Sample space consists of 36 elements. Let X denote the random variable which represents the sum of the numbers on the two dice
∴ X can assume the values 2, 3, 4, ......, or 12.
Now P(X = 2) = P({1, 1}) =

1

36

P(X = 3) = P({1, 2}, {2, 1}) =

2

36

P(X = 4) = P{(1, 3), (2, 2), (3, 1)} =

3

36

P(X = 5) = P{(1, 4), (2, 3), (3, 2) (4, 1)} =

4

36

P(X = 6) = P{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} =

5

36

P(X = 7) = P{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} =

6

36

P(X = 8) = P{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) =

5

36

P(X = 9) = P{(3, 6), (4, 5), (5, 4), (6, 3)} =

4

36

P(X = 10) = P{(4, 6), (5, 5), (6, 4)} =

3

36

P(X = 11) = P{(5, 6), (6, 5)} =

2

36

P(X = 12) = P{(6, 6)} =

1

36

.
Hence, the probability distribution is :Mean =µ=E(X)=∑XP(X)
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

×252=7
Hence the required mean = 7.
Variance σ2X=E(X2)−[E(X)]2
=[22×

1

36

+32×

2

36

+42×

3

36

+52×

4

36

+62×

5

36

+72×

6

36

+82×

5

36

+92×

4

36

+102×

3

36

+112×

2

36

+122×

1

36

]−[7]2
=

1

36

×1974−49=5.838
Standard deviation =√σ2X=σX=2.415