EXAM P SAMPLE SOLUTIONS

44. Solution: C
If k is the number of days of hospitalization, then the insurance payment g(k) is
100k for k=
1, 2, 3
g(k) = { 300+ 50( k− 3) for k=
4, 5.
5
∑
k = 1
1 2 3 4
5
Thus, the expected payment is gk ( ) pk= 100 p+ 200 p+ 300 p+ 350 p+ 400 p =
1
( 100× 5 + 200× 4 + 300× 3+ 350× 2 + 400× 1)
=220
15
--------------------------------------------------------------------------------------------------------
45. Solution: D
Note that ( )
2 2 3
0
3
4
0 4
x x x x 8 64 56 28
E X = ∫ − dx + dx
−2
10 ∫ =− + =− + = =
0 10 30 30 30 30 30 15
−2
0
--------------------------------------------------------------------------------------------------------
46. Solution: D
The density function of T is
1 −t
/3
f () t = e , 0 < t<
∞
3
Therefore,
E[ X] = E⎡⎣max ( T,
2)
⎤⎦
2 2
∞
−t/3 t −t/3
= ∫ e dt+ e dt
0 3 ∫2
3
=−2e | − te | +
∞
e
−t/3 2 −t/3 ∞ −t/3
0 2
2
−2/3 −2/3 −t/3
∞
| 2
=− 2e + 2 + 2e −3e
−2/3
= 2 + 3e
∫
dt
Page 19 of 55