75 1 f075 alookupinappendixtable1gives f075 07734

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: etry = 1 − F(0.5) A look‐up in Appendix Table 1 gives: F(0.5) = 0.6915 The answer is: P( X > μ + 2 ) = 1 − 0.6915 = 0.3085 7 Chapter 7 (b) What is the probability that the sample mean is more than 3 hours below the population mean ? That is, find: P( X < μ − 3) Note: P( X < μ − 3) < P( X < μ − 2) = P( X > μ + 2) This is illustrated in the graph below. PDF of X μ−3 μ−2 μ μ+2 Therefore, the answer must be smaller than the probability calculated for part (a). 8 Chapter 7 The solution method follows similar steps to part (a): ⎛ X − μ (μ − 3) − μ ⎞ P( X < μ − 3) = P⎜ ⎟ ⎜ se( X ) < se( X ) ⎟ ⎠ ⎝ = P Z < − 34 ( ) = 1 − P(Z < 0.75) = 1 − F(0.75) A look‐up in Appendix Table 1 gives: F(0.75) = 0.7734 The answer is: P( X < μ − 3) = 1 − 0.7734 = 0.2266 9 Chapter 7 (c) What is the probability that the sample mean differs from the population mean by more than 4 hours ? That is, find: lower tail upper tail ↓ ↓ 1 − P(μ − 4 < X < μ + 4) = P( X < μ − 4) + P( X...
View Full Document

This note was uploaded on 02/06/2014 for the course ECON ECON 325 taught by Professor Whistler during the Spring '10 term at The University of British Columbia.

Ask a homework question - tutors are online