CA VEE Mathematical Statistics

The net dollar amount in bank transactions for one day follows a normal distribution with mean µ. A random sample of 5 days: 265, -147, -79, 341, 104 p = 0.95 p = 0.975 v = 4 2.132 2.776 v = 5 2.015 2.571 Construct a 95% confidence interval. A. (-165.3, 358.9) B. (-146.0, 339, 6) C. (-132.7, 326.3) D. (-119.2, 312.8) E. (-104.5, 298.1)

A. (-165.3, 358.9)

One claim amount, , is used to test a set of hypotheses regarding . This test is most powerful when the rejection region is derived from L(8) / L(7) <= r Determine the rejection region A. 7.5 + 9ln(r) B. 15 + 18ln(r) C. 7.5 - 9ln(r) D. -56.5 + 9ln(r) E. -56.5 - 9ln(r)

A. 7.5 + 9ln(r)

A contingency table has been created to test the hypothesis that the number of vehicles per policyholder is independent of the policyholder's rating classification. There are four possible rating classifications in the table crossed with the number of vehicles per policyholder. The number of vehicles per policyholder is limited to a value of 1, 2, 3, or 4. Calculate the degrees of freedom to be used in this chi-squared test of independence. A. 9 B. 10 C. 12 D. 14 E. 15

A. 9

Which of the following statements about hypothesis testing are true? A type I error occurs if Ho is rejected when it is true. A type II error occurs if Ho is rejected when it is true. The power of a test is the probability of failing to reject Ho when it is false. A. I B. II C. III D. I and III E. II and III

A. I

Actuaries speculate that when small vehicles are involved in accidents, the chances of serious injury are higher than that for larger sized vehicles. A random sample of 1,000 accidents is classified according to severity of the injuries and the size of the car. The results are: Calculate the chi-square test. A. Less than 6.6 B. 6.6-6.8 C. 6.8-7.0 D. 7.0-7.2 E. At least 7.2

A. Less than 6.6

A insurance company sends a survey to a group of claimants. Out of 52 responses, 23 indicated that their claim was handled to their satisfaction. Assuming the responses were independent, construct a 92% confidence interval for the proportion of claimants who are satisfied with the handling of their claim. A. (0.29, 0.59) B. (0.32, 0.56) C. (0.35, 0.54) D. (0.38, 0.50) E. (0.40, 0.48)

B. (0.32, 0.56)

The following observations were made from a random sample taken from a normal distribution: 2, 0, 4, 4, 6, 3, 1, 5, 6, 9 p = 0.9 p = 0.95 p = 0.975 v = 8 1.397 1.860 2.306 v = 9 1.383 1.833 2.262 Construct a 90% confidence interval. A. (2.43, 5.57) B. (2.45, 5.55) C. (2.82, 5.18) D. (2.83, 5.17) E. (2.84, 5.16)

B. (2.45, 5.55) 4 +/- 1.833 (√7.111 / √10)

The lifetime of a certain brand of lightbulbs are assumed to be normally distributed. A researcher took a random sample of these lightbulbs and observed the following failure times: 17, 19, 24, 24, 25, 34, 40 p = 0.025 p = 0.05 p = 0.95 p = 0.975 v = 6 1.237 1.635 12.592 14.449 v = 7 1.690 2.167 14.067 16.013 Construct the 95% confidence interval for the variance of the lightbulb lifetime. A. (24.9, 236.0) B. (27.6, 322.4) C. (28.4, 184.1) D. (31.7, 243.9) E. (33.5, 317.2)

B. (27.6, 322.4) (n-1)s^2 / X^2

Two independent random samples drawn from normal distributions have equal variances and means and . You are given The sample mean from the first sample is 16.9. The sample mean from the second sample is 22.1. The number of observations from both samples combined is 17. The upper bound of the 90% confidence interval for is 1.44. Determine the lower bound of the 95% confidence interval. A. Less than -14 B. -14 - -12 C. -12 - -10 D. -10 - -8 E. At least -8

B. -14 to -12

You are given the following five independent observations of inter-event times: 8, 9, 9, 10, 11 Calculate the test statistic. A. Less than 0.018 B. 0.018-0.019 C. 0.019-0.020 D. 0.020-0.021 E. At least 0.021

B. 0.018-0.019

A random sample is drawn from a normal distribution with variance 12. Let X2 be the 100th percentile of a chi-squared random variable. p = 0.5 p = 0.6 p = 0.7 v = 8 7.344 8.351 9.524 v = 9 8.343 9.414 10.656 Find the probability that sum(x=1 to 10) (x-Xbar)^2 is greater than 120. A. Less than 0.3 B. 0.3-0.4 C. 0.4-0.5 D. 0.5-0.6 E. At least 0.6

B. 0.3-0.4 P(X2 > 120/sigma2) P(X2 > 120/12) P(X2>10) When v = 9, this occurs between 0.6 and 0.7 that the probability is less than these values. Invert to get the probability greater than.

Claim amounts at an insurance company are independent of one another, modeled by a normal distribution with mean 109 and variance 676. Calculate the probability that a random sample of 25 claim amounts average between 100 and 110. A. 0.48 B. 0.53 C. 0.54 D. 0.67 E. 0.68

B. 0.53 Mean = 109 Variance = 676/25 = 27.04 100 - 109 / √17.04 = 0.19 110 - 109 / √17.04 = 1.73 P(0.19 < Z < 1.73) = 0.5335

A random sample is drawn from a normal distribution with mean µ. - X is the sample mean - S is the unbiased standard deviation - V = X-µ / S/√n - The 20th percentile of V is -0.8583 Calculate P(|V| < 0.8583) A. 0.5 B. 0.6 C. 0.7 D. 0.8 E. 0.9

B. 0.6 V follows a t-distribution. Since the 20th percentile is the given value, 20% of the distribution is below the -0.8583. Since the distribution is symmetric, 20% is above 0.8583. Therefore, the area between these values is the remainder, 0.60.

30,500 residents live in Cape City. A random sample consisting of 3% of the residents was taken. The individuals in the sample were asked about their employment status. 758 reported that they were employed. Find the lower bound of the 98% confidence interval for the proportion of employed residents in Cape City A. Less than 0.79 B. 0.79-0.80 C. 0.80-0.81 D. 0.81-0.82 E. At least 0.82.

B. 0.79 - 0.80

For a random sample of size 13 from a normal distribution with mean µ, you are given the following regarding the observations: Sum (x-xbar)^2 = 77.8 The width of the kth% confidence interval for µ = 2.7005. p = 0.90 p = 0.93 p = 0.96 v = 12 1.356 1.580 1.912 v = 13 1.350 1.572 1.899 Find the value of K A. Less than 0.91 B. 0.91-0.93 C. 0.93-0.95 D. 0.95-0.97 E. At least 0.97

B. 0.91-0.93

An auto insurer has a random sample of 800 claims. The claim amounts have a mean of 2,000 and a variance of 136,875. Using the Central Limit Theorem, approximate the 99th percentile of the total claim amount from the sample. A. 1608000 B. 1624000 C. 1776000 D. 2543000 E. 4867000

B. 1624000 Mean = 2000 Variance = 136875/800 0.99 = P(X <= pi(0.99) / 800) = (x - 2000 / √136875/800) < (pi(99) / 800 - 2000 / √136875/800) A Z score of 2.326 corresponds with 99% 2.326 = p/800 - 2000 / √136875/800 p = 1624340

A claim department has operated under the following assumptions about expected automobile claims: 50% of the claims are for cars 20% of the claims are for motorcycles 15% of the claims are for vans 15% of the claims are for trucks In conducting a chi-squared goodness-of-fit test on the claim department's assumptions, 100 independent claims were recorded as follows: Cars 40 Motorcycles 24 Vans 17 Trucks 19 Total 100 Calculate the chi-squared statistic. A. Less than 2 B. 2-5 C. 5-9 D. 9-15 E. Greater than 15

B. 2-5

During the noon hour, a bank serves 4 personal line customers, 5 business line customers, and 1 new customer. By simple random sampling, 6 of the customers are chosen to take a survey. What is the probability that exactly 2 personal line customers, 3 business line customers, and 1 new customer take the survey? A. 1/5 B. 2/7 C. 3/7 D. 3/5 E. 5/7

B. 2/7 (4C2 x 5C3 x 1C1) / (10C6)

There are 97 men and 3 women in an organization. A simple random sample of 5 people is taken to form a committee. What is the probability that the committee includes all 3 women? A. 2.5 e-5 B. 6.2 e-5 C. 10.3 e -5 D. 30.9 e-5 E. 61.8 e-5

B. 6.2 e-5 (97C2 x 3C3) / (100C5)

You are told that the most powerful test for a certain significance level has a rejection region derived from: L1/L2 <= r Which translates to a critical value of 1.1645. Determine r. A. Less than 65,000 B. 65,000-66,000 C. 66,000-67,000 D. 67,000-68,000 E. At least 68,000

B. 65,000 - 66,000

A class has 8 boys and 7 girls. Using simple random sampling, the teacher selects 3 of the children for a special project. Calculate the probability that two girls and one boy are selected. A. 0.361 B. 0.365 C. 0.369 D. 0.373 E. 0.377

C 0.369 (7C2 c 8C1) / (15C3)

An agency has ten left-handers and twenty right-handers. Five individuals are chosen for a survey using simple random sampling. Calculate the probability all five have the same handedness. A. 0.04 B. 0.07 C. 0.11 D. 0.14 E. 0.22

C. 0.11 Five left-handers plus five right-handers (10C5 + 20C5) / (30C5)

Claims filed under a group of auto insurance policies have normally distributed amounts with mean 19,400 and standard deviation 5,000. A random sample of 25 claims are taken. What is the probability that the average of the 25 claim amounts exceeds 20,000? A. 0.01 B. 0.15 C. 0.27 D. 0.33 E. 0.45

C. 0.27 Mean = 19400 Var = 5000^2/25 = 1000000 P(20000 - 19400 / √1000000) P(Z > 0.6) = 0.2743

Among 12 mice housed in a laboratory, three are male. Five mice are chosen using simple random sampling. Find the probability that exactly two male mice are chosen. A. 0.18 B. 0.28 C. 0.32 D. 0.38 E. 0.43

C. 0.32 (3C2 x 9C3) / (12C5)

A manufacturer makes golf balls with a mean weight of 1.62 ounces and a standard deviation of 0.05 ounces. A random sample of 100 balls are taken. Using the Central Limit Theorem, approximate the probability that the sample mean weight of the 100 balls is between 1.615 and 1.625 ounces. A. 0.18 B. 0.34 C. 0.68 D. 0.94 E. 0.96

C. 0.68 This is within one standard deviation of the mean.

Two independent random samples drawn from normal distributions have equal variances and means and . You are given The first sample of size 3 has a sample mean of 12 and a sample variance of 25.6. The second sample of size 4 has a sample mean of 10 and a sample variance of 19.9. If the upper bound of the k% CI is at least 8.47, find the smallest possible value for k. A. Less than 0.80 B. 0.80-0.86 C. 0.86-0.92 D. 0.92-0.98 E. At least 0.98

C. 0.86-0.92

X1, 2, 3, 4, are independent and normally distributed with a common mean 50 and variance 400. Let psi be the CDF of the standard normal random variable at z. Determine the probability that the sample mean of the four variables is at least 60. A. Psi(1) B. Psi(1/4) C. 1 - Psi(1) D. 1 - Psi(1/4) E. 1 - Psi(-1)

C. 1-Psi(1) Mean = 50 Variance = 400/4 = 100 P(X >= 60) = P(X-50/√100 >= 60-50/√100) P(Z >= 1) = 1 - P(Z<1) = 1 - Psi(1)

A marine scientist has six ocean water samples that are evenly split between the Pacific and Atlantic oceans. Four of the ocean water samples are chosen using simple random sampling. Find the probability that all the Pacific ocean samples are chosen. A. 1/20 B. 1/10 C. 1/5 D. 1/4 E. 1/3

C. 1/5 (3C3 x 3C1) / (6C4)

You are given: A random sample of size 9 Calculated standard deviation is 11.1 Ho: µ = 7 H1: µ > 7 alpha = 10% Calculate c A. Less than 12 B. 12-12.15 C. 12.15-12.30 D. 12.30-12.45 E. At least 12.45

C. 12.15 - 12.30

A random sample of size 10 is drawn from a normal distribution. Ho: sigma2 = 100 Calculate the smallest critical value such that the significance level is no more than 5%. A. Less than 170 B. 170-185 C. 185-200 D. 200-215 E. At least 215

C. 185-200

You are given the following hypothesis test: A random sample of size 10 is drawn from a normal distribution with mean . The calculated sample variance is 9. Find the critical value. A. Less than 51.25 B. 51.25-51.50 C. 51.50-51.75 D. 51.75-52.00 E. At least 52.00

C. 51.50-51.75

Based on a random sample of size 7 from a normal distribution with mean µ, a confidence interval is constructed for µ. The sample standard deviation is calculated as 5.4763. p = 0.9 p = 0.95 p = 0.975 v = 6 1.440 1.943 2.447 v = 7 1.415 1.895 2.365 If we want to be at least 90% confident, determine the smallest possible width of the interval. A. Less than 6 B. 6-8 C. 8-10 D. 10-12 E. At least 12

C. 8-10 Width in one direction: t(s/√n) 1.943 (5.4763/√7) = 4.02

You are given the following information about a hypothesis test that is being performed on a random sample taken from a normal distribution with mean Unbiased variance = 9 Sample size is 16 H0 µ = 80 Calculate the highest possible sample mean that would result in the hypothesis not being rejected at the 1% level. A. Less than 81 B. 81-91.50 C. 81.50-82 D. 82-82.50 E. At least 82.50

C. 81.50 - 82

You are given the following information about a hypothesis test that is being performed on a random sample taken from a normal distribution with mean Variance = 9 Sample size = 16 H0: µ = 80 H1: µ > 80 Calculate the highest possible sample mean that would result in the null hypothesis not being rejected at the 1% significance level. A. Less than 81.00 B. At least 81.00, but less than 81.50 C. At least 81.50, but less than 82.00 D. At least 82.00, but less than 82.50 E. At least 82.50

C. 81.50 - 82.00

For a random sample taken from a normal distribution: 7.6, 4.9, 8.8, 7.6, 6.1, 14.5, 10.3, 5.1 The lower bound of a confidence interval for µ is 6.32. Find the upper bound. A. Less than 8.5 B. 8.5-9.5 C. 9.5-10.5 D. 10.5-11.5 E. At least 11.5

C. 9.5 - 10.5 µ = 8.1125 µ - lower bound = 1.7925 µ + 1.7925 = 9.905

For a random sample of size 8 drawn from a Bernoulli distribution with parameter , the sum of the observations resulted in 5. An analyst proposes to use normal percentiles to construct confidence intervals for . Which of the following statements are true regarding the setup above? Per the analyst, a 90% confidence interval for is (0.3434, 0.9066). Per the analyst, 0.1712 is the calculated standard error used to find a confidence interval for , regardless of the confidence level. It is appropriate to construct the intervals per the analyst, as all the assumptions are met. A. I B. III C. I and II D. I and III E. I, II, and III

C. I and II

Claim sizes follow a normal distribution with mean µ and variance 47.300. Ho: µ = 10,000 One claim of 9600 is observed. Determine the test result. A. Reject at 0.01 B. Reject at 0.02 but not 0.01 C. Reject at 0.05 but not at 0.02 D. Reject at 0.10 but not at 0.05 E. Do not reject

C. Reject at 0.05 but not 0.02

A single observation, , from a normal distribution with mean , variance 4, and PDF. Determine the rejection region. A. x < -1.6 B. x < 3.6 C. x > 3.6 D. x > 6.2 E. x < 6.2

C. x > 3.6

Two teams, Delta and Gamma, handle customer service calls for a corporation. The amount of time a team member spends answering calls on a typical day is normally distributed. Assume both teams are independent and have the same variance for call times. A random sample of four members was taken from each team. Their observed call times are as follows: Delta: 1 2 2 3 Gamma: 2 3 4 5 p = 0.80 p = 0.85 p = 0.90 v = 5. 0.920 1.156 1.476 v = 6 0.906 1.134 1.440 Construct an 80% confidence interval for team Delta's mean call time minus team Gamma's mean call time. A. (-2.72, -0.28) B. (-2.69, -0.31) C. (-2.63, -0.37) D. (-2.60, -0.40) E. (-2.19, -0.81)

D. (-2.60, -0.40)

An actuary considers a portfolio of 40 policyholders which was taken as a random sample from a block of business. She discovers that 11 of these policyholders were each compensated over $10,000 last year. Construct a 94% confidence interval for the proportion of policyholders who were compensated $10,000 or less last year from this block of business. A. (0.14, 0.41) B. (0.17, 0.38) C. (0.20, 0.35) D. (0.59, 0.86) E. (0.62, 0.83)

D. (0.59, 0.86) p +/- Z √[(pq)/n]

The following summarizes the observations from a random sample size of 10 from a normal distribution Sum(x-xbar)^2 = 72 Construct the 90% confidence interval for sigma^2. A. (1.419, 7.218) B. (1.548, 8.782) C. (3.785, 26.667) D. (4.256, 21.654) E. (4.643, 26.345)

D. (4.256, 21.654)

An actuary considers fitting a sample of 200 claims to one of the following models: Gamma distribution with two free parameters, producing a maximized log-likelihood function of -100. Generalized Pareto distribution with three free parameters, producing a maximized log-likelihood function of . Using the Bayesian information criterion, calculate such that the actuary is indifferent between the two models. A. -107.9 B. -102.6 C. -100.0 D. -97.4 E. -92.1

D. -97.4

You are given the following information on a random sample from a normal loss distribution with mean : The sample mean is 42,000. The sample standard deviation is 8,000. There are 25 loss observations in the sample. At which value of alpha would you reject the null? A. 0.01 B. 0.02 but. not 0.01 C. 0.05 but not 0.02 D. 0.10 but not 0.05 E. Do not reject

D. 0.10 but not 0.05

You are given the following information on a random sample from a normal loss distribution with mean : The sample mean is 42,000. The sample standard deviation is 8,000. There are 25 loss observations in the sample. To test the hypothesis h0 = 45000, at which value of alpha would you reject the null. A. 0.01 B. 0.02 but not 0.01 C. 0.05 but not 0.02 D. 0.10 but not 0.05 E. Do not reject

D. 0.10 but not 0.05

Net profits from a movie studio show normal distribution with mean 2.7. A random sample of six movies are examined. p = 0.91 p = 0.92 p = 0.93 p = 0.94 v = 5 1.558 1.649 1.753 1.873 v = 6 1.517 1.603 1.700 1.812 Find the probability that the sample mean net profit is greater than 2.7 minus three-fourths of the sample standard deviation. A. Less than 0.91 B. 0.91-0.92 C. 0.92-0.93 D. 0.93-0.94 E. At least 0.94

D. 0.93 - 0.94 Using the t-distribution P(X-2.7 > -0.75S) P(x - 2.7 / S/√6 > -0.75S / S/√6) P(x - 2.7 / S/√6 > -1.837) Using the table, when V = 5, p = 0.94

You test the following hypotheses by observing one outcome from a normal distribution with mean and standard deviation 1,250: H0: µ = 9000 H1: µ = 12500 Reject H0 if the outcome is greater than 11000. Calculate the probability of a type II error. A. Less than 5% B. 5-7.5% C. 7.5-10% D. 10-12.5% E. At least 12.5%

D. 10-12.5

You are given the following information about two loss severity distributions fit to a sample of 275 closed claims: For the exponential distribution, the natural logarithm of the likelihood function evaluated at the maximum likelihood estimate of its one parameter is -828.369. For the Weibull distribution, the natural logarithm of the likelihood function evaluated at the maximum likelihood estimates of both its parameters is -826.230. The exponential distribution is a subset of the Weibull distribution. The null hypothesis is that the exponential distribution provides a better fit than the Weibull distribution. Calculate the smallest significance level at which the null is rejected. A. Less than 0.5% B. 0.5-1% C. 1-2.5% D. 2.5-5% E. At least 5%

D. 2.5 - 5%

Based on a random sample of size 20 from a normal distribution with variance , the width of the 95% confidence interval for is 150. Determine the unbiased sample variance. A. Less than 25 B. 25-50 C. 50-75 D. 75-100 E. At least 100

D. 75-100

Random sample from a normal distribution with variance sigma2. - X is the sample mean - W = sum(xi-X)^2 / sigma2 - First percentile of W is 1.24 Using a chi-square distribution: p = 0.005 p = 0.015 v = 6 0.676 1.016 v = 7 0.989 1.418 v = 8 1.344 1.860 Determine n A. 5 B. 6 C. 7 D. 8 E. Cannot be determined

D. 8 The first percentile is between the 0.5th and the 1.5th percentile. 1.24 is only between these when v = 7, so there are 7 degrees of freedom. Therefore, n = df + 1 = 7+1 = 8

You assume that actuarial students average at least 6 hours of sleep. You take a random sample to test if they actually average less than 6 hours. The chosen significance level is 0.10. The -value of this test is 0.07 using the random sample. The true average is in fact 6.2 hours of sleep. Determine which of the following statements is true. A. You have correctly rejected the null hypothesis. B. You have correctly failed to reject the null hypothesis. C. You have correctly accepted the alternative hypothesis. D. You committed a type I error. E. You committed a type II error.

D. Type I error

Out of 50 basketball players trying to make a team, 5 will be chosen. 36 of the 50 are seniors and the remaining players are juniors. The manager wants 2 seniors and 3 juniors to make the team. What is the probability of meeting the manager's objectives if players were selected by simple random sampling? A. 0.013 B. 0.030 C. 0.070 D. 0.090 E. 0.108

E. 0.108 Number of ways to select 5 players: 50 C 5 Number of ways to select 2 seniors and 3 juniors: 36 C 2 x 14 C 3 (36C2 x 14C3) / (50C5) = 0.108

IPO returns from the real estate sector are normally distributed with mean µ. An analyst takes a random sample of 12. Percentile of a t-random variable with v degrees of freedom. p = 0.5570 p = 0.6900 p = 0.9475 v = 11 0.1467 0.5102 1.7667 v = 12 0.1464 0.5090 1.7536 What is the probability that X-µ/S is less than 0.51 A. 0.05 B. 0.31 C. 0.56 D. 0.69 E. 0.95

E. 0.95 P(X-µ/S/√n) P(0.51 / 1/√12) P(Z < 1.7667) 0.9475

For a random sample of size for losses following a Weibull distribution, the log-likelihood function is given. Sigma ln = 448.62 Sigma x^2.75 = 25,112,687 Sigma x^3 = 79,409,761 Calculate the test statistic. A. 6.1 B. 7.5 C. 9.0 D. 10.6 E. 12.2

E. 12.2

A random sample of size 21 from a normal distribution yields observations summarized as follows: Mean = 3.5 SD = 0.6156 H0: µ = 3 Calculate the p-value A. Less than 0.002 B. 0.002-0.004 C. 0.004-0.006 D. 0.006-0.008 E. At least 0.008

E. At least 0.008

A company wants to determine whether sick days taken by its employees are evenly distributed throughout the five-day working week. A random sample of the sick days taken by a sample of 100 employees yielded the following data: Sick Days Monday 32 Tuesday 18 Wednesday 18 Thursday 20 Friday 32 Total 120 Determine the p-value for this chi-squared goodness of fit test. A. Less than 0.005 B. 0.005-0.010 C. 0.010-0.025 D. 0.025-0.050 E. At least 0.050

E. At least 0.050

You are given the following information from a random sample: - X is normally distributed with mean 1 and variance 2. - Sample size is 20. - S2 is the unbiased sample variance. Let X2 be the 100th percentile of a chi-squared random variable with v degrees of freedom. The following table lists values of X2 for specific combinations of p and v: p = 0.95 p = 0.975 v = 19. 30.144. 32.852 v = 20. 31.410. 34.170 Calculate c such that P(S2 <= c) = 0.95. A. Less than 2.8 B. Between 2.8 and 2.9 C. Between 2.9 and 3.0 D. Between 3.0 and 3.1 E. At least 3.1

E. At least 3.1 N = 20, df = 20-1 = 19 Using the 19 row X2 = [(n-1)S2] / sigma2 0.95 = P(S2 <= c) = P(19/2 S2 <= 19c/2) c = 3.173

You are given two random samples of paid claim amounts for two Hospitals, A and B, which independently follow normal distributions with equal variances and means and . The results for the paid claim amounts are A: 6.21, 7.34, 5.67, 7.88, 3.89 B: 7.89, 10.12, 9.71, 5.55, 4.33, 12.48 The unbiased standard deviations are 1.56 and 3.04. p = 0.95 p = 0.975 v = 8 1.860 2.306 v = 9 1.833 2.262 v = 10 1.812 2.228 Calculate the upper bound of the 95% CI for the difference B - A. A. Less than 4.9 B. 4.9-5.1 C. 5.1-5.3 D. 5.3-5.5 E. At least 5.5

E. At least 5.5

You are given the following information about a random sample of claim amounts: - Claim severity follows a Pareto distribution with alpha = 3 and theta = 50, with mean and variance given by: mean = theta/alpha - 1 var = (2 theta^2)/[(alpha-2)(alpha-1)] - (theta/(alpha-1))^2 - The sample size is 100. Using the Central Limit Theorem, calculate the probability that the sample mean claim amount will lie between 30 and 35. A. Less than 5% B. 5-6% C. 6-7% D. 7-8% E. At least 8%

E. At least 8% mean = 50/3-1 = 25 var = (2*50^2) / [(3-1)(3-2)] - (50/(3-1))^2 = 1875 sigma = var/n = 1875/100 = 18.75 P(30 < x < 35) = P(35-25/√18.75) - P(30-25/√18.75) 0.1147

The mean time between claims for high-risk policyholders is supposedly 2.6. A researcher believes this mean time is incorrect and conducts a hypothesis test. He observed one high-risk policyholder, which resulted in a time of 4.5. Assume these times are normally distributed with variance 2. What is the result of the test? A. Reject at 0.015 B. Reject at 0.025 but not at the 0.015 C. Reject at the 0.050 but not at the 0.025 D. Reject at the 0.100 but not at 0.050 E. Do not reject

E. Do not reject

A single observation from a distribution with given PDF and CDF is used to test the hypothesis sigma = 0. Determine the rejection region. A. x > √a B. x < a C. x < √1-a D. x > √1-a E. x < √a

E. x < √a