∩B=B as B⊂A.

We also see that B is the union of A1∩A2′∩A3′,A1′∩A2∩A3′,A1′∩A2′∩A3.

We will now find the value of probabilities of these individual events and unite then in order to get.P(B).

For,A1∩A2′∩A3′

we will first draw the Venn diagram as follows.

Using the data given in the question, we get

P(A1∩A2′∩A3′)=

=0.04

P(A1)−P(A1∩A2)−P(A1∩A3)+P(A1∩A2∩A3)

= 0.12−0.06−0.03+0.01

​For, we will first draw the Venn diagram as follows.

Using the data given in the question, we get

P(A1′∩A2∩A3′)

=0.07−0.06−0.02+0.01

=0

=P(A2)−P(A1∩A2)−P(A2∩A3)+P(A1∩A2∩A3)

​For, we will first draw the Venn diagram as follows.

Using the data given in the question, we get

P(A1′∩A2′∩A3)=P(A3)−P(A1∩A3)−P(A2∩A3)+P(A1∩A2∩A3)

=0.05−0.03−0.02+0.01

=0.01

​We will combine all these to get P(B)

P(B)=0.04+0+0.01

=0.05

​Using the definition of conditional probability, we get

P(B∣A)=P(B∩A)/P(A)

=P(B)/P(A)

=0.05/0.14

=0.3571

​Given that the system has at least one type of defect, the probability that it has exactly one type of defect is 0.3571.

Page 80 Problem 8 Answer

Given:  A certain system can experience three different types of defects. Given: that Ai={ event of the type i ;=1,2,3}

Given probabilities:

P(A1)=0.12P(A2)=0.07P(A3)=0.05

P(A1∪A2)=0.13

P(A1∪A3)=0.14

P(A2∪A3)=0.1

P(A1∩A2∩A3)=0.01

​To find the probability that the system does not have third type of defect given that it has both of the first two ty

We need to find P(A3′∣A1∩A2). we will first construct its Venn diagram to get the desired formula.

By using the definition of conditional probability, we get

P(A3′∣A1∩A2)=P(A1∩A2∩A3′)

P(A1∩A2)=P(A1∩A2)−P(A1∩A2∩A3)

P(A1∩A2)=0.06−0.01/0.06

=0.833​

The probability that it does not have the third type of defect given that it has both of the first two types of defects is 0.833.

Page 81 Problem 9 Answer

Given: Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE).

Based on a recent study, suppose that 16% of all ticks in a certain location carry Lyme disease, 10% carry HGE, and 10% of the ticks that carry at least one of these diseases in fact carry both of them.

Given: A randomly selected tick is found to have carried HGE.

To find the probability that the selected tick is also a carrier of Lyme disease.

We will find this using the conditional probability formula: P(A∣B)=P(A∩B)/P(B)

We define,

A={ Carrier of Lyme disease }

B={ Carrier of HGE disease }

​So, we have

P(A)=0.16

P(B)=0.1

P(A∩B∣A∪B)=0.1

​We will calculate the probability of A given that B has already occurred.

We  have​P(A∩B∣A∪B)

=P[(A∩B)∩(A∪B)]/P(A∪B)

=P(A∩B)

P(A∪B) ( as intersection is the subset of union ).

Given, P(A∩B∣A∪B)=0.1.

Putting this in above equation, we get

P(A∩B)/P(A∪B)

=0.1

Applying the formula of union of two sets, we get

P(A∩B)=0.1⋅P(A∪B)

P(A∩B)=0.1⋅[P(A)+P(B)−P(A∩B)]

=0.1⋅[0.16+0.1−P(A∩B)]

=0.016+0.01−0.1⋅P(A∩B)

=0.026−0.1⋅P(A∩B)

​Hence, 1.1⋅P(A∩B)=0.026

P(A∩B)=0.026/1.1

=0.02364

Putting this value in the formula for conditional property,

P(A∣B)

=0.02364/0.1

=P(A∩B)/P(B)

=0.2364

​The probability that is also a carrier of Lymph disease given that it is a carrier of HGE is 0.2364.

Page 81 Problem 10 Answer

Given any events A and B.

To show that: P(A∣B)+P(A′∣B)=1

We will use the conditional probability formula: P(A∣B)=P(A∩B)/P(B),P(B)>0

We have P(B∣A)=P(A∩B)/P(A).

Also, we know that A′represents the complement of A.

Hence, we obtain

P(A∣B)=P(A∩B)

P(B)⋯(1)

P(B∣A)=P(A​′∩B)/P(B)​

Also,A′∩B is that part of B where no element of A is present.

So, P(A′∩B)=P(B)−P(A∩B)

On combining the results,

We get P(A∣B)+P(A′∣B)=P(A∩B)

P(B)+P(A​′∩B)

P(B)=P(A∩B)

P(B)+P(B)−P(A∩B)

P(B)=P(A∩B)+P(B)−P(A∩B)/P(B)

=P(B)/P(B)

=1

Hence, P(A∣B)+P

(A′∣B)=P(B)/P(B)=1

​We proved that P(A∣B)+P(A′∣B)=1

Page 81 Problem 11 Answer

Given: Any three events A,B,C with P(C)>0.

To show that: P(A∪B∣C)=P(A∣C)+P(B∣C)−P(A∩B∣C).

We will use the conditional probability and P(A∪B)=P(A)+P(B)−P(A∩B).

We will find the value of P(A∪B∣C).

P(A∪B∣C)=P[(A∪B)∩C]/P(C)

[Now using the property (A∪B)∩C=(A∩C)∪(B∩C)

P(A∪B∣C)=P[(A∪B)∩C]/P(C)

=P[(A∩C)∪(B∩C)]/P(C)

=P(A∣C)+P(B∣C)+P(A∩B∣C)

=P(A∩C)+P(B∩C)−P(A∩B∩C)/P(C)

= P(A∩C)/P(C)+P(B∩C)

P(C)−P(A∩B)∩C/P(C)

Hence,P(A∪B∣C)=P(A∣C)+P(B∣C)+P(A∩B∣C)

We have shown that P(A∪B∣C)=P(A∣C)+P(B∣C)+P(A∩B∣C)

Page 81 Problem 12 Answer

Given: Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered.

Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have such a locator.

Given : a light aircraft has disappeared.

To find that the aircraft will not be discovered if it has an emergency locator.We will use conditional probability P(A∣B)=P(B∣A)P(A)

P(B),P(B)>0

The given data is, D=0.7,L∣D=0.6 and NL∣ND=0.9

Finding the probability of the airplane being not discovered, and has an emergency locator,

P(ND∣L)=P(L∣ND)P(ND)/P(L)

P(ND)=1−P(D)

P(ND)=1−0.7

P(ND)=0.3

​The probability assembly rule can be used,

P(L)=P(L∣D)P(D)+P(L∣ND)P(ND)

P(L∣ND)=1−P(NL∣ND)

P(L∣ND)=1−0.90

P(L∣ND)=0.10

​Substitute the values in the equation,

P(L)=(0.6)(0.7)+(0.1)(0.3)

P(L)=0.45​

Find the value of P(ND) by subtracting the value of P(D) from 1,

P(ND)=1−P(D)

P(ND)=1−0.45

P(ND)=0.55​

Substitute the values in the conditional probability formula,

P(ND∣L)=(0.1)(0.3)/(0.45)

P(ND∣L)=0.067

​The probability that the aircraft will not be discovered, but has an emergency locator is 0.067.

Page 81 Problem 13 Answer

Given: Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered.

Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have such a locator.

Given : a light aircraft has disappeared.To find the probability that the plane will be discovered given that it does not have an emergency locator.

We will use the conditional probability

P(A|B)=P(B|A)P(A)/P(B),P(B)>0

The formula can be rewritten too suit the situation,

P(D∣NL)=P(NL∣D)P(D)/P(NL)

Find the rest of the unknown values,

P(NL∣D)=1−P(L∣D)

P(NL∣D)=1−0.6

P(NL∣D)=0.4​

Now substitute all the calculated values in the equation,

P(D∣NL)=P(NL∣D)P(D)/P(NL)

P(D∣NL)=(0.4)⋅(0.7)/0.55

P(D∣NL)=0.509

​The probability that it will be discovered and did not have an emergency locator is 0.509

Page 82 Problem 14 Answer

Given:  A company that manufactures video cameras produces a basic model and a deluxe model.

Over the past year, 40% of the cameras sold have been of the basic model.

Of those buying the basic model, 30% purchase an extended warranty, whereas 50% of all deluxe purchasers do so.

A randomly selected purchaser has an extended warranty.

To find the probability that he or she has a basic model.

We will use law of total probability.

Let B be the event that the basic model has been produced.

Let D be the event that the deluxe model has be produces.

Let EW be the evet that the extended warranty is chosen by the customer.

Then we can see:

P(B)/E(D)=0.4,

P(EW∣B)=0.6,

P(EW∣D)=0.3,=0.5.​

The required probability is of the event B∣EW

We get: P(B∣EW)=P(B∩EW)/P(Ew)

Using the property P(A∩B)=P(A∣B)⋅P(B)

we get: P(B∣EW)=P(B∣EW)

Now using law of total probability we get:

P(B∣EW)=P(EW∣B)P(B)

P(EW∣B)P(B)+P(EW∣D)P(D)

=0.3⋅0.4/0.3⋅0.4+0.5⋅0.6

=0.29.

​The probability that the purchaser has a basic model given that he or she has an extended warranty is 0.29.

Page 82 Problem 15 Answer

Given: For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S.,

B be the event that the refrigerator had an ice maker, and  C be the event that the customer purchased an extended warranty.

Given probabilities are :

P(A)=.75

P(B∣A)=.9

P(B∣A′)=.8

P(C∣A∩B)=.8

P(C∣A∩B′)=.6

P(C∣A′∩B)=.7

P(C∣A′∩B′)=.3

To construct a tree diagram consisting of first-, second-, and third-generation branches, and place an event label and appropriate probability next to each branch.

We will first find the required probabilities to draw the diagram and then draw it.

In tree diagram,

In the initial branch there will be first layer of events , and in the diagram the adequate probabilities are given.

In the second branch there will be second layer of events with the given adequate conditional probabilities.

And in the third branch there will be third layer of events with the given adequate conditional probabilities.

The first layer probabilities :

P(A)=0.75

P(A′)=1−P(A)

P(A′)=0.25​

The second layer probabilities :

P(B∣A)=0.9

P(B′∣A)=1−P(B∣A)

P(B′∣A)=0.1

P(B∣A′)=0.8

P(B′∣A′)=1−P(B∣A′)

P(B′∣A′)=0.2

The third layer probabilities : All the events in this will be the probability of intersection of adequate three three events that leads to that  point.

P(C∣A∩B)=0.8

P(C′∣A∩B)=0.2

P(C∣A∩B′)=0.6

P(C′∣A∩B′)=0.4

P(C∣A′∩B)=0.7

P(C′∣A′∩B)=0.3

P(C∣A′∩B′)=0.3

P(C′∣A′∩B′)=0.7

​The tree diagram is :

The tree diagram is :

Page 82 Problem 16 Answer

Given: For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S.,B  be the event that the refrigerator had an ice maker, and C be the event that the customer purchased an extended warranty.

Given probabilities are :

P(A)=.75

P(B∣A)=.9

P(B∣A′)=.8

P(C∣A∩B)=.8

P(C∣A∩B′)=.6

P(C∣A′∩B)=.7

P(C∣A′∩B′)=.3

To compute : P(A∩B∩C)

We will use the multiplication rule P(A∩B)=P(A∣B)⋅P(B)

If there are three events then we will use multiplication rule twice.

Now, using multiplication rule :

P(A∩B∩C)=P(C∣A∩B)⋅P(A∩B)

=P(C∣A∩B)⋅P(B∣A)⋅P(A)∣

=0.75⋅0.9⋅0.8

=0.54

Hence, P(A∩B∩C)=0.54

​We obtain:  P(A∩B∩C)=0.54

Page 82 Problem 17 Answer

Given: For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S.,B  be the event that the refrigerator had an ice maker, and C be the event that the customer purchased an extended warranty.

Given probabilities are :

P(A)=.75

P(B∣A)=.9

P(B∣A′)=.8

P(C∣A∩B)=.8

P(C∣A∩B′)=.6

P(C∣A′∩B)=.7

P(C∣A′∩B′)=.3

​To compute: P(B∩C)

We will use the multiplication rule and the property P(B∩C)=P(B∩C∩A)+P(B∩C∩A′)

for three events.

Using property :

P(B∩C)=P(B∩C∩A)+P(B∩C∩A′)

Now, we will use the multiplication rule twice we get :

=0.54+0.25⋅0.8⋅0.7

=0.68

Hence, P(B∩C)=0.68

​We obtain: P(B∩C)=0.68

Page 82 Problem 18 Answer

Given: For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an ice maker, and C be the event that the customer purchased an extended warranty.

Given probabilities are :

P(A)=.75

P(B∣A)=.9

P(B∣A′)=.8

P(C∣A∩B)=.8

P(C∣A∩B′)=.6

P(C∣A′∩B)=.7

P(C∣A′∩B′)=.3

​To compute P(C).

We will use the property P(C)=P(A∩B∩C)+P(A′∩B∩C)+P(A∩B′∩C)+P(A′∩B′∩C)

Using property ,

P(C)=P(A∩B∩C)+P(A′

∩B∩C)+P(A∩B′∩C)+P(A′∩B′∩C)

=0.75⋅0.9⋅0.8+0.75⋅0.1⋅0.6+0.25⋅0.8⋅0.7+0.25⋅0.2⋅0.3

=0.54+0.045+0.14+0.015

=0.74

Hence, P(C)=0.74

​We obtain:  P(C)=0.74

Page 82 Problem 19 Answer

Given: For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an ice maker, and C  be the event that the customer purchased an extended warranty.

Given probabilities are :

P(A)=.75

P(B∣A)=.9

P(B∣A′)=.8

P(C∣A∩B)=.8

P(C∣A∩B′)=.6

P(C∣A′∩B)=.7

P(C∣A′∩B′)=.3

To compute P(A∣B∩C), the probability of U.S purchase given that an icemaker and an extended warranty are also purchased.

We will use the conditional property given by: P(A∣B)=P(A∩B)

P(B) where P(B)>0.

Using conditional property,

P(A∣B∩C)=P(A∩B∩C)/P(B∩C)

=0.54/0.68

=0.7941

Hence, P(A∣B∩C)=0.7941.

​The probability of a U.S purchase given that an icemaker and an extended warranty are also purchased is: P(A∣B∩C)=0.7941.

Page 82 Problem 20 Answer

Given : The Reviews editor for a certain scientific journal decides whether the review for any particular book should be short (1−2pages), medium (3−4 pages), or long (5−6pages).

Data on recent reviews indicates that 60% of them are short, 30% are medium, and the other 10% are long.

Reviews are submitted in either Word or LaTeX.

For short reviews, 80% are in Word, whereas 50%of medium reviews are in Word and 30% of long reviews are in Word.

Given: a recent review is randomly selected.To find the probability that the selected review was submitted in Word format.

We will draw a tree diagram and then with the help of that use that law total of total probability.

In tree diagram,

In the initial branch there will be first layer of events , and in the diagram the adequate probabilities are given.

In the second branch there will be second layer of events with the given adequate conditional probabilities.

Then, using the multiplication rule to the right of the second generation branches then we will display the product of the particular probabilities that leads to the point which is the probability of intersections.

The first layer of events are given below :

A1={ short book }

A2={ medium book }

A3={ long book }​

Their adequate properties are :

P(A1)=0.6

P(A2)=0.3

P(A3)=0.1​

The second layer of events are given below :

W={ submited in word }

L={ submited in LaTeX }​

Now, the adequate conditional probabilities are :

P(W∣A1)=0.8

P(L∣A1)=1−P(W∣A1)∣

P(L∣A1)=0.2

P(W∣A2)=0.5

P(L∣A2)=1−P(W∣A2)

P(L∣A2)=0.5

P(W∣A3)=0.3

P(L∣A3)=1−P(W∣A3)

P(L∣A3)=0.7

​Now, we will calculate the every of the four intersections using the multiplication rule :

P(A1∩W)=P(A1)⋅P(W∣A1)

=0.6⋅0.8

=0.48

P(A1∩L)=P(A1)⋅P(L∣A1)

=0.6⋅0.2

=0.12​

Similarly, P(A2∩W)=0.15P(A2∩L)=0.15

P(A3∩W)=0.03

P(A3∩L)=0.07​

Now, the tree diagram is :

Now, using the law of total probability :

P(W)=P(W∣A1)P(A1)+P(W∣A2)P(A2)+P(W∣A3)P(A3)

=0.48+0.15+0.03

=0.66

Hence, P(W)=0.66

The probability that the selected review was submitted in Word format is

P(W)=0.66

Page 82 Problem 21 Answer

Given : The Reviews editor for a certain scientific journal decides whether the review for any particular book should be short ( 1−2pages), medium (3−4pages), or long ( 5−6pages).

Data on recent reviews indicates that60% of them are short,  30%are medium, and the other 10% are long.

Reviews are submitted in either Word or LaTeX.

For short reviews, 80% are in Word, whereas 50%

of medium reviews are in Word and 30% of long reviews are in Word.

Given: a recent review is randomly selected.To find the probabilities for the review to be short , medium and long given that it was submitted in Word format.

We will draw the tree diagram and then use Bayes’ theorem in order to get the answer.

In tree diagram,

In the initial branch there will be first layer of events , and in the diagram the adequate probabilities are given.

In the second branch there will be second layer of events with the given adequate conditional probabilities.

Then, using the multiplication rule to the right of the second generation branches then we will display the product of the particular probabilities that leads to the point which is the probability of intersections.

The first layer of events are given below :

A1={ short book }

A2={ medium book

A3={ long book }

​Their adequate properties are :

P(A1)=0.6

P(A2)=0.3

P(A3)=0.1

​The second layer of events are given below :

W={ submited in word }

L={ submited in LaTeX }

​Now, the adequate conditional probabilities are :

P(W∣A1)=0.8

P(L∣A1)=1−P(W∣A1)

P(L∣A1)=0.2

P(W∣A2)=0.5

P(L∣A2)=1−P(W∣A2)

P(L∣A2)=0.5

P(W∣A3)=0.3

P(L∣A3)=1−P(W∣A3)

P(L∣A3)=0.7

​Now, we will calculate the every of the four intersections using the multiplication rule :

P(A1∩W)=P(A1)⋅P(W∣A1)

=0.6⋅0.8

=0.48

P(A1∩L)=P(A1)⋅P(L∣A1)

=0.6⋅0.2

=0.12

​Similarly, P(A2∩W)=0.15

P(A2∩L)=0.15

P(A3∩W)=0.03

P(A3∩L)=0.07

​Now, the tree diagram is :

Now using Bayes’ theorem,

The posterior probabilities are :

P(A1∣W)=P(A​1∩W)/P(W)

=0.48/0.66

=0.727

P(A2∣W)=P(A​2∩W)/P(W)

=0.15/0.66

=0.227

P(A3∣W)=P(A​3∩W)/P(W)

=0.03/0.66

=0.046

​The posterior probability for the review to be short given that it is submitted in Word is:0.727

The posterior probability for the review to be medium given that it is submitted in Word is:0.227

The posterior probability for the review to be long given that it is submitted in Word is:0.046