# 3 roll a die 100 times and let x be the sum of all of

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3. Roll a die 100 times and let X be the sum of all of the rolls. (a) [4 points] What is E ( X ) and SD ( X )? Solution: The mean and standard deviation of a single die roll are 3 . 5 and q 35 12 , respectively. So E ( X ) = 100 · 3 . 5 = 350 and SD ( X ) = 100 q 35 12 = 5 q 35 3 . (b) [4 points] Why can we assume that X is approximately normal? Solution: It is the sum of several (100) independent identically distributed random variables, hence approximately normal by the Central Limit Theorem . (c) [6 points] Approximate P (200 X 400). Solution: With the continuity correction, P (200 X 400) P 200 - 350 - 1 / 2 5 q 35 3 X * 400 - 350 + 1 / 2 5 q 35 3 Φ 400 - 350 + 1 / 2 5 q 35 3 - Φ 200 - 350 - 1 / 2 5 q 35 3 99 . 84% . (I did not require continuity correction, though it is applicable. Without conti- nuity correction, we would have gotten approximately 99.83%.) (d) [8 points] Now let Y be the number of heads from flipping a coin 500 times. Ap- proximate P ( X 3 Y ). Solution: Note that E ( Y ) = 250 and V ar ( Y ) = 500 · 1 2 · 1 2 = 125. Page 2

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Math 431: Final Exam Let Z = X - 3 Y . Then E ( Z ) = E ( X ) - 3 E ( Y ) = 350 - 3 · 250 = - 400 and V ar ( Z ) = V ar ( X ) + 9 V ar ( Y ), because X and Y are independent. So V ar ( Z ) = 100 · 35 12 + 9 · 125 = 4250 3 and SD ( Z ) = 5 q 170 3 . Therefore, with the continuity correction, P ( X 3 Y ) = P ( Z 0) P Z * 400 + 1 / 2 5 q 170 3 Φ 400 + 1 / 2 5 q 170 3 100% . (Again, I did not require continuity correction. I meant to ask for P (2 X 3 Y ), which would have been approximately 85.43%.) 4. Men enter a store at a rate of 7 per hour. Women enter at a rate of 5 per hour. Assume people arrive randomly and independently.
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