PSTAT 5E
Midterm Key
Winter 2006
Question 1
Let
X
be the number of jobs a student has
(a)
P
(a student has 2 jobs) =
P
(2) = 1 – [
P
(0) +
P
(1)] = 1 – [0.5 + 0.3] = 0.2
(b)
P
(a student has at least 1 job) =
P
(1) +
P
(2) = 0.3 + 0.2 = 0.5
(c)
The mean number of jobs is just the
E
(
X
)
E
(
X
) =
∑
=
=
3
1
i
i
i
x
p
µ
= 0(0.5) + 1(0.3) + 2(0.2) = 0.7
Recall that the standard deviation is
)
(
X
Var
.
So we must first find
Var(X).
Var(X)
=
∑
=
−
3
1
2
)
(
i
i
i
x
p
=
2
)
7
.
0
0
(
5
.
0
−
+
2
)
7
.
0
1
(
3
.
0
−
+
2
)
7
.
0
2
(
2
.
0
−
=
=0.245 + 0.027 + 0.338 = 0.61
So, the standard deviation =
)
(
X
Var
=
61
.
0
= 0.781
Question 2
(a)
Let A = {odd number}, then A = {1, 3, 5}.
2
1
)
6
1
(
3
)
5
(
)
3
(
)
1
(
)
(
=
=
+
+
=
P
P
P
A
P
(b)
The sum of the scores is a multiple of 5 if it is equal to 5 or 10. The
outcomes from throwing two dices are:
1,1
2,
13
,1
4,1
5,
16
,1
1,2
2,2
3,2
4,
25
,
26
,2
1,3
2,3
3,
34
,
35
,
36
,3
1,4
2,
43
,
44
,
45
,4
6,4
1,5
2,
53
,
54
,5
5,5
6,5
1,6
2,
63
,6
4,6
5,
66
,6
The circled outcomes correspond to the cases where the sum is a multiple of 5.
Therefore, since the number of circled outcomes is 7 and the total number of
outcomes is 36, P (sum is multiple of 5) = 7/36.