HW3_solution

# HW3_solution - STATS 116 Assignment 3 Solutions 2.4.6...

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STATS 116 Assignment 3 Solutions Author: Yijia Joe Zhou 2.4.6 Drawing Black Balls Remarks: most of you get this problem right. Remember to use Poission approximation to calculate complex binomial distribution probabilities. a) The number of black balls (denoted as X )seen in a series of 1000 draws with replacement has binomial (1000,2/1000) distribution with mean 2 . By Poission approximation, 0 1 2 ( 2) .406 0! 1! ( 2) .271 2! ( 2) 1 ( 2) ( 2) 0.323 P X e e P X e P X P X P X Therefore getting fewer than 2 balls is most likely. b) By independence of the two series, 1000 1000 1 2 1 2 1 2 0 0 4 2 0 ( ) ( , ) ( ) ( ) 4 0.207 ( !) k k k k P X X P X k X k P X k P X k e k 2.5.5 Sample Size Remarks: 1. The critical step in the problem is to approximate the binomial distribution (of how many people vote for A) by normal distribution, using Central Limit Theorem. Some of you blindly used a formula in the textbook without knowing why. Below I provided with a step-by-step solution. 2. Some of you added a continuity correction term of ½ . It is perfectly correct, but this makes your calculation more complicated. In fact, since n is large, 1 2 n is small and can be neglected. Assume a sample size n. Let X be the number of people voting for A. Therefore

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STATS 116 Assignment 3 Solutions Author: Yijia Joe Zhou 0.55 0.5 0.55 ( ) ( ) 2 0.55(1 0.55) 0.55(1 0.55) 0.5 0.55 0.5 0.55 0.5 0.55 1 1 99% 0.55(1 0.55) 0.55(1 0.55) 0.55(1 0.55) n X n n n P A wins election P X P n n n n n n n n P z P z n n n   z follows standard normal distribution. Recall that the variance of a binomial distribution is (1 ) p p n . From the table we find ( 2.326) 0.01   . Solve 0.5 0.55 2.326 0.55(1 0.55) n n n   , we get min 537 n 2.5.8 The Lucky Guys Remarks: This is an interesting problem and more than one method exist because there exist different perspectives in which we may approach the problem. I presented alternative methods in the solution below. a. Method I Think in the perspective of the lucky guy. For fixed i*, the probability that person i* has three winning tickets is 10 3 ( * 3 | * ) 100 3 P person i has winning tickets i fixed . There are 10 possibilities for i*, therefore 10 3 ( ) ( 3 ) 10 0.007421 100 3 P a P someone has winning tickets Method II Think in the perspective of the winning tickets. Label the tickets as 1 2 3 , , w w w . Assume person i* has all three winning tickets. We start from 1 w .
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