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# Let denote the amount of space occupied by an article ISBN: 9780321629111 32

## Solution for problem 15E Chapter 4

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 15E

Let ? ?denote the amount of space occupied by an article placed in a 1-1ft3 packing container. The pdf of ?X i? s a. ? raph the pdf. Then obtain the cdf of X? ? nd graph it. b. ?What is P(X?.5) c. ?Using the cdf from (a), what is P(.25 < X ? .5)? What is P(.25 ? X ? .5)? d. ? hat is the 75th percentile of the distribution? e. ?Compute E? ? ? ? and ? x . f. ?What is the probability that ?X ?is more than 1 standard deviation from its mean value?

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Answer: Step1: Let denote the amount of space. Step2: a). To obtain the cdf of X . F(x) = 0 for x 0 F(x) = 0 for x = 1. Therefore, x F(x) = f(x) dx x 8 = 90x (1 x) dx x F(x) = 90 x (1 x) dx 0 x = 90 x x )dx9 0 x8+1 x9+1 x = 90[ 8+1 - 9+1 )0 x9 x10 x = 90[ 9 10 ]0 10x 9x0 = 90[ 90 ] F(x) = 10x 9x 10 Step3: b). To find P(x 0.5) P(x 0.5)= F(x) = 10x 9x 10 = F(0.5) = 10(0.5) 9(0.5) 10 = 0.01953-0.0088 = 0.0108 Therefore, P(x 0.5) = 0.0108. Step4: c). i) To find P(0.25 < X 0.5). P(0.25 < X 0.5) = P(X > 0.25) P(X 0.5) = [1-P(x <0.25)] - P(X 0.5) = (1-F(0.25)) - F(0.5) 9 10 F(0.5) 10(0.5) 9(0.5) 0.0108 F(0.25) 10(0.25) 9(0.25) 10 0.000029 Then, (1-F(0.25)) - F(0.5) = [ 1-0.000029] - 0.0108 = 0.9892. Therefore, P(0.25 < X 0.5) = 0.9892. ii) Then, to find P(0.25 X 0.5). P(0.25 X 0.5) = P(X 0.5) P(X 0.25) = F(0.5) - F(0.25) = 0.0108 - 0.000029 = 0.01078 Therefore, P(0.25 X 0.5) = 0.01078. Step5: d). To find the 75th percentile in the distribution. That is, P(x x ) = F(x ) = 0.75 0.75 0.75 10x 9 9x 10 = 0.75 0.75 0.75 10x 9x = 0.75 9 x ( 10 - 9x) = 0.75 10...

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##### ISBN: 9780321629111

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