Answer to Question #330176 in Statistics and Probability for Robera

Question

Two dice are rolled. Let X be a random variable denoting the sum of the numbers

on the two dice.

i) Give the probability distribution of X

ii) Compute the expected value of X and its variance

Answer 1

i)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hdashline 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hdashline 3 &4 & 5 & 6 & 7 & 8 & 9 \\ \hdashline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hdashline 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hdashline 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 2 & 1/36 \\ \hdashline 3 & 1/18 \\ \hdashline 4 & 1/12 \\ \hdashline 5 & 1/9 \\ \hdashline 6 & 5/36 \\ \hdashline 7 & 1/6 \\ \hdashline 8 & 5/36 \\ \hdashline 9 & 1/36 \\ \hdashline 10 & 1/12 \\ \hdashline 11 & 1/18 \\ \hdashline 12 & 1/36 \\ \hdashline \end{array}$$

ii)

$$E(X)=2(\dfrac{1}{36})+3(\dfrac{2}{36})+4(\dfrac{3}{36})+5(\dfrac{4}{36})$$

$$+6(\dfrac{5}{36})+7(\dfrac{6}{36})+8(\dfrac{5}{36})+9(\dfrac{4}{36})$$

$$+10(\dfrac{3}{36})+11(\dfrac{2}{36})+12(\dfrac{1}{36})=7$$

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

$$=2^2(\dfrac{1}{36})+3^2(\dfrac{2}{36})+4^2(\dfrac{3}{36})+5^2(\dfrac{4}{36})$$

$$+6^2(\dfrac{5}{36})+7^2(\dfrac{6}{36})+8^2(\dfrac{5}{36})+9^2(\dfrac{4}{36})$$

$$+10^2(\dfrac{3}{36})+11^2(\dfrac{2}{36})+12^2(\dfrac{1}{36})-7^2$$

$$=\dfrac{35}{6}$$