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Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition by Douglas C. Montgomery, George C. Runger

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CHAPTER 3

Section 3-1

3-1.      The range of X is {0,1,2,...,1000}

3-3.      The range of X is {0,1,2,...,99999}

3-5.      The range of X is {1,2,...,491}. Because 490 parts are conforming, a nonconforming part must be selected in 491 selections.

3-7.      The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0,1,2,...}

3-9.      The range of X is {0,1,2,...,15}

3-11.    The range of X is {0,1,2,…,10000}

3-13.    The range of X is {0,1,2,…,40000}

Section 3-2

3-15.    All probabilities are greater than or equal to zero and sum to one.

a)   P(X ≤ 2) = 1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1

b)   P(X > – 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8

c)   P(–1 ≤ X ≤ 1) = 2/8 + 2/8 + 2/8 = 6/8 = 3/4

d)   P(X ≤ –1 or X = 2) = 1/8 + 2/8 + 1/8 = 4/8 = 1/2

3-17.    Probabilities are nonnegative and sum to one.

a)   P(X = 4) = 9/25

b)   P(X ≤ 1) = 1/25 + 3/25 = 4/25

c)   P(2 ≤ X < 4) = 5/25 + 7/25 = 12/25

d)   P(X > –10) = 1

3-19.    X = number of successful surgeries.

P(X = 0) = 0.1(0.33) = 0.033

P(X = 1) = 0.9(0.33) + 0.1(0.67) = 0.364

P(X = 2) = 0.9(0.67) = 0.603

3-21.    X = number of wafers that pass

P(X = 0) = (0.2)3 = 0.008

P(X = 1) = 3(0.2)2(0.8) = 0.096

P(X = 2) = 3(0.2)(0.8)2 = 0.384

P(X = 3) = (0.8)3 = 0.512

3-23.    P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2

3-25.    P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = –0.5 million) = 0.1

3-27.    X = number of components that meet specifications

P(X = 0) = (0.05)(0.02)(0.01) ...

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