EXAM P SAMPLE SOLUTIONS

68. Solution: C
Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the
⎧x
claim benefits paid. Then Y = ⎨
⎩250
for x<
250
and we want to find m such that 0.50
for x ≥ 250
m
−0.004x −0.004x
= 0.004e
dx =−e
m
= 1 – e –0.004m
∫
0
0
This condition implies e –0.004m = 0.5 ⇒ m = 250 ln 2 = 173.29 .
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69. Solution: D
The distribution function of an exponential random variable
T with parameter θ is given by F t
−t
θ
= − e t > 0
( ) 1 ,
Since we are told that T has a median of four hours, we may determine θ as follows:
1
−4
θ
= F( 4) = 1−e
2
1 −4
θ
= e
2
4
− ln ( 2)
=−
θ
4
θ =
ln 2
( )
Therefore, ( ) ( )
( )
5ln 2
−
−5θ4−5 4
Pr T ≥ 5 = 1− F 5 = e = e = 2 = 0.42
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70. Solution: E
Let X denote actual losses incurred. We are given that X follows an exponential
distribution with mean 300, and we are asked to find the 95 th percentile of all claims that
exceed 100 . Consequently, we want to find p95 such that
Pr[100 < x< p95] F( p95) −F(100)
0.95 = =
where F(x) is the distribution function of X .
PX [ > 100] 1 −F(100)
Now F(x) = 1 – e –x/300 .
−p95 /300 −100/300 −1/3−p95
/300
1 −e −(1 −e ) e −e
1/3 − p95
/300
Therefore, 0.95 =
= = 1−e
e
−100/300 −1/3
1 −(1 −e
) e
p95
/300
e −
= 0.05 e –1/3
p95 = −300 ln(0.05 e –1/3 ) = 999
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