An insurance company has three risk classes: low risk, medium risk, and high risk. You are given:i. Claims from the low-risk class have a gamma distribution with parameters a =2 and 0.I. Claims from the medium-risk class have a gamma distribution with parameters a = 3 and 0.iL. Claims from the high-risk class have a gamma distribution with parameters a =iv. Three claims from the low-risk class are observed and they are of sizes 50, 55, and 70.a%3D4 and 0.v. One claim from the medium-risk class is observed and it is of size 100.vi. One claim from the high-risk class is observed and it is of size 200.Calculate the maximum likelihood estimate of 8.31B.3335D3739

Answered: Image /qna-images/answer/811c16d7-ae3a-40a1-8443-14e4cb9dae02.jpg

Answered: An insurance company has three risk…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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5. Let Y be a normally distributed random variable with mean = 0 and variance a=1. Use the method of transformation to find the probability distribution of U=Y². What is the name of this distribution?
b) What is the probability that is less than 2.7 grams? 0.3085 0.6915 0.9982 0.0018
If the average low temperature of a winter month in Rochester, NY is 18∘18∘ and the standard deviation is 3.4∘3.4∘, then according to Chebyshev's theorem, the percentage of average low temperatures in Rochester, NY between 12.56∘12.56∘ and 23.44∘23.44∘ is at least %.
B.13. Table B-6 gives data on the number of new business incorporations (Y) and the number of business failures (X) for the United States from 1984 to 1995. a. What is the average value of new business incorporations?And the variance? b. What is the average value of business failures? And the variance? c. What is the covariance between Y and X? And the correlation coefficient? (X) + var (Y) sr= -0.8 (X+Y) 2X = 16, s2Y = 9Y = 3X2 + 2Y = ax + bY = X>4Y = 0.6X-4 Y = 3X +2 E(X) = 8 and var (X) = 4 E(XY) = a 4X=1 a 2 Y = 1 XIYi f(XI, YI) X = 20 NUMBER OF NEW BUSINESS INCORPORATIONS (Y) AND NUMBER OF BUSINESS FAILURES (X), UNITED STATES, 1984-1995 YEAR Y X 1984 634,991 52,078 1985 664,235 57,253 1986 702, 738 61,616 1987 685,572 61, 111 1988 685,095 57,097 1989 676,565 Source: Economic Report of the President, 2004, Table B-96, p. 395. TABLE B-6 d. Are the two variables independent? e. If there is correlation between the two variables, does this mean that one variable causes the other…
A researcher asks a sample of people about their attitudes toward marijuana legalization and concludes people have a wide variety of opinions, and are equally likely to be anywhere between strongly in favor of it to strongly opposed to it. She reaches this conclusion because the shape of the distribution is
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. (Let 1 = Linda's rats, 2 = Tuan's rats and 3 = Javier's rats.) Weights of Student Lab Rats Linda's rats Tuan's rats Javier's rats 43.0 46.9 51.1 38.5 40.1 40.4 40.7 38.4 37.1 45.8 45.7 44.8 37.8 43.6 47.7 Part (a) State the null hypothesis. H0: ?1 = ?2 = ?3 H0: At least two of the group means ?1, ?2, ?3 are not equal. Part (b) State the alternative hypothesis. Ha: ?1 = ?2 = ?3 Ha: At least two of the group means ?1, ?2, ?3 are not equal. Part (c) Enter an exact number as an…
A marketing analyst has been tracking the attendance of baseball games for two minor league baseball teams. The distribution of each is given in the box plots below. Given these distributions, which of the following are true? Select the two correct answers. The spread of Team 1's distribution is higher than Team 2's. Team 1's distribution is more symmetrical than Team 2's. Team 2's distribution is more symmetrical than Team 1's. The center for Team 2's distribution is higher than Team 1's. The center for Team 1's distribution is higher than Team 2's.
4. According to the information given below, compute the correlation coefficient and the covariance coefficient between stock A and stock B. Standard deviation of the portfolio (stock A and stock B) :0.37 Standard deviation of the stock A : 0.38 Standard deviation of the stock B :0.43 Weight for stock A : 0.60 Weight for stock B :0.40 NOTE: PLEASE SHOW HOW YOU COMPUTE EACH OF THE ITEMS.
QUESTION 8 You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom total? Please do the calculations on the handout provided.
Please let me know if I calculated relative risk correctly. Thank you for your assistance.
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This cumulative review problem uses material from Chapters 3, 5, and 10. Recall that the Poisson distribution deals with rare events. Death from the kick of a horse is a rare event, even in the Prussian army. The following data are a classic example of a Poisson application to rare events. The data represent the number of deaths from the kick of a horse per army corps per year for 10 Prussian army corps over a period of time. Let x represent the number of deaths and f the frequency of x deaths. x 0 1 2 3 or more f 111 61 24 4 (a) First, we fit the data to a Poisson distribution.The Poission distribution states P(x) = e−λλx x! , where λ ≈ x (sample mean of x values). From our study of weighted averages, we get the following. x = Σxf Σf Verify that x = 0.605. Hint: For the category 3 or more, use 3. x = (b) Now we have P(x) = e−0.605(0.605)x x! for x = 0, 1, 2, 3 . Find P(0), P(1), P(2), and P(3 ≤ x). Round to three places…

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