tut_4_sol_updated

tut_4_sol_updated - CollegeofEngineering,QatarUniversity

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College of Engineering, Qatar University Page 1of 5 Probability and Statistics, GENG200 Tutorial - 4 Question Sections 3-97 Page 98: 1) 3-97 SOLUTION: 3-97. X has a hypergeometric distribution N=100, n=4, K=20 a.) ( )( ) () 4191 . 0 3921225 ) 82160 ( 20 ) 1 ( 100 4 80 3 20 1 = = = = X P b.) , the sample size is only 4 0 ) 6 ( = = X P c.) ( )( ) 001236 . 0 3921225 ) 1 ( 4845 ) 4 ( 100 4 80 0 20 4 = = = = X P d.) 8 . 0 100 20 4 ) ( = = = = N K n np X E 6206 . 0 99 96 ) 8 . 0 )( 2 . 0 ( 4 1 ) 1 ( ) ( = = = N n N p np X V Exercises Sections 3-8 Page 98: 2) 3-100 Answer: 3-100. (a) f(x) = / x x 3 12 24 3 36 (b) μ=E(X) = np= 3*24/36=2 V(X)= np(1-p)(N-n)/(N-1) =2*(1-24/36)(36-3)/(36-1)=0.629 (c) P(X 2) =1-P(X=3) =0.717 Exercises for Section 3-9 Page 103: 3) 3-110 SOLUTION: 3-110 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with λ = 10. PX e ! . == = 5 10 5 0 0378 10 5 . b) e eee !!! . ≤= + + + = −−− 3 10 1 10 2 10 3 0 0103 10 10 10 2 10 3 c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with λ = 20. PY e ! . = 15 20 15 00516 20 15 d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with

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College of Engineering, Qatar University Page 2of 5 λ = 5. PW e () ! . == = 5 5 5 01755 55 Exercises for Section 3-4 Page 77: 4) 3-39 SOLUTION: 3-39. Mean and Variance 2 ) 2 . 0 ( 4 ) 2 . 0 ( 3 ) 2 . 0 ( 2 ) 2 . 0 ( 1 ) 2 . 0 ( 0 ) 4 ( 4 ) 3 ( 3 ) 2 ( 2 ) 1 ( 1 ) 0 ( 0 ) ( = + + + + = + + + + = = f f f f f X E
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This note was uploaded on 10/12/2011 for the course STATISTICS 101 taught by Professor Nazim during the Spring '10 term at Qatar University.

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tut_4_sol_updated - CollegeofEngineering,QatarUniversity

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