A random variable X has the following probability distribution:

The mean of the distribution is :

$\frac{39}{16}$

The correct answer is Option (4) → $\frac{39}{16}$

Sum of all probabilities is 1.

$⇒2k+k+4k+6k+3k=1$

$⇒16k=1$

$⇒k=\frac{1}{16}$

Mean of the random variable X.

$E(X)=∑XP(X)$

$E(X)=(0×2k)+(1×k)+(2×4k)+(3×6k)+(4×3k)$

$=39k=39×\frac{1}{16}=\frac{39}{16}$