p chapter 5 solutions 3rd ed fall 2007

p chapter 5 solutions 3rd ed fall 2007 - Probability, Third...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solutions to practice questions Chapter 5 Solution 5.1 Let X be the number of visits (out of the next 10 visits) that result in a referral. Then X follows a binomial distribution with n =10 and p = 0.28 , and the required probability is: 46 Pr (X = 4) =( C4 )(0.28 )(0.72 )= 0.1798 10 Solution 5.2 Let X be the number of people who suffer a side effect. Then X follows a binomial distribution with n = 1,000 and p = 0.005 , and the required probability is: Pr (X = 1)= Pr (X = 0)+ Pr (X = 1) 1,000 1 999 = 0.995 +( C )(0.005 )(0.995 ) 1,000 1 = 0.0067 + 0.0334 = 0.0401 Solution 5.3 The probability that a report is filed is: Pr More than one injury)= 0.23 + 0.17 + 0.09 + 0.04 =
Background image of page 1

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

View Full DocumentRight Arrow Icon
0.53 Let X be the number of reports filed for the next 20 games. Then X follows a binomial distribution with n = 20 and p = 0.53 . The expected number of reports is: [ ]= np = 20 0.53 ( EX = 10.6 The standard deviation of the number of reports is: = =20 0.53 0.47 2.232 npq
Background image of page 2
Probability Solutions to practice questions Chapter 5 Solution 5.4 Using the moment generating function: n t MX ()=(pe +q) t ttn-1 .MX ' ()t =()( n pe )( pe +q) ttn-1 t 2 tn-2 MX ()=()( pe )( +q) +()( n -1)( )( pe +q) . '' tnpe n pe (product rule) (i) The mean is: n-1 [] =MX '()=( )( )( p p +q) =np since EX 0 n p +q =1 (ii) The variance is: ) =EX 2 -(EX ])2 var(X . . [ .. 2 n-12 n-22 EX =M ()=( )( )( p +q) +( )( n - pp +q) =np +nn -1) p .. X '' 0 np n 1)()( ( .. 22 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
. var(X) =np + ( -1)p -( ) =np -np =np (1 -p) =npq nn np Solution 5.5 The number of claims, X , follows the binomial distribution with n =10 and p =0.125 . The expected number of claims is: [ ]=np =10 0.125 EX =1.25 The variance is: var ()=npq =10 0.125 0.875 X =1.09375 The standard deviation is: ()==var 1.09375 1.04583 X The reserve is: 10 (1.25 + 1.04583 ) 2 =33.417 ($million) Solution 5.6 The number of hurricanes, X , can be modeled using a binomial distribution if we treat each year as a Bernoulli
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.

Page1 / 17

p chapter 5 solutions 3rd ed fall 2007 - Probability, Third...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online