var(X) in exponential distribution is 1/(parameter)2 i.e 1/λ2 here so option B
var(3X+5)
⇒3var(X) + var(5)
⇒3*1/λ2 + 0 as var(constant)=0;var(X)=1/λ2
⇒3/λ2
so option B
now you can evaluate the integral to arrive at desired result.you can find this proof in any standard book for variance the formula is var(X) =E[x2]-(E[x])2
var[3x+5]=var[3x]+var[5] =3^2var[x]+0 = 9/λ2 I think this should be answer.
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