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 :

$\frac{337}{83}$

The correct answer is Option (2) →$\frac{337}{83}$

Sum of all probabilities must be 1.

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

$⇒k=\frac{1}{83}$

$E(X)=∑xP(X=x)$

=1(3k)+2(6k)+3(9k)+4(30k)+5(35k)

=3k+12k+27k+120k+175k

= 337k

Substituting $⇒k=\frac{1}{83}$

$E(X)=\frac{337}{83}$