Answer:
When two fair dice are rolled, 6×6=36 observations are obtained.
P(X=2)=P(1,1)= 
36
1
 
P(X=3)=P(1,2)+P(2,1)= 
36
2
 = 
18
1
 
P(X=4)=P(1,3)+P(2,2)+P(3,1)= 
36
3
 = 
12
1
 
P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)= 
36
4
 = 
9
1
 
P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)= 
36
5
 
P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)= 
36
6
 = 
6
1
 
P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)= 
36
5
 
P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)= 
36
4
 = 
9
1
 
P(X=10)=P(4,6)+P(5,5)+P(6,4)= 
36
3
 = 
12
1
 
P(X=11)=P(5,6)+P(6,5)= 
36
2
 = 
18
1
 
P(X=12)=P(6,6)= 
36
1
 
Therefore, the required probability distribution is as follows.
Then, E(X)=∑X 
i
 ⋅P(X 
i
 )
=2× 
36
1
 +3× 
18
1
 +4× 
12
1
 +5× 
9
1
 +6× 
36
5
 +7× 
6
1
 +8× 
36
5
 +9× 
9
1
 +10× 
12
1
 +11× 
18
1
 +12× 
36
1
 
= 
18
1
 + 
6
1
 + 
3
1
 + 
9
5
 + 
6
5
 + 
6
7
 + 
9
10
 +1+ 
6
5
 + 
18
11
 + 
3
1
 
=7
E(X 
2
 )=∑X 
i
2
 ⋅P(X 
i
 )
=4× 
36
1
 +9× 
18
1
 +16× 
12
1
 +25× 
9
1
 +36× 
36
5
 +49× 
6
1
 +64× 
36
5
 +81× 
9
1
 +100× 
12
1
 +121× 
18
1
 +144× 
36
1
 
= 
9
1
 + 
2
1
 + 
3
4
 + 
9
25
 +5+ 
6
49
 + 
9
80
 +9+ 
3
25
 + 
18
121
 +4
= 
18
987
 = 
6
329
 =54.833
Then, Var(X)=E(X 
2
 )−[E(X)] 
2
 
=54.833−(7) 
2
 
=54.833−49
=5.833
∴ Standard deviation = 
Var(X)
 
= 
5.833
 
=2.415