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Salmon A specialty food company sells whole King Salmon to various customers. The mean weight of these salmon is 35 pounds with a standard deviation of 2 pounds. The company ships them to restaurants in boxes of 4 salmon, to grocery stores in cartons of 16 salmon, and to discount outlet stores in pallets of 100 salmon. To forecast costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. a. Find the standard deviations of the mean weight of the salmon in each type of shipment. b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets? Explain.

LSAT The LSAT (a test taken for law school admission) has a mean score of 151 with a standard deviation of 9 and a unimodal, symmetric distribution of scores. A test preparation organization teaches small classes of 9 students at a time. A larger organization teaches classes of 25 students at a time. Both organizations publish the mean scores of all their classes. a. What would you expect the sampling distribution of mean class scores to be for each organization? b. If either organization has a graduating class with a mean score of 160 , they'll take out a full-page ad in the local school paper to advertise. Which organization is more likely to have that success? Explain. c. Both organizations advertise that if any class has an average score below \(145,\) they'll pay for everyone to retake the LSAT. Which organization is at greater risk to have to pay?

Tips A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of \(\$ 9.60\) and a standard deviation of \(\$ 5.40 .\) a. Explain why you cannot determine the probability that a given party will tip him at least \(\$ 20\). b. Can you estimate the probability that the next 4 parties will tip an average of at least \(\$ 15 ?\) Explain. c. Is it likely that his 10 parties today will tip an average of at least \(\$ 15 ?\) Explain.

Groceries A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of \(\$ 32\) and a standard deviation of \(\$ 20\). a. Explain why you cannot determine the probability that the next Sunday customer will spend at least \(\$ 40\). b. Can you estimate the probability that the next 10 Sunday customers will spend an average of at least \(\$ 40 ?\) Explain. c. Is it likely that the next 50 Sunday customers will spend an average of at least \(\$ 40 ?\) Explain.

\(t\) -models, part I Using the \(t\) -tables, software, or a calculator, estimate a. the critical value of \(t\) for a \(90 \%\) confidence interval with \(d f=17\) b. the critical value of \(t\) for a \(98 \%\) confidence interval with \(d f=88\)

\(t\) -models, part IV Describe how the critical value of \(t\) for a \(95 \%\) confidence interval changes as the number of degrees of freedom increases.

Salaries A survey finds that a \(95 \%\) confidence interval for the mean salary of a police patrol officer in Fresno, California, in 2016 is \(\$ 52,516\) to \(\$ 53,509\). A student is surprised that so few police officers make more than \(\$ 53,500\). Explain what is wrong with the student's interpretation.

Cattle Livestock are given a special feed supplement to see if it will promote weight gain. Researchers report that the 77 cows studied gained an average of 56 pounds, and that a \(95 \%\) confidence interval for the mean weight gain this supplement produces has a margin of error of ±11 pounds. Some students wrote the following conclusions. Did anyone interpret the interval correctly? Explain any misinterpretations. a. \(95 \%\) of the cows studied gained between 45 and 67 pounds. b. We're \(95 \%\) sure that a cow fed this supplement will gain between 45 and 67 pounds. c. We're \(95 \%\) sure that the average weight gain among the cows in this study was between 45 and 67 pounds. d. The average weight gain of cows fed this supplement will be between 45 and 67 pounds \(95 \%\) of the time. e. If this supplement is tested on another sample of cows, there is a \(95 \%\) chance that their average weight gain will be between 45 and 67 pounds.

Teachers Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? with \(90.00 \%\) Confidence, \(t\) -interval for \(\mu: 43454<\mu(\) TchPay \()<45398\) a. If we took many random samples of 288 Nevada teachers, about 9 out of 10 of them would produce this confidence interval. b. If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers. c. About 9 out of 10 Nevada teachers earn between \(\$ 43,454\) and \(\$ 45,398 .\) d. About 9 out of 10 of the teachers surveyed earn between \(\$ 43,454\) and \(\$ 45,398\). e. We are \(90 \%\) confident that the average teacher salary in the United States is between \(\$ 43,454\) and \(\$ 45,398\).

Snow Based on meteorological data for the past century, a local TV weather forecaster estimates that the region's average winter snowfall is \(23 "\), with a margin of error of ±2 inches. Assuming he used a \(95 \%\) confidence interval, how should viewers interpret this news? Comment on each of these statements: a. During 95 of the past 100 winters, the region got between \(21 "\) and \(25 "\) of snow. b. There's a \(95 \%\) chance the region will get between \(21 "\) and \(25 "\) of snow this winter. c. There will be between \(21 "\) and \(25 "\) of snow on the ground for \(95 \%\) of the winter days. d. Residents can be \(95 \%\) sure that the area's average snowfall is between \(21 "\) and 25 ". e. Residents can be \(95 \%\) confident that the average snowfall during the past century was between \(21 "\) and \(25 "\) per winter.