Mini Assignment 1 and 2

| September 28, 2018

1. Circuit boards for wireless telephones are etched, in an acid bath, in batches of 100 boards. Asample of seven boards is randomly selected from each lot for inspection. A particular batchcontains two defective boards; and X is the number of defective boards in the sample. P(X=2) is2. You are offered an investment opportunity. Its outcomes and probabilities are presented in the following table.__________.The standard deviation of this distribution is3. Circuit boards for wireless telephones are etched, in an acid bath, in batches of 100 boards. Asample of seven boards is randomly selected from each lot for inspection. A particular batchcontains two defective boards; and X is the number of defective boards in the sample. P(X=1) is4. Ten policyholders file claims with CareFree Insurance. Three of these claims are fraudulent.Claims manager Earl Evans randomly selects three of the ten claims for thorough investigation.If X represents the number of fraudulent claims in Earl’s sample, P(X=0) is5. If X is a binomial random with n=8 and p=0.6, what is the probability that X is equal to 5?6. If X is a normal random variable with mean 80 and standard deviation 5, calculate the Z scoreif X=72.7. Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normaltables to find P(-2.25 < Z < 1.1).8. Let Z be a normal random variable with mean 0 and standard deviation 1. Use the normaltables to find P(Z > 2.4).9. The weights of aluminum castings produced by a process are normally distributed with = 2pounds and= 0.10 pound. Design specifications require the castings to weigh between 1.836and 2.164 pounds, inclusively. The probability that a casting produced by this process willconform to design specifications is10. The expected (mean) life of a particular type of light bulb is 1,000 hours with a standarddeviation of 50 hours. The life of this bulb is normally distributed. What is the probability that arandomly selected bulb would last longer than 1150 hours?11. The weights of aluminum castings produced by a process are normally distributed with = 2pounds and = 0.10 pound. Design specifications require the castings to weigh between 1.836and 2.164 pounds, inclusively. Any casting weighing more than 2.164 pounds is re-worked. Theprobability that a casting produced by this process will be re-worked, due to over-weight, is*************************************************************1. Suppose the alternative hypothesis in a hypothesis test is "the population mean is greater than60". If the sample size is 80 and alpha = .01, the critical value of Z is2. Jennifer Cantu, VP of Customer Services at Tri-State Auto Insurance, Inc., monitors the claimsprocessing time of the claims division. Each week, her staff randomly selects a sample of 64claims and tests the null hypothesis that the "mean processing time is 5 days or less" using a 0.05level of significance. Last week the sample mean and standard deviation were 5.2 days and 0.56days, respectively. The appropriate decision is3. When the rod shearing process at Stockton Steel is "in control" it produces rods with a meanlength of 120 inches. Periodically, quality control inspectors select a random sample of 36 rods.If the mean length of sampled rods is too long or too short, the shearing process is shut down.Sarah Shum, Director of Quality Programs, chose a 0.05 level of significance for this test. Thecritical Z values are4. A researcher is testing a hypothesis of a single mean. The critical Z value for = .01 and aone-tailed test is -2.33. The calculated Z value from sample data is -2.45. The decision made bythe researcher based on this information is to __________ the null hypothesis5. Restaurateur Denny Valentine is evaluating two sites, Raymondville and Rosenberg, for hisnext restaurant. Prevailing images of the two suburbs imply that Raymondville residents(population 1) dine out less often than Rosenberg residents (population 2). Denny plans to testthis hypothesis using a random sample of 81 families from each suburb. His null hypothesis is6. Lucy Baker is analyzing demographic characteristics of two television programs, COPS(population 1) and 60 Minutes (population 2). Previous studies indicate no difference in the agesof the two audiences. (The mean age of each audience is the same.) Lucy plans to test thishypothesis using a random sample of 100 from each audience. Her null hypothesis is7. Lucy Baker is analyzing demographic characteristics of two television programs, COPS(population 1) and 60 Minutes (population 2). Previous studies indicate no difference in the agesof the two audiences. (The mean age of each audience is the same.) Her staff randomly selected100 people from each audience, and reported the following_and= 8. Assuming a two-tail test and= 43 years,= 45 years,= 8,= .05, the appropriate decision is8. A researcher is estimating the average difference between two population means based onmatched-pairs samples. She gathers data on each pair in the study resulting in:Pair Group 1 Group 2110122893111148105912Assume that the data are normally distributed in the population. To obtain a 90% confidenceinterval, the table t value would be9. American First Banks’ policy requires consistent, uniform training of employees at all banks.Consequently, David Desreumaux, VP of Human Resources, orders a survey of mean employeetraining time in the Southeast region (population 1) and the Southwest region (population 2). Hisstaff randomly selected personnel records of 81 employees from each region, and reported thefollowing: = 30 hours, = 27 hours,05, the appropriate decision is= 11, and= 9. Assuming a two-tail test and=.10. Kristen Ashford purchased the subscribers list for Wind Surfing magazine. She plans tosurvey a sample of the subscribers before using the list in her mail order business. She choosesthe first 100 of the 5,000 names. Her sample is a11. Suppose a population has a mean of 90 and a standard deviation of 28. If a random sample ofsize 49 is drawn from the population, the probability of drawing a sample with a mean of morethan 95 is12. Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the servicelevel provided to walk-in customers. Accordingly, he plans a sample of waiting times for walk-incustomers. If the population of waiting times has a mean of 15 minutes and a standard deviationof 4 minutes, the probability that Albert’s sample of 64 will have a mean less than 14 minutes is a13. Suppose a population has a mean of 400 and a standard deviation of 24. If a random sampleof size 144 is drawn from the population, the probability of drawing a sample with a mean ofmore than 404.5 is14. Eugene Gates, Marketing Director of Mansfield Motors Manufacturers, Inc.’s ElectricalDivision, is leading a study to assess the relative importance of product features. An item on asurvey questionnaire distributed to 100 of Mansfield’s customers asked them to rate theimportance of "efficiency of operation" on a scale of 1 to 10 (with 1 meaning "not important"and 10 meaning "highly important"). His staff assembled the following statistics.If Eugene concludes that the average rateof "efficiency of operation" for all customers is 6.0, he is using15. The table t value associated with 12 degrees of freedom and used to compute a 95%confidence interval is16. Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service levelprovided to walk-in customers. Accordingly, his staff recorded the waiting times for 64 randomlyselected walk-in customers, and determined that their mean waiting time was 15 minutes and thatthe standard deviation was 4 minutes. The 95% confidence interval for the population mean ofwaiting times is17. A random sample of 64 items is selected from a population of 400 items. The sample mean is200 and the sample standard deviation is 48. From this data, a 90% confidence interval toestimate the population mean can be computed as18. Suppose a random sample of 36 is selected from a population with a standard deviation of12. If the sample mean is 98, the 99% confidence interval to estimate the population mean is

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