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# 11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245.a. What is the probability that auto...

## Question

###### 11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245.a. What is the probability that automobiles a simple random sample of size 150 insured will have a sample mean insurance cost greater than $9502Answer:What is the probability that a simple random sample of size 50 insured automobiles will have a sample mean insurance cost within$25 of the population mean?Answer:12. The owner of a local restaurant wants to estimate the

11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245. a. What is the probability that automobiles a simple random sample of size 150 insured will have a sample mean insurance cost greater than $9502 Answer: What is the probability that a simple random sample of size 50 insured automobiles will have a sample mean insurance cost within$25 of the population mean? Answer: 12. The owner of a local restaurant wants to estimate the average amount spent per customer for dinner at the establishment Data was collected related to 81 customers_ a. If it is assumed that the population standard deviation for the amount spent is $5, what is the margin of error in estimating at the 90% confidence level the average amount spent? Answer: b. If the sample mean amount spent is$24.83,what is the 90% confidence interval for estimating the average amount spent per customer for everyone who eats dinner at the establishment? Answer: . .

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