This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 4 / Exam C, authored by Mr. Stolyarov. This is Section 31 of the Study Guide. See an index of all sections by following the link in this paragraph.
Some of the problems in this section were designed to be similar to problems from past versions of Exam 4/C, offered jointly by the Casualty Actuarial Society and the Society of Actuaries. They use original exam questions as their inspiration – and the specific inspiration is cited to give students an opportunity to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.
Exam C Sample Questions and Solutions from the Society of Actuaries.
Original Problems and Solutions from The Actuary’s Free Study Guide
Problem S4C31-1. Similar to Question 54 of the Exam C Sample Questions from the Society of Actuaries. The following data is obtained regarding a sample of 1000 gambling losses (X) at a casino:
Range of Losses……….Number of Losses
0 ≤ x ≤ 25……………………..124
You are using an exponential distribution with mean θ to model the data in this sample. Estimate θ by using percentile matching at the 49.3rd percentile. Assume that losses are uniformly distributed within each range specified.
Solution S4C31-1. We note that 124+174+195 = 493, so the 493rd loss of 1000 occurs at x = 100. Thus, we want to find θ such that SX(100) = 1-0.493 = 0.507 = e-100/θ → θ = -100/ln(0.507) = θ = 147.2224406.
Problem S4C31-2. Similar to Question 57 of the Exam C Sample Questions from the Society of Actuaries.
You have collected the following data on a sample of 100 giraffes:
Height of Giraffe (cm.)………….Number of Giraffes
Assume that height is uniformly distributed within each specified range. Estimate the third raw moment of this distribution of giraffe heights.
Solution S4C31-2. For each specified range, we use the following formula from Section 1:
μ’k = E(Xk) = -∞∞∫xk*f(x)dx, with the bounds modified according to each range. Here, k = 3.
The third raw moment will be the probability-weighted average of four integrals, with one integral pertaining to each of the four ranges.
For the 0-300 range the integral is 0300∫x3*(1/300)dx = x4/1200│0300 = 6750000.
For the 300-400 range the integral is 300400∫x3*(1/100)dx = x4/400│300400 = 43750000.
For the 400-500 range the integral is 400500∫x3*(1/100)dx = x4/400│400500 = 92250000.
For the 500-600 range the integral is 500600∫x3*(1/100)dx = x4/400│500600 = 167750000.
Thus, the third raw moment is (34/100)*6750000 + (14/100)*43750000 + (36/100)*92250000 + (16/100)*167750000 = μ’3 = 68,470,000.
Problem S4C31-3. Similar to Question 71 of the Exam C Sample Questions from the Society of Actuaries. Every year, some people come to disbelieve some popular superstitions. You are performing a test of the null hypothesis that the observed frequencies of the numbers of popular superstitions disbelieved this year accord with the historical frequencies. This is the data you gathered.
Number of Superstitions Disbelieved
0: 240 people
1: 23 people
2: 5 people
3+: 2 people
Historically, in any given year, 0.93 of the population will not abandon any of their superstitions in a given year; 0.04 of the population will abandon 1 superstition, 0.02 of the population will abandon 2 superstitions, and 0.01 of the population will abandon 3 or more superstitions.
What is the result of the hypothesis test? Use the table of values for the Chi-square distribution associated with the Exam 4 / C Tables (p. 4).
(a) Reject at the 0.005 significance level.
(b) Reject at the 0.010 significance level, but not at the 0.005 level.
(c) Reject at the 0.025 significance level, but not at the 0.010 level.
(d) Reject at the 0.050 significance level, but not at the 0.025 level.
(e) Do not reject at the 0.050 significance level.
The sample contains a total of 240 + 23 + 5 + 2 = 270 members. Since there are four categories, the Chi-square statistic will have 4-1 = 3 degrees of freedom.
First, we find the Chi-square statistic: χ2 = i=1t∑((Xi – npi)2/npi).
(240 – 0.93*270)2/(0.93*270) + (23 – 0.04*270)2/(0.04*270) + (5 – 0.02*270)2/(0.02*270) + (2 – 0.01*270)2/(0.01*270) = 14.4832736.
We examine p. 4 of the Exam 4 / C Tables in the row corresponding to 3 degrees of freedom, and we find that the Chi-square statistic value for P = 0.995 (significance level 0.005) is 12.838. Since 14.4832736 > 12.838, we reject the null hypothesis at the 0.005 significance level. This means that the correct answer is (a).
Problem S4C31-4. Similar to Question 81 of the Exam C Sample Questions from the Society of Actuaries. There is a 0.36 probability that X follows a Pareto distribution with α = 4 and θ = 99. There is a 0.64 probability that X follows an exponential distribution with θ = 150. You estimate X using the following two-step procedure:
1. A random number from 0 to 1 is generated, indicating to which distribution in this two-point mixture X belongs. The lower random numbers get assigned to the exponential distributions.
2. Another random number from 0 to 1 is generated, where lower numbers correspond to lower values of X, and the percentile of the random number within the [0, 1] interval corresponds to the percentile of X within the distribution.
One such application of this procedure yields the following randomly generated numbers, in order: 0.67, 0.35. What is the estimate x of X corresponding to these random numbers?
Relevant properties for exponential distributions: SX(x) = e-x/θ.
Relevant properties for Pareto distributions: S(x) = θα/(x + θ)α.
Solution S4C31-4. The lower random numbers get assigned to the exponential distributions, and the exponential distribution has a probability of 0.64 of occurring. Since the first randomly generated number is 0.67 > 0.64, X here follows the Pareto distribution.
The second randomly generated number of 0.35 indicates that x is at the 35th percentile of this Pareto distribution. Hence, SX(x) = 1 – 0.35 = 0.65 = 994/(x + 99)4 → (x + 99)4 = 994/0.65 → x = 99/0.650.25 – 99 = x = 11.25717363.
Problem S4C31-5. Similar to Question 84 of the Exam C Sample Questions from the Society of Actuaries. In an attempt to discourage gambling, the Gamblers’ Charity Fund pays gamblers an incentive payment I, consisting of certain fraction p of the amount by which their annual gambling losses are less than $1000. It is known that E(I) = 200. Gambling losses (X) follow an exponential distribution with θ = 1400. Find the value of p.
Relevant properties for exponential distributions: E(X Λ K) = θ(1 – e-K/θ).
I = p(1000 – x) if 0
I = 0 if x ≥ 1000.
Thus, E(I) = p(1000 – E(X Λ 1000)).
E(I) = p(1000 – 1400(1 – e-1000/1400)).
E(I) = 285.3583234p.
200 = 285.3583234p
Thus, p = 0.7008731956.
See other sections of The Actuary’s Free Study Guide for Exam 4 / Exam C.