A random variable X has the following probability distribution:

The variance of the X will be:

1

The correct answer is Option (1) → 1

Given distribution

$X:0,1,2,3$

$P(X):0.1,0.2,0.3,0.4$

Mean

$E(X)=0(0.1)+1(0.2)+2(0.3)+3(0.4)$

$=0+0.2+0.6+1.2=2$

$E(X^2)=0^2(0.1)+1^2(0.2)+2^2(0.3)+3^2(0.4)$

$=0+0.2+1.2+3.6=5$

Variance

$V(X)=E(X^2)-[E(X)]^2$

$=5-4=1$

The variance of $X$ is $1$.