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Point Estimate of Statistics and Probability Calculation

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1. A simple random sample of five months of sales data provided the following information

Month 1 2 3 4 5
Units Sold 94 100 85 94 92

a. Develop a point estimate of the population mean number of units sold per month.
b. Develop a point estimate of the population standard deviation

2. The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009):

Critical Reading 502
Mathematics 515
Writing 494

Assume that the population standard deviation on each part of the test is o = 100.

a. What is the probability a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test?

b. What is the probability a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)?

c. What is the probability a random sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?

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Solution Preview

1. a. The point estimate is the sample mean. The sample mean is (94+100+85+94+92)/5=93

b. The point estimate is the sample standard deviation. The sample standard deviation is sqrt{[(94-93)^2 + (100-93)^2 + (85-93)^2 + (94-93)^2 + (92-93)^2]/(5-1)}=5.385

2. a. What is the probability a random sample of 90 test ...

Solution Summary

The solution gives detailed steps to calculation the point estimate of population mean and standard deviation. Also, the probabilities in a sample with known sample mean and standard deviation are computed step by step.

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