BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: An insurance company offers two kinds of insurance: automobile insurance and homeowners insurance. This year, the company's profit from automobile insurance is normally distributed with mean 200 and standard deviation 400. The company's profit from homeowners insurance is normally distributed with mean 400 and standard deviation 300. The two profits are independent.

Let \(Z\) represent the standard normal random variable.
Determine the probability that the company's overall profit this year is positive.

Options:
A
\(\mathrm{P}[Z>-1.83]\)
B
\(\mathrm{P}[Z>-1.20]\)
C
\(\mathrm{P}[Z>-0.86]\)
D
\(\mathrm{P}[Z>-0.83]\)
E
\(\quad \mathrm{P}[Z>-0.50]\)

Correct answer: B

Solution: Let X - automobile Insurance
Let Y - homeowners insurance
\(\mathrm{E}(X+Y)=200+400=600\)
\(\operatorname{Var}(X+Y)=400^{2}+300^{2}=250,000\)
\(\mathrm{SD}(X+Y)=\sqrt{250,000}=500\)
\(\mathrm{P}[X+Y>0]=\mathrm{P}\left[Z>\frac{0-600}{500}\right]=\mathrm{P}[Z>-1.2]\).
Question: A company's annual profit, in billions, has a normal distribution with variance equal to the cube of its mean. The probability of an annual loss is \(5 \%\).

Calculate the company's expected annual profit.

Options:
A
370 million
B
520 million
C
780 million
D
950 million
E
1645 million

Correct answer: A

Solution: Let \(X\) be the profit random variable. Then, \(0.05=P(X<0)=P(Z<-\mu / \sigma)\) and from the table, \(-\mu / \sigma=-1.645\). From the problem, \(\sigma^{2}=\mu^{3}\). Therefore, \(-1.645=-\mu / \mu^{3 / 2}=-\mu^{-1 / 2}\) and \(\mu=1 / 1.645^{2}=0.37\). in billions, or 370 million.
Question: Math SAT scores are normally distributed and when reported are rounded to multiples of ten.

In 1982 Abby's mother scored at the rounded 93rd percentile on the math SAT exam. In 1982 the mean score was 503 and the variance of the scores was 9604.

In 2008 Abby took the math SAT and got the same rounded numerical score as her mother had received 26 years before. In 2008 the mean score was 521 and the variance of the scores was 10,201.

Calculate the rounded percentile for Abby’s score.

Options:
A
89th
B
90 th
C
\(\quad 91 \mathrm{st}\)
D
\(\quad 92 \mathrm{nd}\)
E
\(\quad 93 r d\)

Correct answer: B

Solution: From the table the 93rd percentile comes from a z-score between 1.47 and 1.56. 1.47 implies a test score of \(503+1.47(98)=647.1\). Similarly, 1.56 implies a score of 655.9 . The only multiple of 10 in this range is 650 . Abby's z-score is then \((650-521) / 101=1.277\). This is at the 90 th percentile of the standard normal distribution.
Question: A motorist is driving without an automobile insurance policy in a region where law requires the motorist to have automobile insurance. The time in months until the motorist is charged for driving without automobile insurance is exponentially distributed with median 1.733 months.

Calculate the probability that the time until the motorist is charged is between the median and mean times until the motorist is charged.

Options:
A
\(0.5-e^{-1}\)
B
\(e^{-0.16}-0.5\)
C
\(e^{-1}\)
D
\(1-e^{-1}\)
E
\(0.5+e^{-1}\)

Correct answer: A

Solution: Let \(X=\) number of months until the motorist is charged, which is exponentially distributed with mean \(\beta\). Then,
\(0.5=e^{-1.733 / \beta}\)
\(\beta=2.5\)
\[
P(1.733<X<2.5)=\left(1-e^{-2.5 / 2.5}\right)-\left(1-e^{-1.733 / 2.5}\right)=1-e^{-1}-1+0.5=0.5-e^{-1}
\]
Question: Under a liability insurance policy, losses are uniformly distributed on \([0, b]\), where \(b\) is a positive constant. There is a deductible of \(b / 2\).

Calculate the ratio of the variance of the claim payment (greater than or equal to zero) from a given loss to the variance of the loss.

Options:
A
\(\quad 1: 8\)
B
\(\quad 3: 16\)
C
\(\quad 1: 4\)
D
\(\quad 5: 16\)
E
\(\quad 1: 2\)

Correct answer: D

Solution: Let \(X\) represent the loss.
The variance for a uniform distribution is the square of the interval length, divided by 12 . Thus, \(\operatorname{Var}(X)=\frac{b^{2}}{12}\).
Let \(C\) represent the claim payment from the loss. Then \(C=0\) for \(X<b / 2\) and \(C=X-b / 2\), otherwise. Then,
\(E(C)=\int_{0}^{b / 2} 0(1 / b) d x+\int_{b / 2}^{b}(x-b / 2)(1 / b) d x=0+(x-b / 2)^{2} /\left.(2 b)\right|_{b / 2} ^{b}=(b / 2)^{2} /(2 b)=b / 8\)
\(E\left(C^{2}\right)=\int_{0}^{b / 2} 0^{2}(1 / b) d x+\int_{b / 2}^{b}(x-b / 2)^{2}(1 / b) d x=0+(x-b / 2)^{3} /\left.(3 b)\right|_{b / 2} ^{b}=(b / 2)^{3} /(3 b)=b^{2} / 24\)
\(\operatorname{Var}(C)=b^{2} / 24-(b / 8)^{2}=5 b^{2} / 192\).
The ratio is \(\left[5 b^{2} / 192\right] /\left[b^{2} / 12\right]=60 / 192=5 / 16\).
Question: An insurer's quarterly profits are independent and normally distributed with mean 8 and standard deviation 24. A quarterly loss counts as a negative profit.

Let \(Z\) represent the standard normal random variable.
Determine the probability that the insurer makes an overall positive profit in a given fourquarter period.

Options:
A
\(\mathrm{P}[\mathrm{Z}>-3.00]\)
B
\(\mathrm{P}[Z>-1.50]\)
C
\(\mathrm{P}[Z>-0.67]\)
D
\(\mathrm{P}[Z>-0.33]\)
E
\(\quad \mathrm{P}[Z>-0.17]\)

Correct answer: C

Solution: The yearly mean \(=4 \times 8=32\)
The yearly variance \(=4 \times 24^{2}=2304\)
Yearly standard deviation \(=\sqrt{2304}\)
Answer \(=P[X>0]=P\left[Z>\frac{0-32}{\sqrt{2304}}\right]=P[Z>-0.67]\)
Question: A renter's insurance policy pays a percentage \(p\) of the amount by which each loss exceeds a deductible \(d\). Each policyholder reports a loss if and only if the loss exceeds the deductible.

The variance of a reported loss is \(v\).
Determine the variance of the claim payment resulting from a reported loss.

Options:
A
\(\quad p v\)
B
\(\quad p^{2} v\)
C
\(\quad p(v-d)\)
D
\(\quad p^{2}(v-d)\)
E
\(\quad(1-p)^{2} v\)

Correct answer: B

Solution: Let \(X\) represent the reported loss; note that \(X\) always exceeds \(d\) and \(\operatorname{Var}[X]=v\). The claim payment is \(p(X-d)\) since \(X\) always exceeds \(d\).
\(\operatorname{Var}[\) claim payment \(]=\operatorname{Var}[p(X-d)]=p^{2} \operatorname{Var}[X-d]=p^{2} \operatorname{Var}[X]=p^{2} v\).
Question: Four distinct integers are chosen randomly and without replacement from the first twelve positive integers. \(X\) is the random variable representing the second smallest of the four selected integers, and \(p\) is the probability function of \(X\).

Determine \(p(x)\) for \(x=2,3, \ldots, 10\).

Options:
A
\(\quad \frac{(x-1)(11-x)(12-x)}{495}\)
B
\(\quad \frac{(x-1)(11-x)(12-x)}{990}\)
C
\(\quad \frac{(x-1)(x-2)(12-x)}{990}\)
D
\(\frac{(x-1)(x-2)(12-x)}{495}\)
E
\(\frac{(10-x)(11-x)(12-x)}{495}\)

Correct answer: B

Solution: For a given \(x\), there are \(x-1\) choices for the smaller of the four integers and \(12-x\) choices for the two larger integers. Thus, there are \((x-1)\binom{12-x}{2}=\frac{(x-1)(12-x)(11-x)}{2}\) triples that satisfy the event. The total number of possible draws is \(\binom{12}{4}=495\) and the probability is \(\frac{(x-1)(12-x)(11-x)}{2} \frac{1}{495}=\frac{x-1)(12-x)(11-x)}{990}\).
Question: The random variables \(X_{1}, X_{2}, X_{3}\) represent the values of three independent investments. Each is normally distributed with \(\mathrm{E}\left(X_{i}\right)=100\) and \(\operatorname{Var}\left(X_{i}\right)=27 \mathrm{i} / 2, i=1,2,3\). Let \(\Phi(x)\) denote the cumulative distribution function of the standard normal variable.

Determine the probability that the average value of the three investments exceeds 106.

Options:
A
\(\quad \Phi(2)\)
B
\(\Phi(2 / 3)\)
C
\(1-\Phi(2 / 3)\)
D
\(1-\Phi(2)\)
E
\(\quad 1-\Phi(2 / \sqrt{3})\)

Correct answer: D

Solution: Let \(Y=\frac{1}{3}\left(X_{1}+X_{2}+X_{3}\right)\). Then \(Y\) is normal, with mean 100 and variance \(\left(\frac{1}{9}\right)\left(\frac{27}{2}\right)(1+2+3)=9\).
Thus \(\mathrm{P}[X>106]=\mathrm{P}\left[\mathrm{Z}>\frac{106-100}{\sqrt{9}}\right]=\mathrm{P}[Z>2]=1-P[Z<2]=1-\Phi(2)\).
Question: Each person in a large population independently has probability \(p\) of testing positive for diabetes, \(0<p<1\). People are tested for diabetes, one person at a time, until a test is positive. Individual tests are independent.

Determine the probability that \(m\) or fewer people are tested, given that \(n\) or fewer people are tested, where \(1 \leq m \leq n\).

Options:
A
\(\frac{m}{n}\)
B
\(\quad(1-p)^{m-n}\)
C
\(\quad 1-(1-p)^{m}\)
D
\(\frac{1-p^{m}}{1-p^{n}}\)
E
\(\quad \frac{1-(1-p)^{m}}{1-(1-p)^{n}}\)

Correct answer: E

Solution: Let \(X=\) number of people tested until a diabetic is found
Then \(X\) is geometric with success parameter \(p\).
\(P(X \leq m \mid X \leq n)=\frac{P(X \leq m)}{P(X \leq n)}=\frac{\sum_{k=1}^{m} P(X=k)}{\sum_{k=1}^{n} P(X=k)}=\frac{\sum_{k=1}^{m} p(1-p)^{k-1}}{\sum_{k=1}^{n} p(1-p)^{k-1}}=\frac{1-(1-p)^{m}}{1-(1-p)^{n}}\).
Question: An auto insurance policy has a deductible of 1 and a maximum claim payment of 5 . Auto loss amounts follow an exponential distribution with mean 2 .

Calculate the expected claim payment made for an auto loss.

Options:
A
\(0.5 e^{-2}-0.5 e^{-12}\)
B
\(2 e^{-\frac{1}{2}}-7 e^{-3}\)
C
\(2 e^{-\frac{1}{2}}-2 e^{-3}\)
D
\(2 e^{-\frac{1}{2}}\)
E
\(3 e^{-\frac{1}{2}}-2 e^{-3}\)

Correct answer: C

Solution: \(\int_{1}^{6}(x-1) 0.5 e^{-0,5 x} d x+\int_{6}^{\infty} 5(0.5) e^{-0.5 x} d x=-\left.(x-1) e^{-0.5 x}\right|_{1} ^{6}+\int_{1}^{6} e^{-0.5 x} d x-\left.5 e^{-0.5 x}\right|_{6} ^{\infty} =-5 e^{-3}+0-\left.2 e^{-0.5 x}\right|_{1} ^{6}+5 e^{-3}=-2 e^{-3}+2 e^{-1 / 2}\)
Question: Let \(X\) be a random variable that takes on the values \(-1,0\), and 1 with equal probabilities.

Let \(Y=X^{2}\).
Which of the following is true?

Options:
A
\(\operatorname{Cov}(X, Y)>0\); the random variables \(X\) and \(Y\) are dependent.
B
\(\operatorname{Cov}(X, Y)>0\); the random variables \(X\) and \(Y\) are independent.
C
\(\quad \operatorname{Cov}(X, Y)=0\); the random variables \(X\) and \(Y\) are dependent.
D
\(\operatorname{Cov}(X, Y)=0\); the random variables \(X\) and \(Y\) are independent.
E
\(\quad \operatorname{Cov}(X, Y)<0\); the random variables \(X\) and \(Y\) are dependent.

Correct answer: C

Solution: \(\operatorname{Cov}(X, Y)=E(X Y)-E(X) E(Y)=E\left(X^{3}\right)-E(X) E\left(X^{2}\right)\)
\(E(X)=E\left(X^{3}\right)=(1 / 3)(-1+0+1)=0\)
\(E\left(X^{2}\right)=(1 / 3)(1+0+1)=2 / 3\)
\(\operatorname{Cov}(X, Y)=0-0(2 / 3)=0\).
They are dependent, because
\(\operatorname{Pr}(X=0, Y=0)=\operatorname{Pr}\left(X=0, X^{2}=0\right)=\operatorname{Pr}(X=0)=1 / 3\)
\(\operatorname{Pr}(X=0) \operatorname{Pr}(Y=0)=(1 / 3)(1 / 3)=1 / 9 \neq 1 / 3\).
Question: A statistician models the size of unemployment claims that range from 0 to 1 using a probability density function proportional to the \(n^{\text {th }}\) root of the size of the claim, for some positive integer \(n\).

Determine the ratio of the \(30^{\text {th }}\) percentile to the \(20^{\text {th }}\) percentile of the size of an unemployment claim.

Options:
A
\(\sqrt[n]{1.5}\)
B
\(\quad(1.5)^{n}\)
C
\(\quad \sqrt[n]{(1.5)^{n+1}}\)
D
\(\quad \sqrt[n+1]{(1.5)^{n}}\)
E
\(\quad(1.5)^{n+1}\)

Correct answer: D

Solution: The density function is \(f(x)=c x^{1 / n}\). Then
\[
1=\int_{0}^{1} c x^{1 / n} d x=\left.\frac{c x^{1 / n+1}}{(1 / n)+1}\right|_{0} ^{1}=\frac{c}{(1 / n)+1}=\frac{c n}{1+n} \Rightarrow c=\frac{1+n}{n}
\]

Let \(x_{p}\) represent the \(p\) th percentile. Then,
\[
p=F\left(x_{p}\right)=\int_{0}^{x_{p}} \frac{(1+n) x^{1 / n}}{n} d x=\left.x^{1+1 / n}\right|_{0} ^{x_{p}}=\left(x_{p}\right)^{1+1 / n} \Rightarrow x_{p}=p^{n /(n+1)}
\]

The desired ratio is
\[
\frac{x_{0.3}}{x_{0.2}}=\frac{0.3^{n /(n+1)}}{0.2^{n /(n+1)}}=\left(1.5^{n}\right)^{1 /(n+1)}=\sqrt[n+1]{1.5^{n}}
\]
Question: The number of automobile accidents on any day in a city has a Poisson distribution with mean 4. The number of accidents on a given day is independent of the number of accidents on any other day.

Calculate the probability that at most one accident occurs in a three-day period.

Options:
A
\(13 e^{-12}\)
B
\(72 e^{-12}\)
C
\(85 e^{-12}\)
D
\(5 e^{-4}\)
E
\(13 e^{-4}\)

Correct answer: A

Solution: Let \(X_{1}, X_{2}, X_{3}\) denote the number of accidents in days 1,2 , and 3 , respectively.
Then, each has Poisson distribution with mean 4. Since the \(X_{i}\) s are assumed to be independent, \(X=X_{1}+X_{2}+X_{3}\) has a Poisson distribution with mean \(3(4)=12\). Then, \(\mathrm{P}[X<2]=\mathrm{P}[X=0]+\mathrm{P}[X=1]=\frac{e^{-12} 12^{0}}{0!}+\frac{e^{-12} 12^{1}}{1!}=13 e^{-12}\).
Question: In a large population of patients, \(20 \%\) have cancer. Of those who have cancer, \(8 \%\) have stage IV cancer.

Patients are tested one at a time, at random, until five patients with stage IV cancer are found.

Let \(N\) represent the number of patients tested. Let \(C\) represent the number of patients tested who have cancer.

Determine the probability function, \(p_{N, C}(n, c)\), for integers \(n\) and \(c\) such that \(5 \leq c \leq n\).

Options:
A
\(\quad p_{N, C}(n, c)=\frac{(n-1)!}{(n-c)!(c-5)!4!}(0.8)^{n-c}(0.184)^{c-5}(0.016)^{5}\)
B
\(\quad p_{N, C}(n, c)=\frac{(n-1)!}{(n-c)!(c-5)!4!}(0.8)^{n-c}(0.12)^{c-5}(0.08)^{5}\)
C
\(\quad p_{N, C}(n, c)=\frac{n!}{(n-c)!(c-5)!5!}(0.8)^{n-c}(0.184)^{c-5}(0.016)^{5}\)
D
\(\quad p_{N, C}(n, c)=\frac{(n-5)!}{(n-c)!(c-5)!}(0.8)^{n-c}(0.184)^{c-5}(0.016)^{5}\)
E
\(\quad p_{N, C}(n, c)=\frac{(n-1)!}{(n-c-1)!(c-4)!4!}(0.8)^{n-c-1}(0.12)^{c-4}(0.08)^{5}\)

Correct answer: A

Solution: A given person in this population has no cancer with probability \(1-0.2=0.8\), cancer other than stage IV with probability \(0.2(1-0.08)=0.184\), and stage IV cancer with probability \(0.2(0.08)=0.016\).

We need to find the probability that i) out of the first \(n-1\) patients, \(n-c\) have no cancer, \(c-5\) have cancer other than stage IV, and 4 have stage IV cancer, and ii) the \(n\)th patient has stage IV cancer.

Keeping in mind that we are given that the patients are independent, we multiply the individual probabilities together. We then multiply this product by the number of possible orders in which these results need to occur, to get \(p_{N, C}(n, c)=\frac{(n-1)!}{(n-c)!(c-5)!4!}(0.8)^{n-c}(0.184)^{c-5}(0.016)^{5}\).
Question: The length of time, \(T\), in months, taken by relatives to file for a death benefit has density function
\[
f(t)= \begin{cases}\frac{4 \beta^{4}}{t^{5}}, & \text { for } t>\beta, 0<\beta<3 \\ 0, & \text { otherwise. }\end{cases}
\]

Calculate the probability that the relatives of a policyholder will not file for the death benefit in the next four months, given that the policyholder died three months ago and the relatives have not yet filed for the death benefit.

Options:
A
\(\frac{81}{2401}\)
B
\(\frac{256}{2401}\)
C
\(\frac{81}{256}\)
D
\(\quad \frac{2401}{81}\left(\frac{81-\beta^{4}}{2401-\beta^{4}}\right)\)
E
\(\quad \frac{2401}{256}\left(\frac{256-\beta^{4}}{2401-\beta^{4}}\right)\)

Correct answer: A

Solution: \(F(t)=4 \beta^{4} \int_{\beta}^{t} r^{-5} d r=1-\left(\frac{\beta}{t}\right)^{4}\), for \(0<\beta \leq t\)
\(P[T>4+3 \mid T>3]=\frac{P[T>7]}{P[T>3]}=\frac{1-F(7)}{1-F(3)}=\frac{\left(\frac{\beta}{7}\right)^{4}}{\left(\frac{\beta}{3}\right)^{4}}=\frac{81}{2401}\)
Question: A continuous random variable, \(X\), has density function \(f(x)\) where \(f(x)= \begin{cases}\frac{x-1}{4}, & \text { for } 1<x<3 \\ \frac{5-x}{4}, & \text { for } 3 \leq x<5 \\ 0, & \text { otherwise } .\end{cases}\)

Determine which of the following expressions equals \(\mathrm{E}(|X-2|)\).

Options:
A
\(\quad \frac{1}{4} \int_{1}^{2}(2-x)(x-1) d x+\frac{1}{4} \int_{2}^{3}(x-2)(x-1) d x+\frac{1}{4} \int_{3}^{5}(2-x)(5-x) d x\)
B
\(\quad \frac{1}{4} \int_{1}^{3}(2-x)(x-1) d x+\frac{1}{4} \int_{3}^{5}(x-2)(5-x) d x\)
C
\(\quad \frac{1}{4} \int_{1}^{2}(x-2)(x-1) d x+\frac{1}{4} \int_{2}^{3}(2-x)(x-1) d x+\frac{1}{4} \int_{3}^{5}(2-x)(5-x) d x\)
D
\(\quad \frac{1}{4} \int_{1}^{3}(x-2)(x-1) d x+\frac{1}{4} \int_{3}^{5}(x-2)(5-x) d x\)
E
\(\quad \frac{1}{4} \int_{1}^{2}(2-x)(x-1) d x+\frac{1}{4} \int_{2}^{3}(x-2)(x-1) d x+\frac{1}{4} \int_{3}^{5}(x-2)(5-x) d x\)

Correct answer: E

Solution: \(E[|X-2|]=\int_{1}^{5}|x-2| f(x) d x=\int_{1}^{2}(2-x) f(x) d x \int_{2}^{3}(x-2) f(x) d x \int_{3}^{5}(x-2) f(x) d x\)
\(=\int_{1}^{2}(2-x) \frac{x-1}{4} d x \int_{2}^{3}(x-2) \frac{x-1}{4} d x \int_{3}^{5}(x-2) \frac{5-x}{4} d x\)
Question: The waiting time for the first claim from a good driver and the waiting time for the first claim from a bad driver are independent and follow exponential distributions with means 6 years and 3 years, respectively.

Calculate the probability that the first claim from a good driver will be filed within 3 years and the first claim from a bad driver will be filed within 2 years.

Options:
A
\(\frac{1}{18}\left(1-e^{-2 / 3}-e^{-1 / 2}+e^{-7 / 6}\right)\)
B
\(\frac{1}{18} e^{-7 / 6}\)
C
\(1-e^{-2 / 3}-e^{-1 / 2}+e^{-7 / 6}\)
D
\(1-e^{-2 / 3}-e^{-1 / 2}+e^{-1 / 3}\)
E
\(1-\frac{1}{3} e^{-2 / 3}-\frac{1}{6} e^{-1 / 2}+\frac{1}{18} e^{-7 / 6}\)

Correct answer: C

Solution: For a good driver, the probability is \(1-e^{-3 / 6}\) and for a bad driver, the probability is \(1-e^{-2 / 3}\). The probability of both is the product, \(\left(1-e^{-3 / 6}\right)\left(1-e^{-2 / 3}\right)=1-e^{-1 / 2}-e^{-2 / 3}+e^{-7 / 6}\).