Question

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

Answer 1

Let, X be the numbers on two dice.

Since pair of dice are thrown.

So, X can be 2,3,4,5,6,7,8,9,10,11,12.

Consider X be the difference between the number, draw a table for the different sample space:

X	Events	Number of Outcomes	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

So the probability distribution is,

X	2	3	4	5	6	7	8	9	10	11	12
P( X )	1 36	2 36	3 36	4 36	5 36	6 36	5 36	4 36	3 36	2 36	1 36

Calculate the value of variance,

Var( X )=E( X 2 )− [ E( X ) ] 2 (1)

Therefore, the mean is given by,

E( X )= ∑ i=1 n X i P 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 ) = 252 36 =7

Calculate the value of E( X 2 ),

E( X 2 )= ∑ i=1 n X i 2 P i =( 2 2 × 1 36 )+( 3 2 × 2 36 )+( 4 2 × 3 36 )+( 5 2 × 4 36 )+( 6 2 × 5 36 )+( 7 2 × 6 36 ) +( 8 2 × 5 36 )+( 9 2 × 4 36 )+( 10 2 × 3 36 )+( 11 2 × 2 36 )+( 12 2 × 1 36 ) = 1974 36 = 329 6

Substitute the values in equation (1),

Var( X )=E( X 2 )− [ E( X ) ] 2 = 329 6 − ( 7 ) 2 =5.833

And standard deviation σ x is,

σ x = Var( x ) = 5.833 =2.415

Thus, the variance is 5.833 and standard deviation is 2.415.