The specified (required) mean and max (90th percentile) elapsed time to repair a
ID: 3135397 • Letter: T
Question
The specified (required) mean and max (90th percentile) elapsed time to repair an item is 3.2 hours and 7.4 hours, respectively. The time to repair probability distribution is not specified. Evaluate this requirement in terms of the mean, median and max (90th, 95th and 99th percentiles) elapsed time to repair using the Exponential, Normal and Lognormal The specified (required) mean and max (90th percentile) elapsed time to repair an item is 3.2 hours and 7.4 hours, respectively. The time to repair probability distribution is not specified. Evaluate this requirement in terms of the mean, median and max (90th, 95th and 99th percentiles) elapsed time to repair using the Exponential, Normal and Lognormal The specified (required) mean and max (90th percentile) elapsed time to repair an item is 3.2 hours and 7.4 hours, respectively. The time to repair probability distribution is not specified. Evaluate this requirement in terms of the mean, median and max (90th, 95th and 99th percentiles) elapsed time to repair using the Exponential, Normal and LognormalExplanation / Answer
Mean = 3.2 and 90th percentile = 7.4 hours
I) Exponential
90th percentile is c such that
0.9 = 1-e-c/3.2
c = 7.368 =7.4
Mean = 3.2
Median = c' such that 0.5 = 1-e-c/3.2
=2.2181
95th percentile = 9.5863
99th percentile = -14.7365
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II. Normal
Mean = 3.2
90th percentile =1.28 in z
Hence corresponding X = 3.2+1.28(std dev) = 7.4
Std dev = 3.28125
From this 95th percentile = 3.2+1.645(3.28125)
= 8.5977
99th percentile = 3.2+2.33(3.28125)
= 10.8454
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Lognormal:
Y = lnx is normal
Mean of ln x =3.2
E(X) = 24.5244
Median = Mean = 24.5244
Max = 24.5244
Var(Y) = Var( ln x) =
90th percentile =
II) NOrmal
III) log normal
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