A random variable X has the following probability distribution

Then the mean of X is :

0.8

Given probability distribution:

$x_i:\ -2,\ -1,\ 0,\ 1,\ 2,\ 3$

$p_i:\ 0.1,\ k,\ 0.2,\ 2k,\ 0.3,\ k$

Sum of probabilities is $1$:

$0.1+k+0.2+2k+0.3+k=1$

$0.6+4k=1$

$4k=0.4 \Rightarrow k=0.1$

Mean of $X$:

$E(X)=\sum x_ip_i$

$=(-2)(0.1)+(-1)(0.1)+(0)(0.2)+(1)(0.2)+(2)(0.3)+(3)(0.1)$

$=-0.2-0.1+0+0.2+0.6+0.3$

$=0.8$

final answer: mean of $X$ is $0.8$