This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STAT 400 Examples for 09/27/2011 (1) Fall 2011 Gamma Distribution : ( ) ( ) x e x x f 1  = , 0 x < OR ( ) ( ) 1 1 x e x x f = , 0 x < If T has a Gamma ( , = 1 / ) distribution, where is an integer, then F T ( t ) = P ( T t ) = P ( X t ) , P ( T > t ) = P ( X t 1 ) , where X t has a Poisson ( t = t / ) distribution. 1. Let Y be a random variable with a Gamma distribution with = 5 and = 3. Find the probability P ( Y > 18 ) a) by integrating the p.d.f. of the Gamma distribution; P ( Y > 18 ) = ( )  18 3 1 5 5 3 5 1 dx e x x =  18 3 4 5,832 1 dx e x x = b) by using the relationship between Gamma and Poisson distributions; P ( Y > 18 ) = P ( X 18 4 ) = 0.285 , where X 18 is Poisson ( 18 / = 6 ). EXCEL: = POISSON( x , , ) gives P( X = x ) = POISSON( x , , 1 ) gives P( X x ) A B A B 1 =POISSON(4,18/3,1) 1 0.285057 2 2 2. During a radio trivia contest, the radio station receives phone calls according to Poisson process with the average rate of five calls per minute. Find the probability that the ninth phone call would arrive during the third minute. X t = number of phone calls in t minutes. Poisson ( t ) T k = time of the k th phone call. Gamma, = k . five calls per minute = 5. a) Find the probability that we would have to wait more than two minutes for the ninth phone call. P ( T 9 > 2 ) = P ( X 2 8 ) = P ( Poisson ( 10 ) 8 ) = 0.333 . OR P ( T 9 > 2 ) = ( )  2 5 1 9 9 9 5 dt t t e =  2 5 8 9 8 5 ! dt t t e = b) Find the probability that the ninth phone call would arrive during the third minute. P ( 2 < T 9 < 3 ) = P ( T 9 > 2 ) P ( T 9 > 3 ) = P ( X 2 8 ) P ( X 3 8 ) = P ( Poisson ( 10 ) 8 ) P ( Poisson ( 15 ) 8 ) = 0.333 0.037 = 0.296 . OR P ( 2 < T 9 < 3 ) = ( )  3 2 5 1 9 9 9 5 dt t t e =  3 2 5 8 9 8 5 !...
View Full
Document
 Fall '08
 Kim
 Statistics, Probability

Click to edit the document details