# 516hw4 - STAT516 Section 2.4 1 FALL 2005 SOLUTION TO HOMEWORK 4 b See text for Poisson(2 histogram c See text and"approximate"

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: STAT516 Section 2.4 1. FALL 2005 SOLUTION TO HOMEWORK 4 b) See text for Poisson(2) histogram. c) See text and "approximate" Poisson(.3284) histogram. 2. a) The number of successes in 500 independent trials with success probability 0.02 has Binomial(500,0.02) distribution with mean = 10. By the Poisson approximation, P(1 success) e - = 0.000454 b) P(2 or fewer successes) =P(0 or 1 or 2 successes) 2 e + + 1 =0.002769 2! - c) P(more than 3 successes) =1-P(3 or fewer successes)= 0.9897 6. a) The number of black balls seen in a series of 1000 draws with replace has Binomial(1000, 2/1000) distribution with mean = 2. By Poisson approximation, - + P(fewer than 2 black balls) = e ( 1 ) = 0.4060 2 = 0.270671 2! P(more than 2 black balls) = 1-0.4060-0.270671=0.3233 P(exactly 2 black balls) = e - So, fewer than two black balls is the most likely event. b) P(both series result in same number of black balls) = = 1000 P (both seies result in k black balls) k =0 1000 P (seies 1 results in k black balls) P (seies 2 results in k black balls) k =0 1000 2 (by 2k = -2 = 0.207 e k! k =0 independence) 7. a) m = int(np + p ) = int(2.6) = 2 . [See page 86 in your text] 2 23 25 1 9 b) P ( S = m) = = 0.266 2 10 10 c) For normal approximation: = 2.5 and = 1.5 . P ( S = m) = 0.2486 d) For Poisson approximation: = 2.5 . P ( S = m) = 0.2565 e) m = 250 . As p is of moderate value, normal would give a better approximation. P ( S = 250) = 0.0268 f) m = 2. As p is very small, Poisson would give a better approximation: P(S = 2) = 0.2565 Section 2.5 20 30 6 4 = P (4 red) = 0.2800 1. a) Without replacement: 50 10 10 4 30 6 20 P (4 red) = = 0.2508 b) With replacement: 4 50 50 2. a) 26 26 25 =13/102 = 0.12745 52 51 30 26 26 2 = [3 times a)] = 0.3823 b) 52 3 26 25 24 =0.8824 c) 1-P(0 red)=152 51 50 40000 60000 k =100 k 100 - k 4. The exact probability is P (at least 45 mean) = 100000 k = 45 100 Using binomial approximation to hyper-geometric and normal approximation to binomial, so 44.5 - 40 = we finally get the probability is approximately 1 - 0.1788 . 4.9 8. a) Suppose a particular person gets all the three winning tickets, the probability is: 10 3 100 3 Since each person can be that particular person, so the final probability is: 10 3 10 =0.007421 100 3 b) Suppose three particular persons are the winners, the probability is: 10 10 10 1 1 1 100 3 10 Since we have combinations of being that three particular persons, the final probability is 3 10 10 10 10 =0.7421 1 1 1 100 3 3 c) 1- a)- b) = 0.2505. To be sure, we the same technique as before, we can obtain the desired probability: 10 10 2 1 10 0.2505 9 = 100 3 Section 3.1 3 3 1 1. X has Binomial(3,1/2) distributions. So P ( X = k ) = for k=0,1,2,3 k 2 a) x P(X=x) 0 1/8 1 3/8 2 3/8 3 1/8 b) Write Y=|X-1| y P(Y=y) 0 3/8 1 1/2 2 1/8 2. a) Joint distribution table for (X,Y) (with replacement) X Y 1 2 3 4 Dist of X 1/16 1/16 1/16 1/16 1/4 1/16 1/16 1/16 1/16 1/4 1/16 1/16 1/16 1/16 1/4 1/16 1/16 1/16 1/16 1/4 1 2 3 4 Dist of Y 1/4 1/4 1/4 1/4 1 1 5 P ( X ) = 10 = Y 16 8 b) Joint distribution table for (X,Y) (with replacement) X Y 1 2 3 4 Dist of X 0 1/12 1/12 1/12 1/4 1/12 0 1/12 1/12 1/4 1/12 1/12 0 1/12 1/4 1/12 1/12 1/12 0 1/4 1/4 1/4 1/4 1/4 1 1 2 3 4 Dist of Y 1 1 P( X ) = 6 = Y 12 2 4. a) Joint distribution table for (X1,X2) X1 X2 1 2 3 4 5 6 Dist of X1 1 1/36 1/36 1/36 1/36 1/36 1/36 1/6 2 1/36 1/36 1/36 1/36 1/36 1/36 1/6 3 1/36 1/36 1/36 1/36 1/36 1/36 1/6 4 1/36 1/36 1/36 1/36 1/36 1/36 1/6 5 1/36 1/36 1/36 1/36 1/36 1/36 1/6 6 1/36 1/36 1/36 1/36 1/36 1/36 1/6 Dist of X2 1/6 1/6 1/6 1/6 1/6 1/6 1 b) Joint distribution table for (Y1,Y2) Y1 1 2 Y2 1 1/36 2/36 2 3 4 5 6 Dist of Y1 0 0 0 0 0 1/36 1/36 0 0 0 0 3/36 3 2/36 2/36 1/36 0 0 0 5/36 4 2/36 2/36 2/36 1/36 0 0 7/36 5 2/36 1/36 1/36 2/36 1/36 0 9/36 6 2/36 2/36 2/36 2/36 2/36 1/36 11/36 Dist of Y2 11/36 9/36 7/36 5/36 3/36 1/36 1 6. There are 8 equally likely outcomes from three fair coin tosses a) X Y 0 1/8 1/8 0 1 1/8 2/8 1/8 2 0 1/8 1/8 b) X and Y are dependent, since, for instance, P(X=2,Y=0)=0, while P(X=2)P(Y=0)=1/4*1/4=1/16 c) z 0 1 2 3 4 P(X+Y=z) 1/8 2/8 2/8 2/8 1/8 0 1 2 10. a) Binomial(n, p) b) Binomial(m, p) c) Binomial(n + m, p)e this fact follows easily from a consideration of the experiment which consists of first tossing a coin m times, and then tossing it n more times. d) Independent; as they are from different blocks of independent trials. ...
View Full Document

## This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at University of California, Berkeley.

Ask a homework question - tutors are online