BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: You are given:}
(i) The number of claims has a Poisson distribution.
(ii) Claim sizes have a Pareto distribution with parameters \(\theta=0.5\) and \(\alpha=6\)
(iii) The number of claims and claim sizes are independent.
(iv) The observed average total payment should be within \(2 \%\) of the expected average total payment \(90 \%\) of the time.

Calculate the expected number of claims needed for full credibility.

Options:
A
Less than 7,000
B
At least 7,000, but less than 10,000
C
At least 10,000, but less than 13,000
D
At least 13,000, but less than 16,000
E
At least 16,000

Correct answer: E

Solution: The standard for full credibility is \(\left(\frac{1.645}{0.02}\right)^{2}\left(1+\frac{\operatorname{Var}(X)}{E(X)^{2}}\right)\) where \(X\) is the claim size variable. For the Pareto variable, \(E(X)=0.5 / 5=0.1\) and \(\operatorname{Var}(X)=\frac{2(0.5)^{2}}{5(4)}-(0.1)^{2}=0.015\). Then the standard is \(\left(\frac{1.645}{0.02}\right)^{2}\left(1+\frac{0.015}{0.1^{2}}\right)=16,913\) claims.
Question: You are given:
(i) Losses follow a single-parameter Pareto distribution with density function:
\[
f(x)=\frac{\alpha}{x^{\alpha+1}}, \quad x>1, \quad 0<\alpha<\infty
\]
(ii) A random sample of size five produced three losses with values 3,6 and 14 , and two losses exceeding 25.

Calculate the maximum likelihood estimate of \(\alpha\).

Options:
A
0.25
B
0.30
C
0.34
D
0.38
E
\(\quad 0.42\)

Correct answer: A

Solution: The distribution function is \(F(x)=\int_{1}^{x} \alpha t^{-\alpha-1} d t=-\left.t^{-\alpha}\right|_{1} ^{x}=1-x^{-\alpha}\). The likelihood function is
\[
\begin{aligned}
L & =f(3) f(6) f(14)[1-F(25)]^{2} \\
& =\alpha 3^{-\alpha-1} \alpha 6^{-\alpha-1} \alpha 14^{-\alpha-1}\left(25^{-\alpha}\right)^{2} \\
& \propto \alpha^{3}[3(6)(14)(625)]^{-\alpha}
\end{aligned}
\]

Taking logs, differentiating, setting equal to zero, and solving:
\(\ln L=3 \ln \alpha-\alpha \ln 157,500\) plus a constant
\(d \ln L / d \alpha=3 \alpha^{-1}-\ln 157,500=0\)
\(\hat{\alpha}=3 / \ln 157,500=0.2507\).
Question: You are given:
(i) Low-hazard risks have an exponential claim size distribution with mean \(\theta\).
(ii) Medium-hazard risks have an exponential claim size distribution with mean \(2 \theta\).
(iii) High-hazard risks have an exponential claim size distribution with mean \(3 \theta\).
(iv) No claims from low-hazard risks are observed.
(v) Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3 .
(vi) One claim from a high-hazard risk is observed, of size 15.

Calculate the maximum likelihood estimate of \(\theta\).

Options:
A
1
B
2
C
3
D
4
E
5

Correct answer: B

Solution: The likelihood function is \(\frac{e^{-1 /(2 \theta)}}{2 \theta} \frac{e^{-2 /(2 \theta)}}{2 \theta} \frac{e^{-3 /(2 \theta)}}{2 \theta} \frac{e^{-15 /(3 \theta)}}{3 \theta}=\frac{e^{-8 / \theta}}{24 \theta^{4}}\). The loglikelihood function is \(-\ln (24)-4 \ln (\theta)-8 / \theta\). Differentiating with respect to \(\theta\) and setting the result equal to 0 yields \(-\frac{4}{\theta}+\frac{8}{\theta^{2}}=0\) which produces \(\hat{\theta}=2\).
Question: You are given:

\begin{tabular}{|r|c|}
\hline Claim Size \((X)\) & Number of Claims \\
\hline\((0,25]\) & 25 \\
\hline\((25,50]\) & 28 \\
\hline\((50,100]\) & 15 \\
\hline\((100,200]\) & 6 \\
\hline
\end{tabular}

Assume a uniform distribution of claim sizes within each interval.
Calculate \(E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right]\).

Options:
A
Less than 200
B
At least 200, but less than 300
C
At least 300 , but less than 400
D
At least 400 , but less than 500
E
At least 500

Correct answer: C

Solution: In general,
\(E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right]=\int_{0}^{200} x^{2} f(x) d x-\int_{0}^{150} x^{2} f(x) d x-150^{2} \int_{150}^{200} f(x) d x\)
\(=\int_{150}^{200}\left(x^{2}-150^{2}\right) f(x) d x\).
Assuming a uniform distribution, the density function over the interval from 100 to 200 is 6/7400 (the probability of 6/74 assigned to the interval divided by the width of the interval). The answer is
\(\int_{150}^{200}\left(x^{2}-150^{2}\right) \frac{6}{7400} d x=\left.\left(\frac{x^{3}}{3}-150^{2} x\right) \frac{6}{7400}\right|_{150} ^{200}=337.84\).
Question: You are given the following information about a commercial auto liability book of business:
(i) Each insured's claim count has a Poisson distribution with mean \(\lambda\), where \(\lambda\) has a gamma distribution with \(\alpha=1.5\) and \(\theta=0.2\).
(ii) Individual claim size amounts are independent and exponentially distributed with mean 5000.
(iii) The full credibility standard is for aggregate losses to be within \(5 \%\) of the expected with probability 0.90 .

Calculate the expected number of claims required for full credibility using limited fluctuated credibility.

Options:
A
2165
B
2381
C
3514
D
7216
E
7938

Correct answer: B

Solution: \(X\) is the random sum \(Y_{1}+Y_{2}+\cdots+Y_{N}\).
\(N\) has a negative binomial distribution with \(r=a=1.5\) and \(\beta=\theta=0.2\).
\(E(N)=r \beta=0.3, \operatorname{Var}(N)=r \beta(1+\beta)=0.36\)
\(E(Y)=5,000, \operatorname{Var}(Y)=25,000,000\)
\(E(X)=0.3(5,000)=1,500\)
\(\operatorname{Var}(X)=0.3(25,000,000)+0.36(25,000,000)=16,500,000\)
Number of exposures (insureds) required for full credibility
\(n_{\text {FULL }}=(1.645 / 0.05)^{2}\left(16,500,000 / 1,500^{2}\right)=7,937.67\).
Number of expected claims required for full credibility
\(E(N) n_{\text {FULL }}=0.3(7,937.67)=2,381\).
Question: You are given:
(i) Losses follow an exponential distribution with mean \(\theta\).
(ii) A random sample of 20 losses is distributed as follows:

\begin{tabular}{|c|c|}
\hline Loss Range & Frequency \\
\hline\([0,1000]\) & 7 \\
\hline\((1000,2000]\) & 6 \\
\hline\((2000, \infty)\) & 7 \\
\hline
\end{tabular}

Calculate the maximum likelihood estimate of \(\theta\).

Options:
A
Less than 1950
B
At least 1950, but less than 2100
C
At least 2100, but less than 2250
D
At least 2250, but less than 2400
E
At least 2400

Correct answer: B

Solution: \(L=F(1000)^{7}[F(2000)-F(1000)]^{6}[1-F(2000)]^{7}\)
\(=\left(1-e^{-1000 / \theta}\right)^{7}\left(e^{-1000 / \theta}-e^{-2000 / \theta}\right)^{6}\left(e^{-2000 / \theta}\right)^{7}\)
\(=(1-p)^{7}\left(p-p^{2}\right)^{6}\left(p^{2}\right)^{7}=p^{20}(1-p)^{13}\)
where \(p=e^{-1000 / \theta}\) The maximum occurs at \(p=20 / 33\) and so \(\hat{\theta}=-1000 / \ln (20 / 33)=1996.90\).
Question: You are given:

\begin{tabular}{|c|c|c|c|}
\hline \begin{tabular}{c} 
Number of \\
Claims
\end{tabular} & Probability & Claim Size & Probability \\
\hline 0 & \(1 / 5\) & & \\
\hline 1 & \(3 / 5\) & 25 & \(1 / 3\) \\
& & 150 & \(2 / 3\) \\
\hline 2 & \(1 / 5\) & 50 & \(2 / 3\) \\
& & 200 & \(1 / 3\) \\
\hline
\end{tabular}

Claim sizes are independent.
Calculate the variance of the aggregate loss.

Options:
A
\(\quad 4,050\)
B
\(\quad 8,100\)
C
10,500
D
12,510
E
15,612

Correct answer: B

Solution: First obtain the distribution of aggregate losses:

\begin{tabular}{|l|l|}
\hline Value & Probability \\
\hline 0 & \(1 / 5\) \\
\hline 25 & \((3 / 5)(1 / 3)=1 / 5\) \\
\hline 100 & \((1 / 5)(2 / 3)(2 / 3)=4 / 45\) \\
\hline 150 & \((3 / 5)(2 / 3)=2 / 5\) \\
\hline 250 & \((1 / 5)(2)(2 / 3)(1 / 3)=4 / 45\) \\
\hline 400 & \((1 / 5)(1 / 3)(1 / 3)=1 / 45\) \\
\hline
\end{tabular}
\(\mu=(1 / 5)(0)+(1 / 5)(25)+(4 / 45)(100)+(2 / 5)(150)+(4 / 45)(250)+(1 / 45)(400)=105\)
\(\sigma^{2}=(1 / 5)\left(0^{2}\right)+(1 / 5)(25)+(4 / 45)\left(100^{2}\right)+(2 / 5)\left(150^{2}\right)+(4 / 45)\left(250^{2}\right)\)
\(+(1 / 45)\left(400^{2}\right)-105^{2}=8100\)
Question: You observe the following five ground-up claims from a data set that is truncated from below at 100 :
\[
\begin{array}{lllll}
125 & 150 & 165 & 175 & 250
\end{array}
\]

You fit a ground-up exponential distribution using maximum likelihood estimation. Calculate the mean of the fitted distribution.

Options:
A
73
B
100
C
125
D
156
E
173

Correct answer: A

Solution: Because the exponential distribution is memoryless, the excess over the deductible is also exponential with the same parameter. So subtracting 100 from each observation yields data from an exponential distribution and noting that the maximum likelihood estimate is the sample mean gives the answer of 73.

Working from first principles,
\[
\begin{aligned}
& L(\theta)=\frac{f\left(x_{1}\right) f\left(x_{2}\right) f\left(x_{3}\right) f\left(x_{4}\right) f\left(x_{5}\right)}{[1-F(100)]^{5}}=\frac{\theta^{-1} e^{-125 / \theta} \theta^{-1} e^{-150 / \theta} \theta^{-1} e^{-165 / \theta} \theta^{-1} e^{-175 / \theta} \theta^{-1} e^{-250 / \theta}}{\left(e^{-100 / \theta}\right)^{5}} \\
& =\theta^{-5} e^{-365 / \theta} .
\end{aligned}
\]

Taking logarithms and then a derivative gives
\(l(\theta)=-5 \ln (\theta)-365 / \theta, l^{\prime}(\theta)=-5 / \theta+365 / \theta^{2}=0\)
\(\hat{\theta}=365 / 5=73\).
Question: You are given the following information about a general liability book of business comprised of 2500 insureds:
(i) \(\quad X_{i}=\sum_{j=1}^{N_{i}} Y_{i j}\) is a random variable representing the annual loss of the \(i\) th insured.
(ii) \(\quad N_{1}, N_{2}, \ldots, N_{2500}\) are independent and identically distributed random variables following a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\).
(iii) \(\quad Y_{i 1}, Y_{i 2}, \ldots, Y_{i N_{i}}\) are independent and identically distributed random variables following a Pareto distribution with \(\alpha=3\) and \(\theta=1000\).
(iv) The full credibility standard is to be within \(5 \%\) of the expected aggregate losses \(90 \%\) of the time.

Calculate the partial credibility of the annual loss experience for this book of business using limited fluctuation credibility theory.

Options:
A
0.34
B
0.42
C
0.47
D
0.50
E
0.53

Correct answer: C

Solution: \(E(N)=r \beta=0.4\)
\(\operatorname{Var}(N)=r \beta(1+\beta)=0.48\)
\(E(Y)=\theta /(\alpha-1)=500\)
\(\operatorname{Var}(Y)=\frac{\theta^{2} \alpha}{(\alpha-1)^{2}(\alpha-2)}=750,000\)
Therefore,
\(E(X)=0.4(500)=200\)
\(\operatorname{Var}(X)=0.4(750,000)+0.48(500)^{2}=420,000\).
The full credibility standard is \(n=\left(\frac{1.645}{0.05}\right)^{2} \frac{420,000}{200^{2}}=11,365, \mathrm{Z}=\sqrt{2,500 / 11,365}=0.47\).
Question: A health plan implements an incentive to physicians to control hospitalization under which the physicians will be paid a bonus \(B\) equal to \(c\) times the amount by which total hospital claims are under \(400(0 \leq c \leq 1)\).

The effect the incentive plan will have on underlying hospital claims is modeled by assuming that the new total hospital claims will follow a Pareto distribution with \(\alpha=2\) and \(\theta=300\).
\(E(B)=100\)
Calculate \(c\).

Options:
A
\(\quad 0.44\)
B
0.48
C
0.52
D
0.56
E
\(\quad 0.60\)

Correct answer: A

Solution: \[
\begin{aligned}
B & =\left\{\begin{array}{cc}
c(400-x) & x<400 \\
0 & x \geq 400
\end{array}\right. \\
100 & =E(B)=c 400-c E(X \wedge 400) \\
& =c 400-c 300\left(1-\frac{300}{300+400}\right) \\
& =c\left(400-300 \frac{4}{7}\right) \\
c & =\frac{100}{228.6}=0.44
\end{aligned}
\]
Question: Computer maintenance costs for a department are modeled as follows:
(i) The distribution of the number of maintenance calls each machine will need in a year is Poisson with mean 3.
(ii) The cost for a maintenance call has mean 80 and standard deviation 200.
(iii) The number of maintenance calls and the costs of the maintenance calls are all mutually independent.

The department must buy a maintenance contract to cover repairs if there is at least a \(10 \%\) probability that aggregate maintenance costs in a given year will exceed \(120 \%\) of the expected costs.

Calculate the minimum number of computers needed to avoid purchasing a maintenance contract using the normal approximation for the distribution of the aggregate maintenance costs.

Options:
A
80
B
90
C
100
D
110
E
120

Correct answer: C

Solution: Let \(N=\) number of computers in department
Let \(X=\) cost of a maintenance call
Let \(S=\) aggregate cost
\[
\begin{aligned}
& \operatorname{Var}(X)=[\operatorname{Standard} \text { Deviation }(X)]^{2}=200^{2}=40,000 \\
& \begin{aligned}
E\left(X^{2}\right) & =\operatorname{Var}(X)+[E(X)]^{2} \\
& =40,000+80^{2}=46,400 \\
E(S)= & N \lambda E(X)=N(3)(80)=240 N \\
\operatorname{Var}(S) & =N \lambda \times E\left(X^{2}\right)=N(3)(46,400)=139,200 N
\end{aligned}
\end{aligned}
\]

We want
\[
\begin{aligned}
& 0.1 \geq \operatorname{Pr}(S>1.2 E(S)) \\
& \quad \geq \operatorname{Pr}\left(\frac{S-E(S)}{\sqrt{139,200 N}}>\frac{0.2 E(S)}{\sqrt{139,200 N}}\right) \Rightarrow \frac{0.2(240) N}{373.1 \sqrt{N}} \geq \Phi^{-1}(0.9)=1.282 \\
& N \geq\left(\frac{1.282(373.1)}{48}\right)^{2}=99.3
\end{aligned}
\]
Question: A towing company provides all towing services to members of the City Automobile Club. You are given:

\begin{tabular}{|c|c|c|}
\hline Towing Distance & Towing Cost & Frequency \\
\hline \(0-9.99\) miles & 80 & \(50 \%\) \\
\hline \(10-29.99\) miles & 100 & \(40 \%\) \\
\hline \(30+\) miles & 160 & \(10 \%\) \\
\hline
\end{tabular}
(i) The automobile owner must pay \(10 \%\) of the cost and the remainder is paid by the City Automobile Club.
(ii) The number of towings has a Poisson distribution with mean of 1000 per year.
(iii) The number of towings and the costs of individual towings are all mutually independent.

Calculate the probability that the City Automobile Club pays more than 90,000 in any given year using the normal approximation for the distribution of aggregate towing costs.

Options:
A
\(3 \%\)
B
\(10 \%\)
C
\(50 \%\)
D
\(90 \%\)
E
\(97 \%\)

Correct answer: B

Solution: First restate the table to be CAC's cost, after the \(10 \%\) payment by the auto owner:

\begin{tabular}{|c|c|}
\hline Towing Cost, \(x\) & \(p(x)\) \\
\hline 72 & \(50 \%\) \\
\hline 90 & \(40 \%\) \\
\hline 144 & \(10 \%\) \\
\hline
\end{tabular}

Then \(E(X)=0.5(72)+0.4(90)+0.1(144)=86.4\).
\(E\left(X^{2}\right)=0.5\left(72^{2}\right)+0.4\left(90^{2}\right)+0.1\left(144^{2}\right)=7905.6\)
\(\operatorname{Var}(X)=7905.6-86.4^{2}=440.64\)
Because Poisson,
\(E(N)=\operatorname{Var}(N)=1000\)
\(E(S)=E(X) E(N)=86.4(1000)=86,400\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=1000(440.64)+86.4^{2}(1000)=7,905,600\)
\(\operatorname{Pr}(S>90,000)+\operatorname{Pr}\left(\frac{S-E(S)}{\sqrt{\operatorname{Var}(S)}}>\frac{90,000-86,400}{\sqrt{7,905,600}}\right)=\operatorname{Pr}(Z>1.28)=1-\Phi(1.28)=0.10\)
Since the frequency is Poisson, you could also have used
\(\operatorname{Var}(S)=\lambda E\left(X^{2}\right)=1000(7905.6)=7,905,600\).
That way, you would not need to have calculated \(\operatorname{Var}(X)\).
Question: You are given:
(i) Losses follow an exponential distribution with the same mean in all years.
(ii) The loss elimination ratio this year is \(70 \%\).
(iii) The ordinary deductible for the coming year is \(4 / 3\) of the current deductible. Calculate the loss elimination ratio for the coming year.

Options:
A
\(70 \%\)
B
\(75 \%\)
C
\(80 \%\)
D
\(85 \%\)
E
\(90 \%\)

Correct answer: C

Solution: \(\operatorname{LER}=\frac{E(X \wedge d)}{E(X)}=\frac{\theta\left(1-e^{-d / \theta}\right)}{\theta}=1-e^{-d / \theta}\)
Last year \(\quad 0.70=1-e^{-d / \theta} \Rightarrow-d=\theta \log (0.30)\)
Next year: \(\quad-d_{\text {new }}=\theta \log \left(1-\mathrm{LER}_{\text {new }}\right)\)
Hence \(\theta \log \left(1-\mathrm{LER}_{\text {new }}\right)=-d_{\text {new }}=\frac{4}{3} \theta \log (0.30)\)
\(\log \left(1-\mathrm{LER}_{\text {new }}\right)=-1.6053\)
\(\left(1-\mathrm{LER}_{\text {new }}\right)=e^{-1.6053}=0.20\)
\(\mathrm{LER}_{\text {new }}=0.80\)
Question: Prescription drug losses, \(S\), are modeled assuming the number of claims has a geometric distribution with mean 4 , and the amount of each prescription is 40 .

Calculate \(E\left[(S-100)_{+}\right]\).

Options:
A
60
B
82
C
92
D
114
E
146

Correct answer: C

Solution: Let \(N=\) number of prescriptions then

\begin{tabular}{|c|c|}
\hline\(n\) & \(f_{N}(n)\) \\
\hline 0 & 0.2000 \\
1 & 0.1600 \\
2 & 0.1280 \\
3 & 0.1024 \\
\hline
\end{tabular}
\(E\left[(S-100)_{+}\right]=E[S]-E[S \wedge 100]\)
\(E[S]=40(4)=160\)
\(E[S \wedge 100]=0(0.2)+40(0.16)+80(0.128)+100(1-0.2-0.16-0.128)=67.84\)
\(E\left[(S-100)_{+}\right]=160-67.84=92.16\)
Question: At the beginning of each round of a game of chance the player pays 12.5 . The player then rolls one die with outcome \(N\). The player then rolls \(N\) dice and wins an amount equal to the total of the numbers showing on the \(N\) dice. All dice have 6 sides and are fair.

Calculate the probability that a player starting with 15,000 will have at least 15,000 after 1000 rounds using the normal approximation.

Options:
A
0.01
B
0.04
C
0.06
D
0.09
E
\(\quad 0.12\)

Correct answer: E

Solution: In each round,
\(N=\) result of first roll, to see how many dice you will roll
\(X=\) result of for one of the \(N\) dice you roll
\(S=\) sum of \(X\) for the \(N\) dice
\(E(X)=E(N)=3.5\)
\(\operatorname{Var}(X)=\operatorname{Var}(N)=2.9167\)
\(E(S)=E(N) E(X)=12.25\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+\operatorname{Var}(N) E(X)^{2}=3.5(2.9167)+2.9167(3.5)^{2}=45.938\)
Let \(S_{1000}\) the sum of the winnings after 1000 rounds
\(E\left(S_{1000}\right)=1000(12.25)=12,250\)
\(S D\left(S_{1000}\right)=\sqrt{1000(45.938)}=214.33\)
After 1000 rounds, you have your initial 15,000 , less payments of 12,500 , plus winnings for a total of \(2,500+S_{1000}\) Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater than 15000-0.5. In this problem, that continuity correction has negligible impact.
\(\operatorname{Pr}\left(2,500+S_{1000}>14,999.5\right)=\operatorname{Pr}\left(S_{1000}>12,499.5\right) \approx \operatorname{Pr}\left(Z>\frac{12,499.5-12,250}{214.33}=1.17\right)=0.12\).
Question: \(X\) is a discrete random variable with a probability function that is a member of the \((a, b, 0)\) class of distributions.

You are given:
(i) \(\quad \mathrm{P}(X=0)=\mathrm{P}(X=1)=0.25\)
(ii) \(\quad \mathrm{P}(X=2)=0.1875\)

Calculate \(\mathrm{P}(X=3)\).

Options:
A
0.120
B
0.125
C
0.130
D
0.135
E
0.140

Correct answer: B

Solution: \(p_{k}=\left(a+\frac{b}{k}\right) p_{k-1}\)
\(0.25=(a+b) 0.25 \Rightarrow a=1-b\)
\(0.1875=\left(a+\frac{b}{2}\right)(0.25) \Rightarrow 0.1875=(1-0.5 b)(0.25) \Rightarrow b=0.5, a=0.5\)
\(p_{3}=\left(0.5+\frac{0.5}{3}\right)(0.1875)=0.125\)
Question: A group dental policy has a negative binomial claim count distribution with mean 300 and variance 800 .

Ground-up severity is given by the following table:

\begin{tabular}{|c|c|}
\hline Severity & Probability \\
\hline 40 & 0.25 \\
\hline 80 & 0.25 \\
\hline 120 & 0.25 \\
\hline 200 & 0.25 \\
\hline
\end{tabular}

You expect severity to increase \(50 \%\) with no change in frequency. You decide to impose a per claim deductible of 100 .

Calculate the expected total claim payment after these changes.

Options:
A
Less than 18,000
B
At least 18,000, but less than 20,000
C
At least 20,000, but less than 22,000
D
At least 22,000 , but less than 24,000
E
At least 24,000

Correct answer: D

Solution: \begin{tabular}{cc} 
Severity & Severity \\
after increase & after increase and deductible \\
60 & 0 \\
120 & 20 \\
180 & 80 \\
300 & 200
\end{tabular}

Expected payment per loss \(=0.25(0)+0.25(20)+0.25(80)+0.25(200)=75\)
Expected payments \(=\) Expected number of losses x Expected payment per loss \(=300(75)=22,500\)
Question: You own a light bulb factory. Your workforce is a bit clumsy - they keep dropping boxes of light bulbs. The boxes have varying numbers of light bulbs in them, and when dropped, the entire box is destroyed.

You are given:
- Expected number of boxes dropped per month: 50
- Variance of the number of boxes dropped per month: 100
- Expected value per box: 200
- Variance of the value per box: 400

You pay your employees a bonus if the value of light bulbs destroyed in a month is less than 8000.

Assuming independence and using the normal approximation, calculate the probability that you will pay your employees a bonus next month.

Options:
A
\(\quad 0.16\)
B
0.19
C
0.23
D
0.27
E
\(\quad 0.31\)

Correct answer: A

Solution: \(\mathrm{E}(S)=\mathrm{E}(N) \mathrm{E}(X)=50(200)=10,000\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=50(400)+200^{2}(100)=4,020,000\)
\(\operatorname{Pr}(S<8,000) \approx \operatorname{Pr}\left(Z<\frac{8,000-10,000}{\sqrt{4,020,000}}=-0.998\right)=0.16\)
Question: For a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1,2 , or 3 , each with probability \(1 / 3\). Loss amounts are independent of the number of losses, and of each other.

An insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2 .

Calculate the expected claim payments for this insurance policy.

Options:
A
\(\quad 2.00\)
B
2.36
C
2.45
D
2.81
E
\(\quad 2.96\)

Correct answer: B

Solution: Let \(S\) denote aggregate loss before deductible.
\(\mathrm{E}(S)=2(2)=4\), since mean severity is 2 .
\(f_{S}(0)=\frac{e^{-2} 2^{0}}{0!}=0.1353\), since must have 0 losses to get aggregate loss \(=0\).
\(f_{s}(1)=\frac{e^{-2} 2^{1}}{1!} \frac{1}{3}=0.0902\), since must have 1 loss whose size is 1 to get aggregate loss \(=1\).
\(E(S \wedge 2)=0 f_{S}(0)+1 f_{S}(1)+2\left[1-f_{S}(0)-f_{S}(1)\right]\)
\(=0(0.1353)+1(0.0902)+2(1-0.01353-0.0902)=1.6392\)
\(E\left[(S-2)_{+}\right]=E(S)-E(S \wedge 2)=4-1.6392=2.3608\)
Question: The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution:
\[
F(x)=1-0.8 e^{-0.02 x}-0.2 e^{-0.001 x}, \quad x \geq 0
\]

The insurance policy pays amounts up to a limit of 1000 per claim.
Calculate the expected payment under this policy for one claim.

Options:
A
57
B
108
C
166
D
205
E
240

Correct answer: C

Solution: Limited expected value \(=\)
\(\int_{0}^{1000}[1-F(x)] d x=\int_{0}^{1000} 0.8 e^{-0.02 x}+0.2 e^{-0.001 x} d x=-40 e^{-0.02 x}-\left.200 e^{-0.001 x}\right|_{0} ^{1000} =-0-73.576+40+200=166.424\)
Question: The random variable for a loss, \(X\), has the following characteristics:

\begin{tabular}{|c|c|c|}
\hline\(x\) & \(F(x)\) & \(E(X \wedge x)\) \\
\hline 0 & 0.0 & 0 \\
\hline 100 & 0.2 & 91 \\
\hline 200 & 0.6 & 153 \\
\hline 1000 & 1.0 & 331 \\
\hline
\end{tabular}

Calculate the mean excess loss for a deductible of 100 .

Options:
A
250
B
300
C
350
D
400
E
450

Correct answer: B

Solution: Mean excess loss \(=\frac{E(X)-E(X \wedge 100)}{1-F(100)}=\frac{331-91}{0.8}=300\)
Note that \(E(X)=E(X \wedge 1000)\) because \(F(1000)=1\).
Question: A dam is proposed for a river that is currently used for salmon breeding. You have modeled:
(i) For each hour the dam is opened the number of salmon that will pass through and reach the breeding grounds has a distribution with mean 100 and variance 900.
(ii) The number of eggs released by each salmon has a distribution with mean 5 and variance 5.
(iii) The number of salmon going through the dam each hour it is open and the numbers of eggs released by the salmon are independent.

Calculate the least number of whole hours the dam should be left open so the probability that 10,000 eggs will be released is greater than \(95 \%\) using the normal approximation for the aggregate number of eggs released.

Options:
A
20
B
23
C
26
D
29
E
32

Correct answer: B

Solution: \(N=\) number of salmon in \(t\) hours
\(X=\) eggs from one salmon
\(S=\) total eggs.
\(E(N)=100 t\)
\(\operatorname{Var}(N)=900 t\)
\(E(S)=E(N) \mathrm{E}(\mathrm{X})=(100 \mathrm{t})(5)=500 \mathrm{t}\)
\(\operatorname{Var}(\mathrm{S})=\mathrm{E}(\mathrm{N}) \operatorname{Var}(\mathrm{X})+\mathrm{E}(\mathrm{X})^{2} \operatorname{Var}(N)=(100 t)(5)+\left(5^{2}\right)(900 t)=23,000 t\)
\(0.95<\operatorname{Pr}(S>10,000)=\operatorname{Pr}\left(Z>\frac{10,000-500 t}{\sqrt{23,000 t}}\right) \Rightarrow \frac{10,000-500 t}{\sqrt{23,000 t}}=-1.645\)
\(10,000-500 t=-1.645(151.66) \sqrt{t}=-249.48 \sqrt{t}\)
\(500 t-249.48 \sqrt{t}-10,000=0\)
\(\sqrt{t}=\frac{249.48 \pm \sqrt{(-249.48)^{2}-4(500)(-10,000)}}{2(500)}=4.73\)
\(t=22.26\)
Round up to 23