The probability distribution of a random variable X is given as:

The mean of the distribution is:

$\frac{14}{5}$

The correct answer is Option (1) → $\frac{14}{5}$

$\sum P(X=x)=1 \Rightarrow k+2k+4k+6k+7k=20k=1$

$k=\frac{1}{20}$

$E(X)=\sum xP(X)$

$=0\cdot k+1\cdot2k+2\cdot4k+3\cdot6k+4\cdot7k$

$=2k+8k+18k+28k=56k$

$=56\cdot\frac{1}{20}=\frac{14}{5}$

The mean is $\frac{14}{5}$.