Isye 2027

# Since x has mean zero the variance varx for a standard

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Unformatted text preview: and mean of T. By deﬁnition of the exponential distribution with parameter λ = 0.01, the pdf of T is fT (u) = (0.01)e−(0.01)u for u ≥ 0 0 for u < 0, where the variable u is a measure of time in seconds. The mean of an exponentially distributed 1 random variable with parameter λ is λ , so E [T ] = 0.1 = 100. 01 [T T The waiting time in minutes is given by S = 60 . By the linearity of expectation, E [S ] = E60 ] = 100 60 = 1.66 . . . . That is, the mean waiting time is 100 seconds, or 1.666 . . . minutes. By the scaling formula with a = 1/60 and b = 0, fS (v ) = fT (60v )60 = (60)(0.01)e−(0.01)60v = (0.6)e−(0.6)v for v ≥ 0 0 for v < 0, where v is a measure of time in minutes. Examining this pdf shows that S is exponentially distributed with parameter 0.60. From this fact, we can ﬁnd the mean of S a second way–it is one over the parameter in the exponential distribution for S, namely 116 = 1.666 . . . , as already noted. . Example 3.6.3 Let X be a uniformly distributed random variable on some interval [a, b]. Find −µ t...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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