# Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X. - Mathematics and Statistics

Sum

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

Concept: Probability Distribution of Discrete Random Variables
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