Question

A discrete random variable x ha s the probability distribution f(m)

f(m)={m/36,m=1,2,3,......8

{ 0,elsewhere

Find the mean and variance of M

Answer 1

$$\begin{matrix} m & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\\ p(m) & \dfrac{1}{36} & \dfrac{2}{36} & \dfrac{3}{36} & \dfrac{4}{36} & \dfrac{5}{36} & \dfrac{6}{36} & \dfrac{7}{36} & \dfrac{8}{36} \end{matrix}$$

$$mean=E(m)=1(\dfrac{1}{36})+2(\dfrac{2}{36})+3(\dfrac{3}{36})+4(\dfrac{4}{36})$$

$$+5(\dfrac{5}{36})+6(\dfrac{6}{36})+7(\dfrac{7}{36})+8(\dfrac{8}{36})=\dfrac{17}{3}$$

$$E(m^2)=1^2(\dfrac{1}{36})+2^2(\dfrac{2}{36})+3^2(\dfrac{3}{36})+4^2(\dfrac{4}{36})$$

$$+5^2(\dfrac{5}{36})+6^2(\dfrac{6}{36})+7^2(\dfrac{7}{36})+8^2(\dfrac{8}{36})=36$$

$$Var(m)=\sigma^2=E(m^2)-(E(m))^2$$

$$=36-(\dfrac{17}{3})^2=\dfrac{35}{9}$$