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