The specified (required) mean and max (90th percentile) elapsed time to repair a

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 Lognormal

Explanation / 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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