6554 1 09452 060 23 chapter 6

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: ) Enter the values: NORMDIST(400, 380, 50 ,1) This returns the probability: 0.6554 19 Chapter 6 (b) & (c) Find P( X > 360) . This gives the probability that a randomly chosen student will spend more than $360 on textbooks in September. Express the problem in the form of a probability statement about the standard normal variable Z: ⎛ X − μ 360 − μ ⎞ P(X > 360) = P⎜ > ⎟ σ⎠ σ ⎝ 360 − 380 ⎞ ⎛ = P⎜ Z > ⎟ 50 ⎝ ⎠ = P(Z > − 0.4) = P(Z < 0.4) by symmetry = F(0.4) This is identical to the probability calculated for part (a). That is, P( X > 360) = 0.6554 The answers in parts (a) and (b) are the same because the normal distribution is symmetric about the mean. 20 Chapter 6 The graph below demonstrates that because of symmetry about the mean: P(X > 360) = P(X < 400) Also, P(X < 360) = P(X > 400) 2 PDF of X ~ N( 380 , 50 ) P(X > 400)= 1 - 0.6554 f(x) P(X < 360) = 1 - 0.6554 360 380 x 400 21 Chapter 6 (d) Find P(300 < X < 400) . This gives the probability that a randomly chosen student will spend between $300 and $400 on textbooks in September. The range probability is calculated as: P(300 < X < 400) = P(X < 400) − P(X < 300) A graph gives a helpful picture of the calculations. 2 PDF of X ~ N( 380 , 50 ) f(x) P(300 < X < 400) P(X < 300) 300 380 x 400 22 Chapter 6 From the previous calculations: P( X < 400) = 0.6554 Now find...
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