hw8_print - STAT416: Probability — Spring 2011...

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: STAT416: Probability — Spring 2011 Assignments 1 Homework #8 (Solution) March 31, 2011 1 Total possible points is 50 1. Ex. 6.1 (10 pts) . Let X = number of heads, then X ∼ B ( n, . 5). n = 10 , P ( X = 5) = 10 5 1 2 10 = 0 . 2461 . n = 30 , P ( X = 15) = 30 15 1 2 30 = 0 . 1445 . n = 50 , P ( X = 25) = 50 25 1 2 50 = 0 . 1123 . In general, n, P X = n 2 = n n 2 1 2 n → , as n → ∞ . 2. Ex. 6.2 (5 pts) . P ( H ) = 0 . 4, P ( H ) = 0 . 5, P ( H ) = 0 . 6. Let X = total number of heads obtained, then X = 0 , 1 , 2 , 3. P ( X = 0) = P ( TTT ) = 0 . 6 × . 5 × . 4 = 0 . 12 . P ( X = 1) = P ( TTH )+ P ( THT )+ P ( HTT ) = 0 . 6 × . 5 × . 6+0 . 6 × . 5 × . 4+0 . 4 × . 5 × . 4 = 0 . 38 . P ( X = 2) = P ( THH ) + P ( HTH ) + P ( HHT ) = 0 . 38 . P ( X = 3) = P ( HHH ) = 0 . 4 × . 5 × . 6 = 0 . 12 . So, P ( X = 0) = P ( X = 3) , and P ( X = 1) = P ( X = 2) . The distribution is not binomial....
View Full Document

This note was uploaded on 04/23/2011 for the course STAT 416 taught by Professor Staff during the Spring '08 term at Purdue University.

Page1 / 2

hw8_print - STAT416: Probability — Spring 2011...

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