Answer to Question #165062 in Statistics and Probability for Fredrick

Question

(a) Let X be a random variable with the following probability distribution x -3 6 9 f(x) 1 6 1 2 1 3 Find the mean and variance of Y = 3X + 2. (b) If X ∼ Bin(6, 1 3 ) and Y ∼ N( 1 2 , 1 4 ) are independent random variables, find the mean and variance of the random variable W = 2X − 4Y + 5.

Answer 1

Solution

a).

$$E(y) = 3E(x)+2$$ $$E(x) =(3*{1 \over 6}) + (6*{1 \over 2})+(9*{1 \over 3})$$

$=6.5$

$$E(y) = 3(6.5)+2$$

Mean =21.5

$$Var(y) = 3^2 var(x) = 9var(x)$$

$$var(x) = \sum x^2 P(X=x) - {(E(x))} ^2$$ $$=(3^2*{1 \over 6}) +(6^2*{1\over 2})+(9^2*{1 \over 3})- 6.5^2$$

$var(x) =4.25$

$$var(y) = 9(4.25) = 38.25$$

b)

$$E(W) = 2E(X)-4E(Y)+5$$ $$E(X) = np = 6({1 \over 3}) =2$$ $$E(Y) ={1 \over 2}$$

$$E(W) =(2*2)- 4({1\over 2})+5$$

E(W) = 7

$$var(w) = 2^2 var(x) + 4^2 var(y) - 2(2*4) cov(xy)$$

But $Cov(xy) =0$

$$\therefore var(w) =4var(x)+16var(y)$$

$$Var(x) =np(1-p) =6*{1 \over 3}*{2 \over 3}={4 \over 3}$$

$Var(y) ={1\over 4}$

$$Var(W) =4({4 \over 3}) +16({1 \over 4})$$

Var(W) =9.33333333