Answer to Question #201852 in Statistics and Probability for Ahmad

Question

a continuous random variable T has the following probability density function.

f(t) = { 32t²/128, -4≤t<4

0, otherwise

a. find E(t)

b. find Var(T)

c. Var(2T+3)

Answer 1

Solution:

$f(t) = \{ \dfrac{32t^2}{128}, -4≤t<4 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0, otherwise$

(a):

$E(t)=\int_{-4}^4t(\dfrac{32t^2}{128})=\int_{-4}^4(\dfrac{t^3}{4}) \\=[\dfrac{t^4}{16}]_{-4}^4 \\=\dfrac{4^4}{16}-\dfrac{(-4)^4}{16} \\=0$

(b):

$E(t^2)=\int_{-4}^4(\dfrac{t^4}{4}) \\=[\dfrac{t^5}{20}]_{-4}^4 \\=\dfrac{4^5}{20}-\dfrac{(-4)^5}{20} \\=2(\dfrac{4^5}{20})=102.4$

Now, $Var[T]=\sigma^2=E(X^2)-[E(X)]^2=102.4-0^2=102.4$

(c):

$Var[2T+3]=4\ Var[T]=4(102.4)=409.6$