EXAM P SAMPLE SOLUTIONS

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81. Solution: C
1
Let X 1, . . . , X25
denote the 25 collision claims, and let X = (X 1 + . . . +X25)
. We are
25
given that each Xi (i = 1, . . . , 25) follows a normal distribution with mean 19,400 and
standard deviation 5000 . As a result X also follows a normal distribution with mean
19,400 and standard deviation 1
(5000) = 1000 . We conclude that P[ X > 20,000]
25
⎡X −19,400 20,000 −19,400⎤ ⎡X
−19,400
⎤
= P⎢ > = P
0.6
1000 1000
⎥ ⎢ >
⎣ ⎦ ⎣ 1000
⎥ = 1 − Φ(0.6) = 1 – 0.7257
⎦
= 0.2743 .
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82. Solution: B
Let X 1, . . . , X1250
be the number of claims filed by each of the 1250 policyholders.
We are given that each Xi follows a Poisson distribution with mean 2 . It follows that
E[X i] = Var[X i] = 2 . Now we are interested in the random variable S = X 1 + . . . + X1250
.
Assuming that the random variables are independent, we may conclude that S has an
approximate normal distribution with E[S] = Var[S] = (2)(1250) = 2500 .
Therefore P[2450 < S < 2600] =
⎡2450 −2500 S−2500 2600 −2500⎤ ⎡ S−2500
⎤
P⎢ < < ⎥ = P 1 2
2500 2500 2500
⎢
− < <
50 ⎥
⎣ ⎦ ⎣ ⎦
⎡S−2500 ⎤ ⎡S−2500 ⎤
= P
⎢
< 2 − P 1] = P[Z < 2] + P[Z < 1] – 1 ≈ 0.9773 + 0.8413 – 1 = 0.8186 .
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83. Solution: B
Let X1,…, Xn denote the life spans of the n light bulbs purchased. Since these random
variables are independent and normally distributed with mean 3 and variance 1, the
random variable S = X 1 + … + Xn is also normally distributed with mean
μ = 3n
and standard deviation
σ = n
Now we want to choose the smallest value for n such that
⎡ S−3n 40−3n⎤ 0.9772 ≤ Pr[ S > 40] = Pr ⎢ > ⎥
⎣ n n ⎦
This implies that n should satisfy the following inequality:
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