If m and `sigma^2` are the mean and variance of the random variable X whose distribution is given by

Options

A. `m=sigma^2=2`
B. `m=1,sigma^2=2`
C. `m=sigma^2=1`
D. `m=2,sigma^2=1`

Answer

C

Explanation

Given distribution is
`therefore" Mean",m=sum_(i=1)^4p_ix_i=0xx1/3+1xx1/2+2xx0+3xx1/6`
`=0+1/2+01/2=1`
Variance , `sigma^2=sum_(i=1)^4p_i(x_i-m)^2`
`=1/3(0-1)^2+1/2(1-1)^2+0(2-1)^2+1/6(3-1)^2`
`=1/3+0+0+2/3=1`
`therefore" "m=sigma^2=1`