For the probability distribution of a discrete random variable X as given below, the mean of X is
For the probability distribution of a discrete random variable X as given below, the mean of X is
$\frac{3}{5}$
$\frac{4}{5}$
$\frac{6}{5}$
$\frac{8}{5}$
Solution
$\sum_{x=-2}^3 P(X=x)=1$
$\Rightarrow \frac{1}{10}+k+\frac{2}{10}+k+\frac{3}{10}+k+\frac{3}{10}+k+\frac{4}{10}+k+\frac{2}{10}=1$
$\Rightarrow 5 k+\frac{15}{10}=1 \Rightarrow k=\frac{-1}{10}$
So, probability distribution becomes
$\operatorname{Mean}(\mu)=\frac{-2}{10}+\frac{-1}{10}+\frac{2}{10}+\frac{6}{10}+\frac{3}{10}=\frac{4}{5}$
