A cubical die is thrown. Find the mean and variance of $X$, giving the number on the face that shows up.

Let $S$ be the sample space
$X$ be the number on the face that shows up
when a cubical die is thrown

mean $\mu =1$

$P(x=1)+2P(x=2)+3P(x=3)+4P(x=4)+5P(x=5)+6P(x=6)$

$=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right)$

$=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}$

$\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}\Rightarrow \mu =\frac{7}{2}$

variance ${\sigma }^2=\sum x_i^{2}P\left(X=x_i\right)-{\mu }^2$

$$\begin{gathered} =1^{2}P(x=1)+2^{2}P(x=2)+3^{2}P(x=3)+4^{2}P(x=4) \\ +5^2(x=5)+6^{2}P(x=6)-{\left(\frac{7}{2}\right)}^2 \end{gathered}$$

$=1\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+9\left(\frac{1}{6}\right)+16\left(\frac{1}{6}\right)+25\left(\frac{1}{6}\right)+36\left(\frac{1}{6}\right)-\frac{49}{4}$

$=\frac{1+4+9+16+25+36}{6}-\frac{49}{4}$

$\frac{91}{6}-\frac{49}{4}=\frac{182-147}{12}=\frac{35}{12}$