a.
Compute the
a.
Answer to Problem 42E
The probability distribution of d1 is:
d1 | 0 | 650 | 2,000 |
p(d1) | 13 | 13 | 13 |
Explanation of Solution
It is given that there are three cities, namely City B, City C, and City D. The meeting is going to be held in one of the three cities. The shortest highway distance from City B to City C passes through City D. The highway distance from City B to City D is 1,350 miles, and the distance from City D to City C is 650 miles.
The random variable d1 represents the driving distance from City C to the place of meeting.
Therefore, the random variable d1 can take values 0, 650, and 2,000.
The probability distribution of d1 is as given below:
d1 | 0 | 650 | 2,000 |
p(d1) | 13 | 13 | 13 |
b.
Obtain the mean of d1.
b.
Answer to Problem 42E
The mean for the random variable d1 is 883.333.
Explanation of Solution
Calculation:
The mean for the random variable d1 is obtained as given below:
μd1=∑d1⋅p(d1)=(0)(13)+(650)(13)+(2,000)(13)=26503=883.333
Thus, the mean for the random variable d1 is 883.333.
c.
Obtain the standard deviation of d1.
c.
Answer to Problem 42E
The standard deviation for the random variable d1 is 834.5.
Explanation of Solution
Calculation:
The variance for the random variable d1 is obtained as given below:
σ2d1=∑(d1−μd1)2p(d1)=[(0−833.333)2(13)+(650−833.333)2(13)+(2,000−833.333)2(13)]=[(−833.333)2(13)+(−183.333)2(13)+(−1,166.667)2(13)]=694,443.89+33,610.99+1,361,111.893=2,089,166.773=696,388.923
Thus, the variance for the random variable d1 is 696,388.923.
The standard deviation for the random variable d1 is obtained as given below:
σd1=√σ2d1=√696,388.923=834.5
Thus, the standard deviation for the random variable d1 is 834.5.
d.
Check either the probability distributions of d2 and d3 are same as the probability distribution of d1.
d.
Answer to Problem 42E
Neither the probability distributions of d2 and d3 are same as the probability distribution of d1.
Explanation of Solution
Calculation:
The random variable d2 represents the driving distance from City D to the place of meeting.
Therefore, the random variable d2 can take values 0, 650, and 1,350.
The probability distribution of d2 is as given below:
d2 | 0 | 650 | 1,350 |
p(d2) | 13 | 13 | 13 |
The random variable d3 represents the driving distance from City B to the place of meeting.
Therefore, the random variable d3 can take values 0, 650, and 1,350.
The probability distribution of d2 is as given below:
d3 | 0 | 1,350 | 2,000 |
p(d3) | 13 | 13 | 13 |
Therefore, neither the probability distributions of d2 and d3 are same as the probability distribution of d1.
e.
Derive the probability distribution of t.
e.
Answer to Problem 42E
The probability distribution of t is,
t | 650 | 3,350 |
p(t) | 23 | 13 |
Explanation of Solution
Calculation:
The random variable t defined as t=d1+d2.
The values of random variable t can be obtained as follows:
Place of meeting | d1 | d2 | t=d1+d2 | Probability |
City C | 0 | 650 | 650 | 13 |
City D | 650 | 0 | 650 | 13 |
City B | 2,000 | 1,350 | 3,350 | 13 |
The probability distribution of t is as given below:
t | 650 | 3,350 |
p(t) | 23 | 13 |
f.
Check whether the below statements are true or false.
- i. E(t)=E(d1)+E(d2)
- ii. σ2t=σ2d1+σ2d2
f.
Answer to Problem 42E
- i. True.
- ii. False.
Explanation of Solution
Calculation:
From Part (b), E(d1)=883.333 and from Part (c), σ2d1=696,388.923.
The mean for the random variable d2 is obtained as given below:
μd2=E(d2)=∑d2⋅p(d2)=(0)(13)+(650)(13)+(1,350)(13)=20003=666.666
Thus, E(d2)=666.666.
The variance for the random variable d2 is obtained as given below:
σ2d2=∑(d2−μd2)2p(d2)=[(0−666.666)2(13)+(650−666.666)2(13)+(1,350−666.666)2(13)]=[(−666.666)2(13)+(−16.666)2(13)+(683.334)2(13)]=444,443.555+277.755+466,945.3553=911,666.6653=303,888.888
Thus, σ2d2=303,888.888
The mean for the random variable t is obtained as given below:
μt=E(t)=∑t⋅p(t)=(650)(23)+(3,350)(13)=4,6503=1,550
Thus, E(t)=1,550.
The variance for the random variable t is obtained as given below:
σ2t=∑(t−μt)2p(t)=[(650−1,550)2(23)+(3,350−1,550)2(13)]=[(−900)2(23)+(1800)2(13)]=1,620,000+3,240,0003=4,860,0003=1,620,000
Thus, σ2t=1,620,000
i.
From the above results, it can be obtained as follows:
E(t)=1,550.
E(d1)+E(d2)=883.333+666.666=1,549.999≈1550
Therefore, E(t)=E(d1)+E(d2).
Therefore, the given statement is true.
ii.
From the above results, it can be obtained as follows:
σ2t=1,620,000
σ2d1+σ2d2=696,388.923+303,888.888=1,000,277.811
Therefore, σ2t≠σ2d1+σ2d2
Therefore, the given statement is false.
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Chapter 7 Solutions
Introduction to Statistics and Data Analysis
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