show work Chapter04 sec4-5 HW: Problem 6 Previous Problem List Next (2 points) A
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Chapter04 sec4-5 HW: Problem 6 Previous Problem List Next (2 points) According to the article " Predictive Model for Pitting Corrosion in Buried Oil and Gas Pipelines" (Corrosion, 2009: 332-342), the lognormal distribution has been reported as the best option for describing the distribution of maximum pit depth data from cast iron pipes in soil. The authors suggest that a lognormal distribution with 0.353 and = 0.765 is appropriate for maximum pit depth ( mm) of buried pipelines. For this distribution, the mean value and variance or pit depth are a. E(X)1.598574 b. Varx)2.402356 I c.The probability that maximum pit depth is between 1 and 3 is 5068 d. What value c is such that only 1% of all specimens have a maximum pit depth exceeding c? c 8.247 Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 4 times. Your overall recorded score is 25% You have unlimited attempts remainingExplanation / Answer
Here = 0.353
= 0.765
(a) Here mean value E(X) = EXP ( + 2/2) = EXP( 0.353 + 0.7652/2) = 1.9072
Var(X) = [Exp (2) - 1] Exp (2 + 2) = [exp (0.7652 ) - 1] * exp[2 * 0.353 + 0.7652]
Var(X) = 0.6605 * 3.6372 = 2.4024
(c) Pr (1 < X < 3) = Pr(X < 3) - Pr(X < 1) = 1/2 [ erf [(ln X2 - )/sqrt(22) - erf [(ln X1 - )/sqrt(22)]
Pr(1 < X < 3) = 1/2 [ erf [(ln 3 - 0.353)/sqrt(2 * 0.7652) - erf [(ln 1 - 0.353)/sqrt(2 * 0.7652)]
Pr(1 < X < 3) = 1/2 [erf (0.6892) - erf (-0.3263)]
Pr(1 < X <3) = 1/2 [ 0.6703 + 0.3555] = 0.5129
(d) Here the value is X0 above which there are only 1% of all specimens have maximum depth.
Pr(X < X0 ) = 0.01
1/2 + 1/2 erf [ (ln X - )/ sqrt(22) ] = 0.99
1/2 erf [ (ln X - )/ sqrt(22) ] = 0.49
erf [ (ln X - )/ sqrt(22) ] = 0.98
(ln X - )/ sqrt(22) = erf-1 (0.98) = 1.645
(ln X - ) = 1.645 * sqrt (2 * 0.7652 ) = 1.7797
ln X = 1.7797 + 0.353 = 2.1327
X = 8.4376
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