10_18ans - STAT 400 1. Fall 2011 Examples for 10/18/2011...

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

View Full Document Right Arrow Icon
STAT 400 Examples for 10/18/2011 ( Answers ) Fall 2011 1. Binomial distribution, n = 25, p = 0.50. Normal approximation: mean = n × p = 25 × 0.50 = 12.5. n × p × ( 1 – p ) = 25 × 0.50 × 0.50 = 6.25. SD = 25 . 6 = 2.5. a) P(X = 17) = PMF @ 17 = 0.0322 . b) P(X = 17) = P(16.5 X 17.5) - - 5 . 2 5 . 12 5 . 17 Z 5 . 2 5 . 12 5 . 16 P = P(1.60 Z 2.00) = 0.9772 – 0.9452 = 0.0320 .
Background image of page 1

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

View Full DocumentRight Arrow Icon
c) P(X 11) = 1 – CDF @ 10 = 1 – 0.2122 = 0.7878 . d) P(X 11) = P(X 10.5) - 5 . 2 5 . 12 5 . 10 Z P = P(Z 0.80) = 1 – 0.2119 = 0.7881 . e) P(10 X 14) = CDF @ 14 – CDF @ 9 = 0.7878 – 0.1148 = 0.6730 . f) P(10 X 14) = P(9.5 X 14.5) - - 5 . 2 5 . 12 5 . 14 Z 5 . 2 5 . 12 5 . 9 P = P(– 1.20 Z 0.80) = 0.7881 – 0.1151 = 0.6730 .
Background image of page 2
2. Let X = number of passengers who do not cancel their reservations. Then X has Binomial distribution,
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 11/07/2011 for the course STAT 400 taught by Professor Kim during the Fall '08 term at University of Illinois, Urbana Champaign.

Page1 / 5

10_18ans - STAT 400 1. Fall 2011 Examples for 10/18/2011...

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