# a3s - P Z ≥ 2 08 = 0 0188 less then 2 of the time we...

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MATH 1131 3.0 - Fal 2010 Solution for Assignment 3 Question 1: a. X Binomial (5 , 0 . 8). Therefore P ( X = 5) = 5! 5!0! (0 . 8) 5 (0 . 2) 0 = 0 . 32768 b. Given n = 250 ,p = 0 . 8 and we have np 5 and n (1 - p ) 5) P p 0 . 75) = P ± ˆ p - . 8 0 . 8(0 . 2) / 250 0 . 75 - . 8 0 . 8(0 . 2) / 250 ² P ( Z ≥ - 1 . 98) = 0 . 9761 Question 2: Let Y = Isaac’s net gain, and W = number of games played. We know that W Geometric (0 . 35) and hence E ( W ) = 1 0 . 35 . a. Y = 100 - 50 W E ( Y ) = 100 - 50 E ( W ) = 100 - 50(1 / 0 . 35) = - 42 . 86 b. Isaac’s loss is the casino’s gain. Therefore, for 100 diﬀerent players, the casino sees X 1 ,...X 100 where each one behaves like X = 50 W - 100 which has mean 42.86 and variance 50 2 var ( W ) = 50 2 (1 - 0 . 35) /. 35 2 = 13265 . 31 We want P ( X i 1) = P ( ¯ X 0). Since n is large, we can apply CLT and thus the required probability is: P ( ¯ X 0) P ( Z ≤ - 3 . 72) 0 Question 3: Given n = 81 ,p = 0 . 25 a. P (0 . 19 ˆ p 0 . 22) P ( - 1 . 25 Z ≤ - 0 . 62) = 0 . 1620 b. Want to ﬁnd x such that P p x ) = 0 . 90 1 . 285 = x - 0 . 25 0 . 25(0 . 75) / 81 x = 0 . 3118. c. P p 0 . 35)
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Unformatted text preview: P ( Z ≥ 2 . 08) = 0 . 0188. less then 2% of the time we would see ˆ p as large or larger then 0.35. Which implies it is an unusual value. Question 4: a. P ( μ-1 . 96 σ < X < μ + 1 . 96 σ ) = P (-1 . 96 < Z < 1 . 96) = 0 . 95 b. P ( μ-2 . 5 σ < X < μ + βσ ) = P (-2 . 5 < Z < β ) = 0 . 95 ⇒ P ( Z ≤ β )-P ( Z ≤ -2 . 5) = . 95 ⇒ P ( Z ≤ β ) = 0 . 95 + 0 . 0062 = 0 . 9562 ⇒ β = 1 . 71 Length of the interval in part (a) = 2(1 . 96 σ ) = 3 . 92 σ Length of the interval in part (b) (1 . 71-(-2 . 5)) σ = 4 . 21 σ c. Draw the picture and see how you move one bound of the interval will aﬀect the other bound when the total area is kept constant....
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