The probability distribution of the random variable X is given by
Then, the value of V(X) is equal to

Options

A. 0
B. 1
C. `1/2`
D. `2/3`

Answer

B

Explanation

`E(X)=Sigmax_iP(x_i)=1/8(1)+1/2(2)+1/8(3)+1/4(4)`
`=1/8+1+3/8+1=5/2`
Now, `V(X)=E(X^2)-[E(X)]^2`
`=Sigmax_i^2.P(x_i)-(5/2)^2`
`=1/8(1)^2+1/2(2)^2+1/8(3)^2+1/4(4)^2-(25)/4`
`=1/8+2+9/8+4-(25)/4`
`=(1+16+9+32-50)/8=(58-50)/8=8/8=1`