Question

Let A be a continuous random variable with probability density function

f_{A}(a) = \frac{1}{9}a^{2} \ \ \ \ \ \ \ 0\leq A\leq 3

Random variable D is given by

D = 1 + 3A

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(a) What is the probability density function of D? specify the domain of D.

Answer is f_{D}(d) = \frac{(D-1)^{2}}{243} \ \ \ \ \ \ \ \ \ \ if \ \ 1\leq D\leq 10

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(b) Find E(D) and Var(D).

Answer \ is \ E(D) = \frac{31}{4}, Var(D) = \frac{243}{80}

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