solutions08

solutions08 - Math 302.102 Fall 2010 Solutions to...

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Math 302.102 Fall 2010 Solutions to Assignment #8 1. Since X has a binomial distribution with parameters n = 100 and p = 4 / 7, we can use the central limit theorem to approximate P { X 50 } . That is, we know that if X Bin( n,p ), then X - np p np (1 - p ) has an approximately normal N (0 , 1) distribution as long as both np and n (1 - p ) are not too small. Solution 1 . Using the continuity correction, we find P { X 50 } = P { X > 49 . 5 } = P X - 100 · 4 7 q 100 · 4 7 · 3 7 > 49 . 5 - 100 · 4 7 q 100 · 4 7 · 3 7 P { Z > - 1 . 544 } where Z ∼ N (0 , 1). Using a table of normal probabilities, we find P { Z > - 1 . 544 } . = 0 . 9382. Thus, P { X 50 } ≈ 0 . 9382 . Solution 2 . Observe that the largest X could be is 100. Thus, P { X 50 } is really equivalent to P { 50 X 100 } , and so using the continuity correction, we find P { X 50 } = P { 50 X 100 } = P { 49 . 5 < X < 100 . 5 } = P 49 . 5 - 100 · 4 7 q 100 · 4 7 · 3 7 < X - 100 · 4 7 q 100 · 4 7 · 3 7 < 100 . 5 - 100 · 4 7 q 100 · 4 7 · 3 7 P {- 1 . 544 < Z < 8 . 761 } = P { Z < 8 . 761 } - P { Z < - 1 . 544 } where Z ∼ N (0 , 1). Using a table of normal probabilities, we find P { Z < 8 . 761 } = 1 and P { Z < - 1 . 544 } . = 0 . 0618. Thus, P { X 50 } ≈ 1 - 0 . 0618 = 0 . 9382 . Solution 3 . Without using the continuity correction, we find P { X 50 } = P X - 100 · 4 7 q 100 · 4 7 · 3 7 50 - 100 · 4 7 q 100 · 4 7 · 3 7 P { Z > - 1 . 443 } where Z ∼ N (0 , 1). Using a table of normal probabilities, we find P { Z > - 1 . 443 } . = 0 . 9251. Thus, P { X 50 } ≈ 0 . 9251 . Remark. The exact probability can be determined using a computer. It is 0 . 9381732. Note that the approximations provided by Solutions 1 and 2 are remarkably accurate.
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2. Let X j denote the outcome of the j th game so that Y j = X j - 4 denotes the net winnings on the j th game. The total net winnings are therefore Y 1 + ··· + Y 100 = X 1 + ··· + X 100 - 400 . Let Y = Y 1 + ··· + Y 100 100 and X = X 1 + ··· + X 100 100 so that Y = Y 1 + ··· + Y 100 100 = X 1 + ··· + X 100 - 400 100 = X 1 + ··· + X 100 100 - 400 100 = X - 4 . It is equivalent to work with either
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solutions08 - Math 302.102 Fall 2010 Solutions to...

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