Question

The P.M.F of a r.v. $X$ is $P\left(x\right)=\frac{1}{5}$ for $x=1,2,3,\dots 14,15;$ $P\left(x\right)=0$ otherwise.
Find (i) $E\left(X\right)$ (ii) $Var\left(X\right)$

Answer 1

i)E(X)=$\sum xP(X=x)=1\ast \frac{1}{5}+2\ast \frac{1}{5}+3\ast \frac{1}{5}...+15\ast \frac{1}{5}=\frac{15\ast 16}{2\ast 5}=24$

ii)Var(X)=$=E\left(X^2\right)-{E\left(X\right)}^2$

$\Rightarrow E\left(X^2\right)=\sum x^{2}P(X=x)=1^2\ast \frac{1}{5}+2^2\ast \frac{1}{5}+3^2\ast \frac{1}{5}...+{15}^2\ast \frac{1}{5}=\frac{15\ast 16\ast 31}{6\ast 5}=31\ast 8=248$

Var(X)=248-576=-328