Solution Found!
Refer to Chebyshev’s inequality given in Exercise 44.
Chapter 4, Problem 67E(choose chapter or problem)
Refer to Chebyshev’s inequality given in Exercise 44. Calculate P( | X - ? | ? k?) for k = 2 and k = 3 when X ? Bin (20,. 5) , and compare to the corresponding upper bound. Repeat for X ? Bin (20,. 75) Reference exercise 44 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 – ?| ? 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 ? and ? for the distribution of Exercise 13. Then evaluate P(| X - ? | ? 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/ 18, respectively. What is P( |X - ?| ? 3?), and how does it compare to the corresponding bound? d.? ?Give a distribution for which P( |X - ?| ? 5?) = . 04
Questions & Answers
QUESTION:
Refer to Chebyshev’s inequality given in Exercise 44. Calculate P( | X - ? | ? k?) for k = 2 and k = 3 when X ? Bin (20,. 5) , and compare to the corresponding upper bound. Repeat for X ? Bin (20,. 75) Reference exercise 44 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 – ?| ? 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 ? and ? for the distribution of Exercise 13. Then evaluate P(| X - ? | ? 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/ 18, respectively. What is P( |X - ?| ? 3?), and how does it compare to the corresponding bound? d.? ?Give a distribution for which P( |X - ?| ? 5?) = . 04
ANSWER:Answer : Step 1 of 2 : Chebyshev’s inequality states that for any probability distribution of an random variable X and any number k that is at least 1, P(|X – | k) 1 k2 Where, X~B(20, 0.5) The claim is to compare to the corresponding upper bound. Repeat for X Bin (20,. 75) 1 Using Chebyshev’s inequality P(|X – | k) k2 Where, k = 2 and k = 3 For k = 2 P(|X – | k) 1 4 For k = 3 P(|X – | k) 1 9