# 6554 1 09452 060 23 chapter 6

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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...
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## 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.

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