Are sampling without replacement 3 roll a die 100

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are sampling without replacement . 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 result of rolling a die 50 times and summing the num- bers. Approximate P ( X 2 Y + 60). Solution: Let Z = X - 2 Y . Then E ( Z ) = E ( X ) - 2 E ( Y ) = 350 - 2 · 175 = 0 and V ar ( Z ) = V ar ( X ) + 4 V ar ( Y ), because X and Y are independent. So Page 2
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Math 431: Final Exam V ar ( Z ) = 100 35 12 + 4 · 50 35 12 = 25 · 35 and SD ( Z ) = 5 35. Therefore, with the continuity correction, P ( X 2 Y + 60) = P ( Z 60) P Z * 60 + 1 / 2 5 35 Φ 60 + 1 / 2 5 35 97 . 96% .
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