HW3_solution - STATS 116 Assignment 3 Solutions 2.4.6...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern