AMA470 Midterm exam March 5‚ 2010 Please show full working out in order to obtain full marks. 1. Suppose that: • The number of claims per exposure period follows a Poisson distribution with mean λ = 110. • The size of each claim follows a lognormal distribution with parameters µ and σ 2 = 4. • The number of claims and claim sizes are independent. (a) Give two conditions for full credibility that can be completely determined by the information above. Make sure to define all terms in your definition
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II. STATEMENT OF THE PROBLEM As part of a long-term study of individuals 65 years of age or older‚ sociologists and physicians at the Wentworth Medical Center in upstate New York investigated the relationship between geographic location‚ health status ( healthy or one or more comorbidities)‚ and depression. Random samples of 20 healthy individuals were selected from three geographic locations: Florida‚ New York‚ and North Carolina. Then‚ each was given a standardized test to measure depression
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Examining Stock Returns for Normal Distributions July11‚ 2012 Part A. A1 (CRSP 2000-2008) | VW Daily | EW Daily | VW Monthly | EW Monthly | Mean | 0.00% | 0.05% | -0.12% | 0.50% | σ | 1.35% | 1.12% | 4.66% | 6.14% | Table A1 shows return means and standard deviations for the CRSP market portfolio from 2000-2008. In comparing daily vs monthly returns in both cases‚ equally weighted (EW) and value weighted (VW)‚ Table A1 shows the mean and standard deviation are
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Explain why the distribution B(n‚p) can be approximated by Poisson distribution with parameter if n tends to infinity‚ p 0‚ and = np can be considered constant. 2. Show that – and + are the turning points in the graph of the p.d.f. of normal distribution with mean and standard deviation . 3. What is the relationship between exponential distribution and Poisson distribution? II. Computation of probability (7%) 1. Let the random variable X follow a Binomial distribution
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The line between normal and abnormal cognitive changes with age remains indistinct. Normal aging is due to physiological processes over a person’s lifetime‚ in which the biological clock controls development and survival of nerve cells. That does not exclude a spectrum of variable levels of health or a continuum within normal aging‚ as well as between normal and pathological aging. At one end there are individuals with “successful aging” [34]. At the other end‚ we find frail‚ easily incompensated
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Study Set for Midterm II‚ Chapters 7 & 8 ESSAY. Write your answer in the space provided or on a separate sheet of paper. 1) The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 71. 2) At a computer manufacturing company‚ the actual size of computer chips is normally distributed with a mean of 1 centimeter
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Components of a Table with an example 2 2 10 6 a) Describe the characteristics of Normal probability distribution. b) In a sample of 120 workers in a factory‚ the mean and standard deviation of wages were Rs. 11.35 and Rs.3.03 respectively. Find the percentage of workers getting wages between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally distributed. Characteristics of Normal probability distribution Formula/Computation/Solution to the problem 3 4 10
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Statistics Math 1342 Final Exam Review Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a heart. 17 7 3 4 A) B) C) D) 52 52 13 13 Answer: D 2) The events A and B are mutually exclusive. If P(A) = 0.7 and P(B) = 0.2‚ what is P(A or B)? A) 0.5 B) 0.9 C) 0.14 D)
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by N. W IENER in about 1920. ´ His proof was simplified by P. L E VY; we shall outline L´ vy’s construction in section ?? e below. But notice that properties (3) and (4) are compatible. This follows from the following elementary property of the normal distributions: If X‚ Y are independent‚
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> data=read.table("d:/111113/1.txt"‚header=T) > model1=lm(S~u_direction+mx+my+mz‚data) > summary(model1) Call: lm(formula = S ~ u_direction + mx + my + mz‚ data = data) Residuals: Min 1Q Median 3Q Max -11.8430 -0.3962 0.3252 0.7887 18.3963 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.50372 0.12738 -3.955 7.93e-05 *** u_direction -0.40368 0.07996 -5.048 4.85e-07 *** mx -0.40573 0
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