This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution: Continuing from (b), V ar ( H ) = n G N ( N G ) N N n N 1 = 15 16 Â· 47 51 â‰ˆ . 863 . This is the variance that we would get if we chose cards with replacement (15 / 16), multiplied by the correction factor (47 / 51) that we use because we 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 Â· 3 . 5 = 350 and SD ( X ) = âˆš 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 â‰¤ X â‰¤ 400). Solution: With the continuity correction, P (200 â‰¤ X â‰¤ 400) â‰ˆ P 200 350 1 / 2 5 q 35 3 â‰¤ X * â‰¤ 400 350 + 1 / 2 5 q 35 3 â‰ˆ Î¦ 400 350 + 1 / 2 5 q 35 3  Î¦ 200 350 1 / 2 5 q 35 3 â‰ˆ 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 â‰¤ 2 Y + 60). Solution: Let Z = X 2 Y . Then E ( Z ) = E ( X ) 2 E ( Y ) = 350 2 Â· 175 = 0 and V ar ( Z ) = V ar ( X ) + 4 V ar ( Y ), because X and Y are independent. So Page 2 Math 431: Final Exam V ar ( Z ) = 100 35 12 + 4 Â· 50 35 12 = 25 Â· 35 and SD ( Z ) = 5...
View
Full Document
 Spring '12
 Miller
 Probability, Probability theory, continuity correction

Click to edit the document details