Category Archives: Practice Problems

Exam P Practice Problem 98 – flipping coins

By Dan Ma on November 11, 2017 | Leave a comment

Problem 98-A

Coin 1 is an unbiased coin, i.e. when flipping the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when flipping the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.

Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2856$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3060$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.3295$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.3564$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.3690$

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Problem 98-B

Box 1 contains 3 red balls and 1 white ball while Box 2 contains 2 red balls and 2 white balls. The two boxes are identical in appearance. One of the boxes is chosen at random. A ball is sampled from the chosen box with replacement until a white ball is obtained.

Determine the probability that the chosen box is Box 1 if the first white ball is observed on the 6th draw.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.7530$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.7632$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.7825$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7863$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7915$

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Posted in: Practice Problems | Tagged: Actuarial Exam, Bayes' Theorem, CAS Exam 1, Exam P, Exam P Practice Problems, Geometric distribution, SOA Exam P, SOA General Probability

Exam P Practice Problem 97 – Variance of Claim Sizes

By Dan Ma on November 6, 2017 | Leave a comment

Problem 97-A

For a type of insurance policies, the following is the probability that the size of claim is greater than $x$ .

$\displaystyle P(X>x) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ x \le 0 \\ \text{ } & \text{ } \\ \displaystyle \biggl(1-\frac{x}{10} \biggr)^6 &\ \ \ \ \ \ 0<x<10 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ x \ge 10 \end{array} \right.$

Calculate the variance of the claim size for this type of insurance policies.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \frac{10}{7}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \frac{75}{49}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \frac{95}{49}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \frac{15}{7}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \frac{25}{7}$

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Problem 97-B

For a type of insurance policies, the following is the probability that the size of a claim is greater than $x$ .

$\displaystyle P(X>x) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ x \le 0 \\ \text{ } & \text{ } \\ \displaystyle \biggl(\frac{250}{x+250} \biggr)^{2.25} &\ \ \ \ \ \ x>0 \\ \end{array} \right.$

Calculate the expected claim size for this type of insurance policies.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 200.00$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 203.75$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 207.67$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 217.32$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 232.74$

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Posted in: Insurance and Risk Management, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, Expected Value, Probability, SOA Exam P, Variance

Exam P Practice Problem 96 – Expected Insurance Payment

By Dan Ma on April 6, 2016 | Leave a comment

Problem 96-A

An insurance policy is purchased to cover a random loss subject to a deductible of 1. The cumulative distribution function of the loss amount $X$ is:

$\displaystyle F(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ x<0 \\ \text{ } & \text{ } \\ \displaystyle \frac{3}{25} \ x^2 - \frac{2}{125} \ x^3 &\ \ \ \ \ \ 0 \le x<5 \\ \text{ } & \text{ } \\ 1 &\ \ \ \ \ \ 5<x \end{array} \right.$

Given a random loss $X$ , determine the expected payment made under this insurance policy.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.50$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.54$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.72$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 4.63$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 6.26$

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Problem 96-B

An insurance policy is purchased to cover a random loss subject to a deductible of 2. The density function of the loss amount $X$ is:

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{8} \biggl(1- \frac{1}{4} \ x + \frac{1}{64} \ x^2 \biggr) &\ \ \ \ \ \ 0<x<8 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

Given a random loss $X$ , what is the expected benefit paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.51$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.57$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.63$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.60$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.00$

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Posted in: Insurance and Risk Management, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, Expected Insurance Payment, Expected Value, Insurance and risk management, Probability, SOA Exam P

Exam P Practice Problem 95 – Measuring Dispersion

By Dan Ma on March 8, 2016 | Leave a comment

Problem 95-A

The lifetime (in years) of a machine for a manufacturing plant is modeled by the random variable $X$ . The following is the density function of $X$ .

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{2500} \ (100x-20x^2+ x^3) &\ \ \ \ \ \ 0<x<10 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

Calculate the standard deviation of the lifetime of such a machine.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.0$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.7$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.0$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.0$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.9$

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Problem 95-B

The travel time to work (in minutes) for an office worker has the following density function.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{1000} \ (50-5x+\frac{1}{8} \ x^2) &\ \ \ \ \ \ 0<x<20 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ \text{otherwise} \end{array} \right.$

Calculate the variance of the travel time to work for this office worker.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.87$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.00$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6.50$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.75$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15.00$

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$\copyright \ 2016 \ \ \text{Dan Ma}$

Posted in: Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, SOA Exam P, Standard Deviation, Variance

Exam P Practice Problem 94 – Tracking High School Students

By Dan Ma on January 1, 2016 | 2 Comments

Problem 94-A

A researcher tracked a group of 900 high school students taking standardized tests in math and chemistry. Some of the students were given after-school tutoring before the tests (in both subjects) and the rest of the students received no tutoring. The following information is known about the test results:

510 of the students passed math test and 475 of the students passed chemistry test.
Of the students who failed both subjects, there were 20% more students who did not receive tutoring than there were students who received tutoring.
Of the students who failed chemistry and had tutoring, there were 99 more students who failed math than there were students who passed math.
Of the students who failed chemistry and had no tutoring, there were 4 more students who failed math than there were students who passed math.
There were 126 students who failed math and passed chemistry.
249 students passed math and had received no tutoring.

Determine the probability that a randomly selected student from this group had tutoring given that the student passed both subjects.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.6810$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.6828$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.6859$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.6877$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.6989$

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Problem 94-B

An insurance company tracked a group of 800 insureds for 2 years. It was found that 560 of the insureds had no claims in year 1 and 380 of the insureds had no claims in year 2. Of the insureds who had no claims in both years, there were four times as many male insureds than there were female insureds. Furthermore, there were 230 male insureds who had no claims in year 2 and there were 53 females insureds who had claims in both years. It is also known that there were 85 male insureds who had claims in year 1.

Determine the number of insureds who had no claims in year 1 but had claims in year 2.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 320$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 347$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 369$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 420$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 560$

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Posted in: General Probability, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, CAS General Probability, Exam P, Exam P Practice Problems, Probability, SOA Exam P, SOA General Probability

Exam P Practice Problem 93 – Determining Average Claim Frequency

By Dan Ma on October 30, 2015 | Leave a comment

Problem 93-A

An actuary performs a claim frequency study on a group of auto insurance policies. She finds that the probability function of the number of claims per week arising from this set of policies is $P(N=n)$ where $n=1,2,3,\cdots$ . Furthermore, she finds that $P(N=n)$ is proportional to the following function:

$\displaystyle \frac{e^{-2.9} \cdot 2.9^n}{n!} \ \ \ \ \ \ \ n=1,2,3,\cdots$

What is the weekly average number of claims arising from this group of insurance policies?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.900$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 3.015$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3.036$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.069$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.195$

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Problem 93-B

Let $N$ be the number of taxis arriving at an airport terminal per minute. It is observed that there are at least 2 arrivals of taxis in each minute. Based on a study performed by a traffic engineer, the probability $P(N=n)$ is proportional to the following function:

$\displaystyle \frac{e^{-2.9} \cdot 2.9^n}{n!} \ \ \ \ \ \ \ n=2,3,4,\cdots$

What is the average number of taxis arriving at this airport terminal per minute?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.740$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 2.900$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3.339$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.489$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.692$

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Posted in: Insurance and Risk Management, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Conditional Mean, Exam P, Exam P Practice Problems, Insurance and risk management, Poisson distribution, Probability, SOA Exam P

Exam P Practice Problem 92 – Expected Claim Payment

By Dan Ma on October 29, 2015 | Leave a comment

Problem 92-A

The size of a claim that an auto insurance company pays out is modeled by a random variable with the following density function:

$\displaystyle f(x)=\frac{1}{5000} \ (100-x) \ \ \ \ \ \ \ 0<x<100$

By subjecting the insured to a deductible of 12 per claim, what is the expected reduction in claim payment?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 9.50$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 10.6$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 11.1$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 11.8$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 12.0$

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Problem 92-B

The size of a claim that an auto insurance company pays out is modeled by a random variable with the following density function:

$\displaystyle f(x)=\frac{1}{3200} \ (80-x) \ \ \ \ \ \ \ 0<x<80$

By subjecting the insured to a deductible of 10 per claim, by what percent is the expected claim payment reduced?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 10 \%$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 15 \%$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 22 \%$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 25 \%$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 33 \%$

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Posted in: Insurance and Risk Management, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, Expected Insurance Payment, Insurance and risk management, Probability, SOA Exam P

Exam P Practice Problem 91 – Reviewing a Group of Policyholders

By Dan Ma on July 8, 2014 | Leave a comment

Problem 91-A

A life insurance actuary reviewed a group of policyholders whose policies or contracts were inforce as of last year. The actuary found that 12% of the policyholders who had only a life insurance policy did not survive to this year and that 7.5% of the policyholders who had only an annuity contract did not survive to this year. The actuary also found that 5.9% of the policyholders who had both a life insurance policy and an annuity contract did not survive to this year.

In this group of policyholders, 65% of the policyholders had a life insurance policy and 57% of the policyholders had an annuity contract. Furthermore, each policyholder in this group either had a life insurance policy or an annuity contract.

What is the percentage of the policyholders that did not survive to this year?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 7.8 \%$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 9.0 \%$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 12.0 \%$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 13.4 \%$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 25.4 \%$

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Problem 91-B

A sport coach in a university tracks a group of athletes. The coach finds that 36% of the athletes who play soccer only are first year university students and that 20% of the athletes who are involved only in track and field are first year university students. The coach also discovers that 27% of the athletes participates in both soccer and track and field are first year university students.

According to university records, 45% of the athletes in this group play soccer and 68% of the athletes in this group participate in track and field. Each of the athletes in this group either plays soccer or participates in track and field.

Out of this group of athletes, what is the percentage of the athletes that are not first year university students?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 64 \%$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 67 \%$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 70 \%$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 74 \%$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 80 \%$

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Posted in: General Probability, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, CAS General Probability, Exam P, Exam P Practice Problems, Probability, SOA Exam P, SOA General Probability

Exam P Practice Problem 90 – Insurance Benefits

By Dan Ma on July 5, 2014 | Leave a comment

Problem 90-A

A random loss follows an exponential distribution with mean 20. An insurance reimburses this random loss up to a benefit limit of 30.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 4.5$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 5.1$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 6.3$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 8.5$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 11.2$

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Problem 90-B

A random loss follows an exponential distribution with mean 100. An insurance reimburses this random loss up to a benefit limit of 200.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 12.6$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 13.5$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 24.6$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 40.6$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 40.7$

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Posted in: Exponential distribution, Insurance and Risk Management, Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, Expected Insurance Payment, Exponential distribution, Insurance and risk management, Median, Probability, SOA Exam P

Exam P Practice Problem 89 – Finding Median

By Dan Ma on July 4, 2014 | 3 Comments

Problem 89-A

The random variables $X$ and $Y$ have the following joint density function.

$\displaystyle f(x,y)=\frac{1}{32} \ (4-x) \ \ \ \ \ \ \ 0<x<4,\ 0<y<4$

Suppose that $m$ is the median of $X+Y$ . Which of the following is true about $m$ ?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.5<m<3$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 2<m<3$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3<m<4$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.5<m<4$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.5<m<4.5$

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Problem 89-B

The random variable $X$ has the following density function.

$\displaystyle f(x)=\frac{3}{16000} \ (400-x^2) \ \ \ \ \ \ \ 0<x<20$

Suppose that $m$ is the median of $X$ . Which of the following is true about $m$ ?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 6<m<7$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 5.5<m<6$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 5.5<m<6.5$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 7<m<8$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 7<m<9$

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Posted in: Practice Problems | Tagged: Actuarial Exam, CAS Exam 1, Exam P, Exam P Practice Problems, Median, Probability, SOA Exam P

SOA Exam P / CAS Exam 1

Category Archives: Practice Problems

Exam P Practice Problem 98 – flipping coins

Exam P Practice Problem 97 – Variance of Claim Sizes

Exam P Practice Problem 96 – Expected Insurance Payment

Exam P Practice Problem 95 – Measuring Dispersion

Exam P Practice Problem 94 – Tracking High School Students

Exam P Practice Problem 93 – Determining Average Claim Frequency

Exam P Practice Problem 92 – Expected Claim Payment

Exam P Practice Problem 91 – Reviewing a Group of Policyholders

Exam P Practice Problem 90 – Insurance Benefits

Exam P Practice Problem 89 – Finding Median

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