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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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 A pollster selected 4 of 7 available people. How many different groups of 4 are possible?10%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 a mean of 70 inches and a standard deviation of 3 inches. What percentage of uniforms will have to fit staff shorter than 67inches? What percentage will 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 vehicles require warranty service within the first year. A dealer sold 20 Nissans yesterday. Use equation for Binomial Probability for part a) and Table II for part b) & c). Show work!a) What is the probability that none of these vehicles requires warranty service? Use the Binomial equation for P(X=0).b) What is the probability that exactly one of these vehicles requires warranty service? c) Determine the probability 3 or more of these vehicles require warranty 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 have an 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 the equation 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 Poisson Tables.(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 their employees. A sample of 10 employees reveals the following amounts spent last week in dollars.101 97 93 103 100 93 99 90 102 96Develop a 95% confidence interval for the population mean. Interpret the result.x =??, S=??, t / 2 =??, Range of  = (??)15%7. The National Safety Council reported that 56 % of American turnpike drivers are men. A sample of 256 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 165 were driven by men. At the .01 significance level, can we conclude that a larger proportion 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 of five resulted in the following values: 17, 18, 20, 16, & 15. Using the .01 significance 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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