A random variable X has the following probability distribution:

then, which of the following is correct?

mean = 2 variance

The correct answer is Option (3) → mean = 2 variance **

Given distribution:

X = 0, 1, 2

P(X) = \frac{1}{4}, \frac{1}{2}, \frac{1}{4}

Mean:

$E(X)=0\cdot\frac14+1\cdot\frac12+2\cdot\frac14=\frac12+ \frac12 =1$

$E(X^{2})=0^{2}\cdot\frac14+1^{2}\cdot\frac12+2^{2}\cdot\frac14=\frac12+1=\frac32$

Variance:

$V(X)=E(X^{2})-[E(X)]^{2}=\frac32-1=\frac12$

Hence:

Mean $=1$, Variance $=\frac12$

Therefore:

Mean = 2 × Variance

i.e., $1 = 2\cdot\frac12$

Correct option: mean = 2 variance