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

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Sum

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

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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
  Is there an error in this question or solution?
Chapter 7: Probability Distributions - Exercise 7.1 [Page 233]

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