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Question
Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean or expectation of X and variance of X
Sum
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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_("i" = 1)^"n"x"P"(x_"i") = 252/36` = 7 | `sum_("i" = 1)^"n"x_"i"^2"P"(x_"i") = 1974/36` |
∴ E(X) = `sum_("i" = 1)^11x_"i""P"(x_"i")` = 7
E(x2) = `sum_("i" = 1)^"n"x_"i"^2"P"(x_"i") = 1974/36`
Var(X) = E(X2) – [E(X)]2 = `1974/36 - (7)^2`
= `1974/36 - 49`
= `35/6`
= 5.83
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