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Question
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution 1
If two fair dice are rolled then the sample space S of this experiment is
S = {(1,1), (1,2),(1,3),(1,4),(1,5),(1,5),(1,6),(2,1),(2,2),(2,3),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
| sum of Nos. (x) | Favourable events | No of favourable | P (x) |
| 2 | (1,1) | 1 | `1/36` |
| 3 | (1, 2), (2, 1) | 2 | `2/36` |
| 4 | (1, 3), (2, 2), (3, 1) | 3 | `3/36` |
| 5 | (1, 4), (2, 3), (3, 2), (4, 1) | 4 | `4/36` |
| 6 | (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) | 5 | `5/36` |
| 7 | (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) | 6 | `6/36` |
| 8 | (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) | 5 | `5/36` |
| 9 | (3, 6), (4, 5), (5, 4), (6, 3) | 4 | `4/36` |
| 10 | (4, 6), (5, 5), (6, 4) | 3 | `3/36` |
| 11 | (5, 6), (6, 5) | 2 | `2/36` |
| 12 | (6,6) | 1 | `1/36` |
∴ the probability distribution of X is given by
| X=xi | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P[X=xi] | `1/36` | `2/36` | `3/36` | `4/36` | `5/36` | `6/36` | `5/36` | `4/36` | `3/36` | `2/36` | `1/36` |
Expected value = E (X) = Σxi · P (xi)
= `2(1/36)+3(2/36)+4(3/36)+5(4/36)+6(5/36)+7(6/36)+8(5/36)+9(4/36)+10(3/36)+11(2/36)+12(1/36)`
=`1/36 (2+6+12+20+30+42+40+36+30+22+12)`
`1/ 36 xx 252 = 7.`
Also, Σxi2 · P (xi)
= `4xx1/36 + 9xx2/36 + 16xx3/36 + 25xx4/36 + 36xx5/36 + 49xx6/36 + 64xx5/36 + 81xx4/36 + 100xx3/36 + 121xx2/36 + 144xx1/36`
= `1/36[4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144]`
= `1/ 36` (1974) = 54.83
∴ variance = V(X) = Σxi2 · P (xi) - [E(X)]2
= 54·83 - 49
= 5.83
∴ standard deviation = `sqrt(V(X))`
= `sqrt(5.83)=2.41`
Solution 2
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 | `7/36` | `40/36` | `320/36` |
| 9 | `8/36` | `36/36` | `324/36` |
| 10 | `9/36` | `30/36` | `300/36` |
| 11 | `10/36` | `22/36` | `242/36` |
| 12 | `11/36` | `12/36` | `144/36` |
| `sum_("i" = 1)^"n"x_"i""P"(x_"i")` = 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`
∴ Standard deviation = `sqrt("Var"("X"))`
= `sqrt(35/6)`
= 2.415
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