Homework_9_Solution - Stat 414 Homework 9 Fall 2008 (45...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Stat 414 Homework 9 Fall 2008 (45 points) 1. 3.4 16 (5 points) By using the table IV in the page 685, we have P(X > 30.14) = .05 where X denotes a single observation. Since 10 observations are independent and are identically distributed from the chi-square(df =19), the random variable W (the number out of 10 observations that exceed 30.14) is Bin(n=10,p=.05). Therefore, P(W = 2) = .9885 - .9139 = .0746. 2. 3.5 9 (9 points 3 points for each part) (a) If we set 1 = 0 or 2 = 1, < < = = x e e x f and e x F x x e x e , ) ( 1 ) ( y = e x so y > 0 and x = g(y) = lny. g(y) = 1/y < < = = = y e y ye y e e y f y y e y y , 1 1 ) ( ln ln Therefore Y has an exponential distribution with the parameter = 1. (b) If 1 0 or 2 1, < < = = x e e x f and e x F x x e x e , 1 ) ( 1 ) ( 2 / ) 1 ( 2 1 2 / ) 1 ( / ) ( 2 < < = = = = + = = > = y e y e y y ye y f y...
View Full Document

This note was uploaded on 11/10/2008 for the course STAT 414 taught by Professor Senturk,damla during the Fall '07 term at Pennsylvania State University, University Park.

Page1 / 4

Homework_9_Solution - Stat 414 Homework 9 Fall 2008 (45...

This preview shows document pages 1 - 2. Sign up to view the full document.

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