What is the variance of a probability distribution function of the form: #f(x)=ke^(-2x)#?

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What is the variance of a probability distribution function of the form: #f(x)=ke^(-2x)#?

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Answer 1 · 2015-10-27T06:54:58

The distribution is an exponential distribution. k = 2 and E (x) = 1/2 ,
E ( #x^2# )= 1/2 #=># V (x) = E ( #x^2#) - #{E (x)}^2# - 1/2 - #(1/2) ^2# = 1/2 - 1/4 = 1/4.

Explanation:

The limit of the distribution is (0, #oo# ) To find k,
#int_0^B# k #e^- (2x)# dx = k #Gamma# (1)/ 2 = 1 #=># k/2 = 1 #=># k = 2.
E (x) = #int_0^Bx

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What is the variance of a probability distribution function of the form: #f(x)=ke^(-2x)#?

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Explanation:

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