The probability distribution of the random variable X is given by X 0…

The correct answer is Option (1) → $\frac{764}{625}$

Given probability distribution

$X:\;0,\;1,\;2,\;3$

$P(X=0)=0.2,\;P(X=1)=k,\;P(X=2)=2k,\;P(X=3)=2k$

Using total probability equal to $1$

$0.2+k+2k+2k=1$

$0.2+5k=1$

$5k=0.8$

$k=0.16$

Mean

$E(X)=0(0.2)+1(k)+2(2k)+3(2k)$

$=k+4k+6k$

$=11k$

$E(X)=11(0.16)=1.76$

Second moment about origin

$E(X^2)=0^2(0.2)+1^2(k)+2^2(2k)+3^2(2k)$

$=k+8k+18k$

$=27k$

$E(X^2)=27(0.16)=4.32$

Variance

$\text{Var}(X)=E(X^2)-[E(X)]^2$

$=4.32-(1.76)^2$

$=4.32-3.0976$

$=1.2224$

The variance of the random variable $X$ is $1.2224$.