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
