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f(x) = 1/3 e^(-x/3) for x > 0

Find the variance of Y.

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I found the expected value of X to be 14 minutes, which was correct. This means the variance for X will be: E(X^2) - E(X) = 18 - 14 = 4. Will the following work for the variance?

Var(Y) = (4^2)(4) = 64.

If so, what does that mean to have a "negative" take off time?

Thanks for your help!! :)