The probability distribution of a discrete random variable X is defined as :

$P(X=x)=\left\{\begin{matrix}3kx & \text{for x = 1, 2, 3}\\5k(x+2) & \text{for x = 4, 5}\\0 & otherwise \end{matrix}\right.$

The mean of the distribution is :

$\frac{413}{113}$

The correct answer is Option (2) → $\frac{413}{113}$

Sum of all probabilities must be 1.

$3k+5k(2+2)+5k(3+2)+5k(4+2)+5k(5+2)=1$

$⇒k=\frac{1}{113}$

$E(X)=∑xP(X=x)$

$=1×3×\frac{1}{113}+2×20×\frac{1}{113}+3×25×\frac{1}{113}+4×30×\frac{1}{113}+5×35×\frac{1}{113}$

$=\frac{413}{113}$