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Unformatted text preview: Solutions to Homework 9 May 1, 2007 6.6: 2 Z = X + Y , where X,Y are independent binary variables with p = 1 / 2. Thus E ( Z ) = 2( 1 2 ) = 1, and so by definition, Var( Z ) = E ([ Z 1] 2 ) = 1 4 (0 1) 2 + 1 2 (1 1) 2 + 1 4 (2 1) 2 = 1 2 Finally, 2 = q 1 2 . 707 Comments The possible numbers of heads you get when tossing a fair coin twice are 0, 1, and 2 with probability 1 / 4, 1 / 2 and 1 / 4, respectively. 6.6: 4 Var(X + k) = E((X + k) E(X + k)) 2 ) = E((X + k) E(X) k) 2 = E ( X = E ( X )) 2 = Var(X) . 6.6: 9(a) Let X be the random variable that represents the number of times you see 4 dots when you toss a fair die 600 times. X is distributed as B 600 , 1 / 6 , so E ( X ) = 100 and X = q 600(1 / 6)(5 / 6) = 10 q 5 / 6. By Chebyshevs Inequality, Pr(  X 100  10 q 5 / 6) > 1 1 / 2 . We want to bound Pr(  X 100  > 10). Taking = q 6 / 5, it follows that Pr(  X 100  > 10) = 1 Pr(  X 100  10) < 1 / q 6 / 5 2 = 5 / 6. Alternatively, you can use the fact that Pr(  X 100  > 10) = Pr(  X 100  11). Since X = 10 q 5 / 6, it follows that 11 = (11 / 10) q 6 / 5 X . Thus, Pr(  X 100  11) = Pr(  X 100  (11 / 10 q 6 / 5 X . Chebysheve Inequality also says that Pr( X E ( X )  X ) 1 / 2 , so Pr(  X 100  11) = Pr(  X 100  (11 / 10) q 6 / 5 X ) 1 / ((11 / 10) q 6 / 5) 2 = 250 / 363 . This alternate approach actually gives a better estimate....
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 Spring '07
 

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