HW 8 (Sect 60.2) Req Solutions

# HW 8 (Sect 60.2) Req Solutions - Problem Solutions for...

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Problem Solutions for Required Homework #8 for ASW ESBE (5e) Chapter 6, Section 6.2 Required Homework #8: Problems 45 and 47, in the end-of chapter Supplementary Exercises on pages 251 to 252. Problem 6.45. Let x be the continuous r.v. of the number of fatal crashes per year. Given: μ = 1550 accidents/year σ = 300 accidents/year x is normal. a. Want prob( x < 1000) For x = 1000, 1000 1550 1.83 300 x z μ σ - - = = = - From negative z table, down the z column to -1.8, across the row to under the column for .03; the table entry for -1.83 is 0.0336. prob( x < 1000) = prob( z < -1.83) (by equivalence) = 0.0336 b. Want prob(1000 < x < 2000) For x = 2000 2000 1550 1.50 300 z - = = At x = 1000, z = -1.83, as already calculated in part a above. Then prob(1000 < x < 2000) = prob(-1.83 < z <1.50) (by equivalence) = prob ( z <1.50) – prob( z < -1.83) ASW 5e Req HW #8 for section 6.2 1

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= 0.9332 – 0.0336 = 0.8996 c. Find the x value cutting off 5% or 0.05 of the total area in the upper tail. In other words, the upper tail area is 0.05 for what x value? Denote this special x value x *. The procedure is to calculate the area to the right of the mean, find the
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## This note was uploaded on 02/19/2011 for the course OM 210 taught by Professor Singer during the Spring '08 term at George Mason.

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HW 8 (Sect 60.2) Req Solutions - Problem Solutions for...

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