(a) Calculate the median of an Exponential(λ) random variable.
m - median value
"F(m) = P(X \\le m) = 1 - e^{-\\lambda m} = 1\/2"
From here
"m = ln(2)\/\\lambda"
(b) IfX∼Exponential(λ), what isP(X≤E(X))?
"P(X\\le E(x)) = P(X \\le 1\/\\lambda) = 1-e^{-\\lambda * 1\/ \\lambda} = 1-e^{-1}=1-1\/e = 0.63"
(c) My opinion that it is possible. Because the average value is not always equal to median
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