NCERT Solutions
Class 12 Maths
Probability

Ex.13.4 Q.13
Let X denotes the sum of the numbers obtained when two fair dice are rolled.
Find the variance and standard deviation of X.
When two fair dice are rolled, 6 * 6 = 36 observations are obtained.
P(X = 2) = P(1, 1) = 136
P(X = 3) = P (1, 2) + P(2, 1) = 236=118
P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 336=112
P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 436=119
P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 536
P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 636=116
P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 536
P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 436 = 19
P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 336 = 112
P(X = 11) = P(5, 6) + P(6, 5) = 236 = 118
P(X = 12) = P(6, 6) = 136
Therefore, the required probability distribution is as follows:
|
X |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
|
P(X) |
136 |
118 |
112 |
19 |
536 |
16 |
536 |
19 |
112 |
118 |
136 |
Then, E(X) = Σ Xi * P(Xi)
= 2 * 136+ 3 * 118+ 4 * 112+ 5 * 19+ 6 * 536+ 7 * 16+ 8 * 536+ 9 * 19+ 10 * 112+ 11 * 118+ 12 * 136
= 118+16+13+59+56+76+109+ 1 +56+1118+13
= 7
Now, E(X) = Σ X2i * P(Xi)
=
=
=
=
= 54.833
Again, Var(X) = E[X2] – [E(X)]2
= 54.833 – 72
= 54.833 – 49
= 5.833
So, Standard deviation = √[Var(X)]
= √(5.833)
= 2.415