A result called Chebyshev’s inequality states that for any

Chapter 4, Problem 44E

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QUESTION:

A result called ?Chebyshev’s inequality ?states that for any probability distribution of an rv X and any number k that is at least 1, P( | X - µ | k ?) ? 1/k2 . In words, the probability that the value of X lies at least k standard deviations from its mean is at most 1/k2. a.? hat is the value of the upper bound for k = 2? K + 3? K = 4? K= 5? K = 10? b. ?Compute µ and ? for the distribution of Exercise 13. Then evaluate P(|X - µ| ? k?) for the values of k given in part (a). What does this suggest about the upper bound relative to the corresponding probability? c. ?Let X have possible values -1, 0, and 1, with probabilities 1/18, 8/9 and 1/8 , respectively. What is P(|X - µ|? 3?), and how does it compare to the corresponding bound? d.? ?Give a distribution for which P(|X - µ|? 5?) = .04. Reference exercise -13 A mail-order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. Calculate the probability of each of the following events. a.? at most three lines are in use} b.? ?{fewer than three lines are in use} c.?? at least three lines are in use} d.? ?{between two and five lines, inclusive, are in use} e.?? between two and four lines, inclusive, are not in use} f.? ?{at least four lines are not in use}

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QUESTION:

A result called ?Chebyshev’s inequality ?states that for any probability distribution of an rv X and any number k that is at least 1, P( | X - µ | k ?) ? 1/k2 . In words, the probability that the value of X lies at least k standard deviations from its mean is at most 1/k2. a.? hat is the value of the upper bound for k = 2? K + 3? K = 4? K= 5? K = 10? b. ?Compute µ and ? for the distribution of Exercise 13. Then evaluate P(|X - µ| ? k?) for the values of k given in part (a). What does this suggest about the upper bound relative to the corresponding probability? c. ?Let X have possible values -1, 0, and 1, with probabilities 1/18, 8/9 and 1/8 , respectively. What is P(|X - µ|? 3?), and how does it compare to the corresponding bound? d.? ?Give a distribution for which P(|X - µ|? 5?) = .04. Reference exercise -13 A mail-order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. Calculate the probability of each of the following events. a.? at most three lines are in use} b.? ?{fewer than three lines are in use} c.?? at least three lines are in use} d.? ?{between two and five lines, inclusive, are in use} e.?? between two and four lines, inclusive, are not in use} f.? ?{at least four lines are not in use}

ANSWER:

Answer : Step 1 of 4 : Given, A result called Chebyshev’s inequality states that for any probability distribution of an rv X and any number k that is at least 1, P( | X - µ | k ) 1/k2 . In words, the probability that the 2 value of X lies at least k standard deviations from its mean is at most 1/k . a) The claim is to find the upper bound for k = 2, k = 3, k = 4, k = 5 and k = 10. This is a simple problem of evaluating 2 1 for each given k. k 1 For k = 2 , k2 = 0.25 k = 3 , 1 =0.11 k2 k = 4 , 12 =0.0625 k 1 k = 5 , k2 = 0.04 k = 10 , 1 = 0.01 k2

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