Course 1 May 2000 Multiple Choice Exams - Society of Actuaries

18. Answer: C
Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the claim
⎧x
for x<
250
benefits paid. Then Y = ⎨
and we want to find m such that
⎩250 for x ≥ 250
m
−0.004x
−0.004x
m
0.50 = ∫ 0.004e dx=−e
0
= 1 – e –0.004m
0
This condition implies e –0.004m = 0.5 fi m = 250 ln 2 = 173.29 .
19. Answer: D
Let X denote the difference between true and reported age. We are given X is uniformly
distributed on (-2.5,2.5) . That is, X has pdf f(x) = 1/5, -2.5 < x < 2.5 . It follows that
µ = E[X] = 0
x
2.5 2 3 3
s 2 x = Var[X] = E[X 2 x x 2.5 2(2.5)
] = ∫ dx =
−2.5
= =2.083
2.5
5 15 15
−
s x =1.443
Now X
48
, the difference between the means of the true and rounded ages, has a distribution that
is approximately normal with mean 0 and standard deviation 1.443 = 0.2083 . Therefore,
48
⎡ 1 1 ⎤ ⎡ −0.25 0.25 ⎤
P
⎢
− ≤ X48
≤ = P ≤Z
≤
⎣ 4 4⎥ ⎦
⎢
⎣0.2083 0.2083⎥
= P[-1.2 £ Z £ 1.2] = P[Z £ 1.2] – P[Z £ –1.2]
⎦
= P[Z £ 1.2] – 1 + P[Z £ 1.2] = 2P[Z £ 1.2] – 1 = 2(0.8849) – 1 = 0.77 .
20. Answer: C
The joint pdf of X and Y is f(x,y) = f 2 (y|x) f 1 (x) = (1/x)(1/12), 0 < y < x, 0 < x < 12 .
Therefore,
12 x
12 12 2
1 y x x x 12
E[X] = ∫∫x⋅ dydx= dx dx
12x
∫ =
12 0
∫ = = 6
12 24 0
E[Y] =
E[XY] =
0 0 0 0
12 x
12 2 x 12
2 12
y ⎡ y ⎤ x x 144
dydx = ⎢ ⎥ dx = dx = =
∫∫ ∫ ∫ = 3
12x
0
0 0 0⎣24x⎦
24 48 48
0 0
12 x
12 2 x 12 2 3 3
12
y ⎡y ⎤ x x (12)
dydx = ⎢ ⎥ dx = dx = =
∫∫ ∫ ∫ = 24
12 0
0 0 0⎣24⎦
24 72 72
0 0
Cov(X,Y) = E[XY] – E[X]E[Y] = 24 - (3)(6) = 24 – 18 = 6 .
Course 1 8 May 2000