Statistics- A pollster selected 4 of 7 available people

QUESTION

Quiz #2`Name________________________Section No. _____________(5%)1. A pollster selected 4 of 7 available people. How many different groups of 4 arepossible?10%2. Your firm has a contract to make 2000 staff uniforms for a fast –food retailer. Theheights of the staff are normally distributed with a mean of 70 inches and a standarddeviation of 3 inches. What percentage of uniforms will have to fit staff shorter than67inches? 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 servicewithin the first year. A dealer sold 20 Nissans yesterday. Use equation for BinomialProbability 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 theBinomial 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. Recordsshow that the probability a motorist will have an accident during the weekend and willfile a claim is quite small (.0005). Suppose Alden wrote 400 policies for the forthcomingweekend. Compute the probability that exactly two claims will be filed using the equationfor Poisson Probability.Note: The symbol λ is the mean (expected value) which we used as μ = np. So λ isnothing more than the mean number of occurrences (successes = np) in a particularinterval.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 Valuesused 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 toestimate the mean weekly child-care cost of their employees. A sample of 10 employeesreveals 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 aremen. A sample of 256 cars traveling southbound on the New Jersey Turnpikeyesterday revealed that 165 were driven by men. At the .01 significance level, can weconclude that a larger proportion of men were driving on the New Jersey Turnpikethan the national statistics indicate? First, state H0 & HaHO: ??, Ha: ??a) Is this a Z or t test?b) Test Statistic = ?c) Critical value = ?d) p-value = ?e) Reject Ho: (yes or no)15%8. Given the hypothesis: H0: μ≥18 & Ha: μ<18, a random sample of five resulted inthe following values: 17, 18, 20, 16, & 15. Using the .01 significance level, can weconclude 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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