Question

Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Answer 1

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.
P(X=2)=P(1,1)=$\frac{1}{36}$

P(x=3) = p(1,2)+p(2,1)=$\frac{2}{36}$=$\frac{1}{18}$

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = $\frac{3}{36}=\frac{1}{12}$

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) =$\frac{4}{36}=\frac{1}{9}$

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = $\frac{5}{36}$

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = $\frac{6}{36}=\frac{1}{6}$

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = $\frac{5}{36}$

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = $\frac{4}{36}=\frac{1}{9}$

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) =$\frac{3}{36}=\frac{1}{12}$

P(X = 11) = P(5, 6) + P(6, 5) = $\frac{2}{36}$=$\frac{1}{18}$

P(X = 12) = P(6, 6) = $\frac{1}{36}$

x	2	3	4	5	6	7	8	9	10	11	12
p(x)	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

Then, E(X)=$\sum X_i.P(X_i)$

=2 × $\frac{1}{36}$+3× $\frac{1}{18}$+4 × $\frac{1}{12}$+5 × $\frac{1}{9}$+6 × $\frac{5}{36}$+7× $\frac{1}{6}$+8× $\frac{5}{36}$+9×$\frac{1}{9}$+10 ×$\frac{1}{12}$+11×$\frac{1}{18}$+12×$\frac{1}{36}$

= 7

$E(X^2)= \sum X^2_i.P(X_i)$

= 4×$\frac{1}{36}$+9× $\frac{1}{18}$+16 × $\frac{1}{12}$+25 × $\frac{1}{9}$+36 × $\frac{5}{36}$+49× $\frac{1}{6}$+64× $\frac{5}{36}$+81×$\frac{1}{9}$+100 ×$\frac{1}{12}$+121×$\frac{1}{18}$+144×$\frac{1}{36}$

= $\frac{987}{18}$ = 54.833

Then Var(X) = $E(X^2)-[E(X)]^2$

=54.833-$(7)^2$

= $54.833-49$

=$5.833$

Therefore Standard deviation

= $\sqrt{Var (X)}$

= $\sqrt{5.833}$

= 2.415