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 : 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 Step 2 of 4 : b) We have to 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 P(x=0) = 0.10 P(x=1) = 0.15 P(x= 2) = 0.20 P(x = 3) = 0.25 P(x = 4) = 0.20 P(x = 5) = 0.06 P(x = 6) = 0.04 We have to find the mean and standard deviation = (0)(0.1) +(1)(0.15) + (2)(0.20) + (3)(0.25)+ (4)(0.20) + (5)(0.06) + (6)(0.04) = 2.64 2 Variance = = (0 - 2.64)(0.1) +(1 - 2.64) 2 (0.15) + (2 - 2.64) 2 (0.20) + (3 - 2.64) 2 (0.25)+ (4 - 2.64) 2 (0.20) + (5 - 2.64 ) 2 (0.06) + (6 - 2.64) 2 (0.04) = 2.3074 Standard deviation = variance = 2.3074 = 1.54 Now we have to find P(|X - µ| k) for the give k in part (a). we are trying to find is the range of values for X for which the equation |X - µ| k For k = 2 P( | X – 2.64 | k ) = P( x -2 or x +2 ) = P( x-0.44) + P( x 5.72) = P(X = 6) =0 .04 For k = 3 P( | X – 2.64 | k ) = P( x -3 or x -3 ) = P( x-1.98) + P( x 7.26) = 0 For k = 4 P( | X – 2.64 | k ) = P( x -4 or x -4 ) = P( x-3.52) + P( x 8.8) = 0 For k = 5 P( | X – 2.64 | k ) = P( x -5 or x -5 ) = P( x-5.06) + P( x 10.34) = 0 For k = 10 P( | X – 2.64 | k ) = P( x -10 or x -10 ) = P( x-12.76) + P( x 18.04) = 0 The Chebyshev bound is overly conservative. Notices of the other values of k the probability x is within k standard deviations of the mean is 0.