Chebyshev’s inequality, (see Exercise 44, Chapter 3), is valid for continuous as well as discrete distributions. It states that for any number k satisfying k ? 1,P(|X – ? ?k?) ?1/k2 (see Exercise 44 in Chapter 3 for an interpretation). Obtain this probability in the case of a normal distribution for , 2, and 3, and compare to the upper bound. 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 ?- µ | k ?) ? 1/?k?2 . In words, the probability that the value of ?X ?lies at least ?k ?standard deviations from its mean is at most 1/?k?2. 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.

Answer : Step 1 of 3 : Let the Chebyshev’s inequality be P( X k| ) | 12 k We have to find the probability values for normal distribution for k = 1, 2 and 3 By normal distribution, For k = 1 P( | |) 1 = 1 - P(- < (x - )< ) k2 = 1 - P(-1< (x - )<1) 2 1 = 1 - ( (z 1) - (z-1) ) Where, (z 1) = 0.8413 ( from the standard normal table) (z - 1) = 0.1587 ( from the standard normal table) Therefore, P( | |) = 1 - ( 0.8413 - 0.1587 ) = 1 - 0.6826 = 0.3174 Hence the equation satisfies Step 2 of 3 : For k = 2 P( | |) 1 = 1 - P(- < (x - )< ) k2 = 1 - P(-2< (x - )<2) 12 2 = 1 - ( (z 2) - (z-2) ) Where, (z 2) = 0.9772 ( from the standard normal table) (z - 1) = 0.0228 ( from the standard normal table) Therefore, P( | |) = 1 - ( 0.9772 - 0.0228 ) = 1 - 0.9544 = 0.0456 Hence the equation satisfies