Question

Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.

Answer 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) = Σx_i · P (x_i)

= `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, Σx_i² · P (x_i)

= `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) = Σx_i² · P (x_i) - [E(X)]²

= 54·83 - 49

= 5.83

∴ standard deviation = `sqrt(V(X))`

= `sqrt(5.83)=2.41`

Answer 2

The sample space of the experiment consists of 36 elementary events in the form of ordered pairs (x_i, y_i), where x_i = 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