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A POLLSTER SELECTED 4 OF 7 AVAILABLE PEOPLE. HOW MANY DIFFERENT GROUPS OF 4 ARE POSSIBLE?

A POLLSTER SELECTED 4 OF 7 AVAILABLE PEOPLE. HOW MANY DIFFERENT GROUPS OF 4 ARE POSSIBLE?

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Quiz #2`Name________________________ Section No. _____________ (5%) 1. A pollster selected 4 of 7 available people. How many differentgroups of 4 are possible?nCr = n! / ( r! (n - r)! )nCr = 7! / ( 4! (7-4)! )nCr = 3510%2. Your firm has a contract to make 2000 staff uniforms for a fast –food retailer. The heights of the staff are normally distributed with amean of 70 inches and a standard deviation of 3 inches. What percentageof uniforms will have to fit staff shorter than 67inches? What percentagewill have to be suitable for staff taller than 76 inches.?a) 16% & 2.5%b) 68% & 95%c) 32% & 5%(15%) 3. The industry standards suggest that 20% of new vehiclesrequire warranty service within the first year. A dealer sold 20 Nissansyesterday. Use equation for Binomial Probability for part a) and Table IIfor part b) & c) . Show work!a) What is the probability that none of these vehicles requires warrantyservice? Use the Binomial equation for P(X=0) .b) What is the probability that exactly one of these vehicles requireswarranty service?c) Determine the probability 3 or more of these vehicles requirewarranty service.d) Compute the mean and std. dev. of this probability distribution.(15%) 4. Allen & Associates write weekend trip insurance at a very nominal charge. Records show that the probability a motorist will havean accident during the weekend and will file a claim is quite small(.0005) . Suppose Alden wrote 400 policies for the forthcoming weekend.Compute the probability that exactly two claims will be filed using theequation for Poisson Probability.Note: The symbol λ is the mean (expected value) which we used as μ =np. So λ is nothing more than the mean number of occurrences(successes = np) in a particular interval.Get the probability that the number of claims is at least 3 from PoissonTables.(10%) 5. Given a standard normal distribution, determine the following.Show Table Values used in each part.a) P(Z<1.4) b) P(Z>1.4) c) P(Z< -1.4) d) P( - 0.50<Z<1.0) e) P(0.50<Z<1.5) 15%6. A company is considering offering child care for their employees.They wish to estimate the mean weekly child-care cost of theiremployees. A sample of 10 employees reveals the following amountsspent last week in dollars.101 97 93 103 100 93 99 90 102 96Develop a 95% confidence interval for the population mean. Interpretthe result.x =??, S=??, t/ 2=??, Range of = (??) 15%7. The National Safety Council reported that 56 % of Americanturnpike drivers are men. A sample of 256 cars traveling southbound onthe New Jersey Turnpike yesterday revealed that 165 were driven bymen. At the .01 significance level, can we conclude that a largerproportion of men were driving on the New Jersey Turnpikethan the national statistics indicate? HO: ??, Ha: ??.a) Is this a Z or t test?.b) Test Statistic = ?.c) Critical value = ?.d) p-value = ?.e) Reject Ho: (yes or no) First, state H0 & Ha15%8. Given the hypothesis: H0: μ≥18 & Ha: μ<18, a random sample offive resulted in the following values: 17, 18, 20, 16, & 15. Using the .01significance level, can we conclude the population mean is less than 18?.a) Is this a Z or t test?.b) Test statistic = ?.c) Critical value = ?.d) Reject Ho: (yes or no)

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