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2008hw3_solutions

# 2008hw3_solutions - ST 314 HW#3 Solutions 1 Text p 133...

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ST 314 HW #3 Solutions 1. Text, p. 133, exercise 3.40. The r.v. Y is distributed exponential with λ = 0.01. a. P(Y < 30) = 1 – e - λ (30) = 1 – e -.01(30) = 1- e -.30 = .2592 b. P(Y > 15) = 1 – [1 – e - λ (15) ] = e -.15 = .8607 c. P(Y = 100) = 0 d. P(50 Y 150) = [1 – e - λ (150) ] – [1 – e - λ (50) ] = -e -1.5 + e -.5 = .3834 2. Text, p. 141, exercise 3.47. Let Y = thickness of bolts. Then Y is normally distributed with mean μ = 10.0 mm and standard deviation σ = 1.6 mm. a. P(9.2 Y 10.8) = P( (9.2-10)/1.6 (Y- μ )/ σ (10.8-10)/1.6 ) = P (-.5 Z .5) = P(Z < .5) – P(Z < -.5) = .6915 – .3085 = .3830 b. P(Y 9.2) = P ((Y- μ )/ σ (9.2-10)/1.6) = P(Z -.5) = .3085 3. Text, p. 142, exercise 3.53. Let Y = volume percent. Then Y is normally distributed with μ = 70 and σ = 2. a. P(Y > 75) = P((Y- μ )/ σ > (75-70)/2) = P(Z > 2.5) = 1 - .9938 = .0062 b. P(67 < Y < 73) = P((67-70)/2 < Z < (73-70)/2) = P(-1.5 < Z < 1.5) = P(Z < 1.5) – P(Z < -1.5) = .9332 – 0.668 = 0.8664 c. We want to find μ so that P(Y < 70) < .02. But P(Y < 70) = P((Y- μ )/ σ < (70- μ )/ σ )) = P(Z < (70- μ )/2) and from Table 1 we see (70- μ )/2 = -2.06 so that μ = 70 – 2(-2.06) = 74.12 4. Text, p. 152, exercise 3.56. Let Y = concentricity of an engine oil seal groove.

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