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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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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?

Step-by-Step Solution:
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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 - 9x = 0.75 x 0.75 -9x = x9 10 9 -9x(x ) = -9.25 9x = -9.25 10 9.25 x = 9 9.25 10 x = ( 9 ) Therefore, x = 1.0027. Therefore,the 75th percentile in the distribution is 1.0027. Step6: e). To compute E(x) and x. E(x) = xf(x) dx 1 = x90 x (1 x) dx 0 1 = 90 (1 x)dx 0 1 1 = 90 dx x dx 0 0 x10 1 x11 1 = 90[ 10] 0 [ 11 ]0 110 010 110 010 = 90 [ 10 10 ] [11 11] = 90[ 1 1 ] 10 11 1110 = 90[ 110 ] = 0.8182 E(x) = 0.8182. Then, 2 2 E(x) = x f(x) dx 1 2 8 = x 90 x (1 x) dx 0 1 10 = 90 (1 x)dx 0 11 12 = 90[ x x ] 1 11 12 0 90 = 12×11 = 0.68. 2 Therefore, V(x) = = E(x )- [E(x)] = 0.68 - (0.8182) 2 = 0.012. Standard deviation = 0.012 = 0.1095. Therefore, E(x) = 0.8182 and = 0.1095. Step7: e). To find the probability that X is more than 1 standard deviation from its mean value. That is, ± 1 0.8182±0.1095 (0.9277,0.7087) Therefore , (0.7087,0.9277) Therefore, P(- < X< + ) = F(0.9277) - F(0.7087) = 10(0.927) 9(0.709) 10(0.709) + 9(0.709) 10 = 0.69 P(- < X< + ) = 0.69. Therefore,the probability that X is more than 1 standard deviation from its mean value is 0.69.

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Let denote the amount of space occupied by an article