Basic Business Statistics, Student Value Edition
14th Edition
ISBN: 9780134685113
Author: Mark L. Berenson, David M. Levine, David F. Stephan, Kathryn Szabat
Publisher: PEARSON
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Textbook Question
Chapter 5, Problem 1PS
Given the following
a. Compute the expected value for each distribution.
b. Compute the standard deviation for each distribution.
c. What is the probability that x will be at least 3 in Distribution A and Distribution B?
d. Compare the results of distributions A and B.
a.
Expert Solution
To determine
Find the expected value for distribution A and distribution B.
Answer to Problem 1PS
The expected value of distribution A is 1, and distribution B is 3.
Explanation of Solution
Calculation:
The table of probability distributions for distribution A and B is given.
The formula for calculating expected value is
E(X)=N∑i=1xiP(X=xi)
Here, xi is the ith value of X , P(X=xi) is the probability of occurrence of ith value of X , and N is the number of values in the discrete variable X .
The expected value of distribution A is calculated by substituting the values in the above formula. It is calculated as,
E(X)=(0×0.50)+(1×0.20)+(2×0.15)+(3×0.10)+(4×0.05)=0+0.20+0.30+0.30+0.20=1
The expected value of distribution B is
E(X)=(0×0.05)+(1×0.10)+(2×0.15)+(3×0.20)+(4×0.50)=0+0.10+0.30+0.60+2=3
Thus, the expected value of distribution A and distribution B are 1 and 3, respectively.
b.
Expert Solution
To determine
Find the standard deviation of distribution A and B.
Answer to Problem 1PS
The standard deviation for both distribution A and B is 1.22474.
Explanation of Solution
Calculation:
The standard deviation is calculated using the formula,
σ=√N∑i=1(xi−E(X))2P(X=xi)
Here, xi is the ith value of X , P(X=xi) is the probability of occurrence of ith value of X , and N is the number of values in the discrete variable X .
For distribution A, E(X)=1 . So, the standard deviation for distribution A is calculated as,
σ=√[(0−1)2×0.50]+[(1−1)2×0.20]+[(2−1)2×0.15]+[(3−1)2×0.10]+[(4−1)2×0.05]=√1.5=1.22474
For distribution B, E(X)=3 . So, the standard deviation for distribution B is calculated as,
σ=√[(0−3)2×0.05]+[(1−3)2×0.10]+[(2−3)2×0.15]+[(3−3)2×0.20]+[(4−3)2×0.50]=√1.5=1.22474
Thus, the standard deviation of distribution A and B is same, that is, 1.2247.
d.
Expert Solution
To determine
Compare the results of the two distributions.
Answer to Problem 1PS
The mean value for distribution B is higher than distribution A but the standard deviation is same for both the distributions.
Explanation of Solution
From part a, the expected value for distribution A and B are obtained as 1 and 3, respectively. From part b, the standard deviation for both the distributions is obtained as 1.2247.
Thus, it can be said that the mean value or the expected value of distribution B is greater than that of distribution A. However, both of them have the same spread as the value of standard deviation is same.
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Chapter 5 Solutions
Basic Business Statistics, Student Value Edition
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