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Question
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Solution
The sample space of the experiment consists of 36 elementary events in the form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6.
The random variable X, i.e., the sum of the numbers on the two dice takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.
| X = xi | P(xi) | xiP(xi) | xi2P(xi) |
| 2 | `(1)/(36)` | `(2)/(36)` | `(4)/(36)` |
| 3 | `(2)/(36)` | `(6)/(36)` | `(18)/(36)` |
| 4 | `(3)/(36)` | `(12)/(36)` | `(48)/(36)` |
| 5 | `(4)/(36)` | `(20)/(36)` | `(100)/(36)` |
| 6 | `(5)/(36)` | `(30)/(36)` | `(180)/(36)` |
| 7 | `(6)/(36)` | `(42)/(36)` | `(294)/(36)` |
| 8 | `(5)/(36)` | `(40)/(36)` | `(320)/(36)` |
| 9 | `(4)/(36)` | `(36)/(36)` | `(324)/(36)` |
| 10 | `(3)/(36)` | `(30)/(36)` | `(300)/(36)` |
| 11 | `(2)/(36)` | `(22)/(36)` | `(242)/(36)` |
| 12 | `(1)/(36)` | `(12)/(36)` | `(144)/(36)` |
| \[\sum\limits_{i=1}^{n} x_i\text{P}(x_i)\] = `(252)/(36)` = 7 | \[\sum\limits_{i=1}^{n} x_i^2\text{P}(x_i)\] = `(1974)/(36)` |
E(X2) = \[\sum\limits_{i=1}^{11} x_i^2\text{P}(x_i)\] = `(1974)/(36)`
Var(X) = E(X2) – [E(X)]2
= `(1974)/(36) - (7)^2`
= `(1974)/(36) - 49`
= `(35)/(6)`
= 5.8333
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