We roll a standard die 100 times Let X be the sum of the num

We roll a standard die 100 times. Let X be the sum of the numbers that come up in

these 100 rolls. Use the Chebyshev’s inequality to bound Pr(|X 350| 50).

Solution

Let Xi be the number on the face of the die for roll i. Let X be the sum of the dice rolls. Therefore X = P100 i=1 Xi . By linearity of expectation, we write E[X] = P100 i=1 E[Xi ]. We can compute E[Xi ] = X 6 j=1 jP[Xi = j] = X 6 j=1 j(1/6) = (1/6)6(7) 2 = 7/2, where we use the fact that Pn j=1 j = n(n+1) 2 . Then we have E[X] = 100(7/2) = 350. To use Chebyshev’s inequality, the only remaining value we need to compute is the variance of X. By the independence of the dice rolls we have Var(X) = Var X i Xi ! = X 100 i=1 Var(Xi) To compute the variance of a single dice roll, we use Var(Xi) = E[X2 i ] + E[X] 2 E[X2 i ] = X 6 j=1 j 2P[Xi = j] = X 6 j=1 j 2 (1/6) = 1 6 · 6(7)(13) 6 = 91/6 where we use the fact that Pn j=1 j 2 = n(n+1)(2n+1) 6 . Now we can finish computing the variance of Xi as Var(Xi) = E[X2 i ] E[X] 2 = 91/6 (7/2)2 = 35/12. And the variance of X is Var(X) = 100(231/4). Finally, we can by Chebyshev’s inequality we have P[|X 350| 50] 100(35/12) 502 = 7/60