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