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    •H.P.Gautam The purpose of this article is not to explain any more the usefulness of normal distribution in decision-making process no matter whether in social sciences or in natural sciences. Nor is the purpose of making any discussions on the theory of how it can be derived. The only objective of writing this article is to acquaint the enthusiastic readers (specially students) with the simple procedure ( iterative procedure) for finding the numerical value of a normally distributed variable. The

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    skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately

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    points) 3. Suppose that a random sample of size 64 is to be selected from a population having [pic] and standard deviation 5. (a) What are the mean and standard deviation of the [pic] sampling distribution? Can we say that the shape of the distribution is approximately normal? Why or why not? (10 points) (b) What is the probability that [pic] will be within 0.5 of the population mean? (5 points) (c) What is the probability that [pic] will differ from the population

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    Unit 6. Normal Distribution Solution to problems Statistics I. International Group Departamento de Economa Aplicada Universitat de Valncia May 20‚ 2010 Problem 35 Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000‚ 180) Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 Take 5 weeks at random. Find the probability of weekly sales not exceeding 850 tickets in more than two weeks Ticket price is 4.5 Euros

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    at Weston Materials‚ Inc.‚ a national manufacturer of unattached garages‚ reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect? z(29) = (29-32)/2 = -3/2 z(34) = (34-32)/2 = 1 z(32) = 0 P(32 < x < 34) = P(0< z < 1) = 0.34 b. What percent of

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    York City |   |   |   | Population  | (b) | To know how many different private airlines provide service in the United States |   |   |   | Sample  | (c) | To estimate weekly earnings of computer programmers in United States |   |   |   | Sample  | (d) | To find the marks of a class of 20 students in a physics exam |   |   |   | Population  | eBook Link award: 0 out of 16.66 points     From the data in the Statistical Abstract of United States represented below

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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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    summarize the data from the two studies‚ what are your preliminary observations about the depression scores? b. Using ANOVA on both data sets‚ what is the hypotheses being tested in both cases? What are your conclusions? c. Use inferences about individual treatment means where appropriate. What are your conclusions? d. Discuss extensions of this study or other analyses that you feel might be helpful III. DESCRIPTIVE STATISTICS 1. Data Table 1.1 shows the data of 60 individuals

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    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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    are listed below. Find the mean speed. 181.1 201.2 202.2 193.2 190.1 201.2 201.4 194.5 191.3 199.2 201.4 196.0 192.2 196.2 7) A) 210.9 Answer: B B) 195.8 C) 201.2 D) 196.1 1 8) Calculate the correlation coefficient‚ r‚ for the data below. x y -10 -8 4 -13 A) -0.549 -1 1 -4 -10 -6 -7 -9 -6 B) -0.132 -5 -2 -3 -12 -2 -1 -9 0 C) -0.581 8) D) -0.104 Answer: D 9) Given H0

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