A result called Chebyshev’s inequality states that for any probability distribution of a...

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 σ) ≤ 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. What is the value of the upper bound for k = 2? K + 3? K = 4? K= 5? K = 10?

b. Compute µ and σ for the distribution of Exercise 13. Then evaluate P(|X - µ| ≥ kσ) 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|≥ 3σ), and how does it compare to the corresponding bound?

d. Give a distribution for which P(|X - µ|≥ 5σ) = .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}