Sections 5.4 %26 5.5 Lecture Problems (1)

Sections 5.4 %26 5.5 Lecture Problems (1) - Solutions for...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions for Lecture Problems from Chapter 5 34. What kind of problem is this? What kind of problem is this? Your choices are either binomial or Poisson. People aged 25 and older either have completed four years of college or not. One or the other, can’t be both, and no third choice. So this is success vs. failure problem. For all parts of this problem, the “success” is selecting an individual who have completed four years of college. (Looking at what is asked in the questions.) So this is a binomial problem. Let X be the r.v. that counts the number of individuals who have completed four years of college in a sample of 15 people. Given: p = 0.28, because 28% of all people aged 25 and older have completed four years of college (from first sentence of paragraph). Binomial with p = 0.28; n and x are given in the questions. (Note if p = 0.28, then 1 – p = 1 – 0.28 = 0.72) Then Prob ( X = x | n , p ) is calculated by ( ) ( ) x x . . ! x ! x ! . p , n | x prob - 5 - 5 1 72 0 × 28 0 × 1 15 = 28 0 = 15 = a. Want Prob ( x =4 | n =15, p =0.28). Chap 5 lecture problems 1 © Harvey Singer 2009 ( ) ( ) 22616335 0 = 026956 0 00614656 0 1365 = 72 0 × 28 0 × 13 × 7 × 15 = 72 0 × 28 0 × 1 × × 11 × 1 × 2 × 3 × 4 1 × × 11 × 12 × 13 × 14 × 15 = 72 0 × 28 0 × 11 4 15 = 72 0 × 28 0 × 1 4 15 = 28 0 = 15 = 4 = 11 4 11 2 11 4 4 15 4 . ... . × . × . . . . ... ... . . ! ! ! . . ! ! ! . p , n | x prob - 4 - 5
Background image of page 1

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

View Full DocumentRight Arrow Icon
So Prob ( x =2 | n =6, p =0.23) = 0.2262. b. Want Prob ( x ≥3 | n =15, p =0.28). Prob ( x ≥3 | n =15, p =0.28) = Prob ( x =3 | n =15, p =0.28) + Prob ( x =4 | n =15, p =0.28) + Prob ( x =5 | n =15, p =0.28) + … More clever way is to use complements. Prob ( x ≥3 | n =15, p =0.28) = 1 – [ Prob ( x =0 | n =15, p =0.28) + Prob ( x =1 | n =15, p =0.28) + Prob ( x =2 | n =15, p =0.28)] Prob ( x =0 | n =15, p =0.28) = 0.0072 Prob ( x =1 | n =15, p =0.28) = 0.0423 0 15 15! (0) (.28) (1 .28) .0072 0!(15)! f = - = Prob ( x =2 | n =15, p =0.28) = 0.1150 Prob ( x ≥3 | n =15, p =0.28) = 1 – [0.0072 + 0.0423 + 0.1150] = 1 – [0.0072 + 0.0423 + 0.1150] = 1 – 0.1645 = 0.8355 Chap 5 lecture problems 2 © Harvey Singer 2009
Background image of page 2
35. What kind of problem is this? What kind of problem is this? Your choices are either binomial or Poisson. Students either withdraw from the course without completing the course or not. One or the other, can’t be both, and no third choice. So this is success vs. failure problem. For all parts of this problem, the “success” is selecting an individual who has withdrawn from the course. (Looking at what is asked in the questions.) So this is a binomial problem. Let X be the r.v. that counts the number of individuals who have withdrawn from the course in a sample of 20 students. Given:
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/26/2011 for the course OM 210 taught by Professor Singer during the Fall '08 term at George Mason.

Page1 / 16

Sections 5.4 %26 5.5 Lecture Problems (1) - Solutions for...

This preview shows document pages 1 - 4. Sign up to view the full document.

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