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6. [1/3 Points] DETAILS PREVIOUS ANSWERS ASWSBE14 7.TB.3.018. A simple random sample of 8 employees of a corporation provided the following information. Employee 1

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6. [1/3 Points] DETAILS PREVIOUS ANSWERS ASWSBE14 7.TB.3.018. A simple random sample of 8 employees of a corporation provided the following information. Employee 1 2 3 4 5 6 7 8 Age 21 35 27 43 50 50 23 22 Gender W M M W W W M W (a) Determine the point estimate for the average age of all employees. 33.88 MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Enter a number. (b) What is the point estimate for the standard deviation of the population? (Round your answer to four decimal places.) 11.60 (c) Determine a point estimate for the proportion of all employees who are female. 0.625 Need Help? Read It 3/5 Points] DETAILS PREVIOUS ANSWERS ASWSBE14 7.TB.3.020. An experimental diet to induce weight loss was followed for one week by a randomly selected group of 12 students with the following results. Student Loss in Pounds 1 1.9 2 2.9 3 0.4 4 1.7 5 0.0 6 1.5 7 5.5 8 3.5 9 4.5 10 4.1 11 1.1 12 2.9 (a) Find a point estimate for the average amount lost after one week on this diet. 2.5 Is this an unbiased estimate of the population mean? Explain. Yes, this an unbiased estimate of the population mean, since E(x) + . Yes, this an unbiased estimate of the population mean, since E(X) = . No, this is not an unbiased estimate of the population mean, since E(X) = . No, this is not an unbiased estimate of the population mean, since the sample size is too small. No, this is not an unbiased estimate of the population mean, since E(X) + . (b) Find a point estimate for the variance of the amount lost on this diet. (Round your answer to four decimal places.) 2.864 Is this an unbiased estimate of the population variance? Explain. Yes, this an unbiased estimate of the population variance, since E(s) = . 2 17. [2/5 Points] DETAILS PREVIOUS ANSWERS ASWSBE14 7.TB.5.040. You may need to use the appropriate appendix table or technology to answer this question. A population of 1,000 students spends an average of $11.40 a day on dinner. The standard deviation of the expenditure is $4. A simple random sample of 64 students is taken. (a) What is the expected value (in dollars) of the sampling distribution of the sample mean? $ 11.40 What is the standard deviation (in dollars) of the sampling distribution of the sample mean? (Round your answer to four decimal places.) $ 0.5 What is the shape of the sampling distribution of the sample mean? skewed to the left uniform normal skewed to the right exponential (b) What is the probability that these 64 students will spend a combined total of more than $776.43? (Round your answer to four decimal places.) 0.0721 (c) What is the probability that these 64 students will spend a combined total between $763.81 and $791.07? (Round your answer to four decimal places.) 0.1149 Need Help? Read It

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