MECH215-Chapter04-SolutionstoSampleProblems

# MECH215-Chapter04-SolutionstoSampleProblems - PROBLEM 4.2...

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4-1 PROBLEM 4.2 KNOWN: N > 10 000; x = 204 units ; S x = 18 units FIND: x' - z 1 σ x x' + z 1 σ at P = 90% ASSUMPTIONS Measurand follows a normal density function Data set sufficiently large such that x x’ and S x σ SOLUTION Using the definition z 1 = (x 1 - x')/ σ we find the z 1 value corresponding to the one-sided probability integral p(z 1 ) = 0.45 from Table 4.3. This gives, z 1 = 1.65 Then, 1.65 = (x 1 - x')/ σ or 1.65 σ = x 1 - x' or x 1 = x' + 1.65 σ But a normal (Gaussian) distribution is symmetric about the mean value. Hence, for 90% probability (2)(0.45), x' - 1.65 σ x x' + 1.65 σ (90%) or 174.3 x 233.7 units (90%)

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4-2 PROBLEM 4.3 KNOWN: x = 121.6 psi S x = 14 psi N is very large FIND: P(x > 150 psi) ASSUMPTIONS: Normal distribution S x σ ; x x' SOLUTION The z variable is defined by z 1 = -(x 1 - x')/ σ = -(121.6 -150)/14 = 2.028 We look up P(2.028) from Table 4.3. Interpolation gives P(2.028) = 0.4786 This expresses the probability that 121.6 x 150 psi. Then, the probability that x > 150 psi is 0.5 - 0.4786 = 0.0214 or there is a 2.14% probability that any measurement will yield a value in excess of 150 psi.
4-3 PROBLEM 4.9 KNOWN: Three datasets from Table 4.8 Column 1,2, and 3 N = 20 FIND: F , S F , and F S SOLUTION The mean value for each dataset is 1 F = 50.5 N

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## This note was uploaded on 10/31/2009 for the course MECH 420 taught by Professor M.s during the Spring '09 term at American University of Beirut.

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MECH215-Chapter04-SolutionstoSampleProblems - PROBLEM 4.2...

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