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1.) The mean television viewing time for Americans is 15 hours per week (Money, November 2003). Suppose a sample of 40 Americans is taken to further investigate viewing habits. Assume the population standard deviation for weekly viewing time is = 3 hours. A. What is the probability the sample mean will be within 1 hour of the population mean (to 4 decimals)? B. What is the probability the sample mean will be within 45 minutes of the population mean (to 4 decimals)? A) Let x be the sample mean. Then x is normally distributed with mean 15 hours and standard deviation / n = 3/ 40 = 0.4743. Therefore Z = ( x -15)/0.4743 follows a Standard normal distribution. Now the probability the sample mean will be within 1 hour of the population mean is given by, P[14 < x < 16] = P[(14 -15)/0.4743 < ( x -15)/0.4743 < (16-15)/0.4743] = P[ -2.1082 < Z < 2.1082] = P[ Z < 2.1082] – P[Z < -2.1082] = 0.9825 – 0.0175 = 0.9650 B) Now the probability the sample mean will be within 45 minutes of the population mean is given by, P[14.25 < x < 15.75] = P[(14.25 -15)/0.4743 < ( x -15)/0.4743 < (15.75-15)/0.4743] = P[ -1.5811 < Z < 1.5811] = P[ Z < 1.5811] – P[Z < -1.5811] = 0.9431 – 0.0569 = 0.8862 2.) Americans have become increasingly concerned about the rising cost of Medicare. In 1990, the average annual Medicare spending per enrollee was $3267; in 2003, the average annual Medicare spending per enrollee was $6883 (Money, Fall 2003). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was $2,000 A. Calculate the standard error of the mean amount of Medicare spending for a sample of fifty 2003 enrollees (to 2 decimals). B. What is the probability the sample mean will be within +/- $300 of the population mean (to 4 decimals)? C. What is the probability the sample mean will be greater than $7500 (to 4 decimals)? A) The standard error of the mean amount of Medicare spending for a sample of fifty 2003 enrollees is given by, S.E = / n = 2,000/ 50 = 282.8427 = $282.84 B) Let x be the sample mean. Then x is normally distributed with mean $6883 and standard deviation / n = 2,000/ 50 = $282.84 Therefore Z = ( x - 6883)/282.84 follows a Standard normal distribution. Now, the probability the sample mean will be within +/- $300 of the population mean is given by, P[6583 < x < 7183] = P[(6583- 6883)/282.84 < ( x - 6883)/282.84 < (7183- 6883)/282.84] = P[ -1.067 < Z < 1.067] = P[ Z < 1.067] – P[Z < -1.067] = 0.8556 – 0.1444 = 0.7112 C) Now the probability the sample mean will be greater than $7500 is given by, P[ x > 7500] = P ( x - 6883)/282.84 > (7500- 6883)/282.84] = P[ Z > 2.1814] = 1- P[Z < 2.1814] = 1 – 0.9854 = 0.0146 3.) The president of Doerman Distributors, Inc., believes that 34% of the firm's orders come from first-time customers. A simple random sample of 120 orders will be used to estimate the proportion of first-time customers. A. What is the probability that the sample proportion will be between .20 and .40 (to 4 decimals)? B. What is the probability that the sample proportion will be between .25 and .35 (to 4 decimals)? A) Here the population proportion is p = 0.34. We have the sample proportion p follows a normal distribution with mean p = 0.34 and p(1 p) 0.34(1 0.34) = 0.0432 n 120 Thus Z = ( p - 0.34)/0.0432 follows a Standard normal distribution. Now, the probability that the sample proportion will be between .20 and .40 is given by P[ 0.20 < p < 0.40] = P[(0.20- 0.34)/0.0432 < ( p - 0.34)/0.0432 < (0.40- 0.34)/0.0432] =P[-3.2375 < Z < 1.3875] = P[ Z < 1.3875] – P[Z < -3.2375] = 0.9174 – 0.0006 = 0.9168 B) The probability that the sample proportion will be between .25 and .35 is given by P[ 0.20 < p < 0.40] = P[(0.25- 0.34)/0.0432 < ( p - 0.34)/0.0432 < (0.35- 0.34)/0.0432] =P[-2.0812 < Z < 0.2312] = P[ Z < 0.2312] – P[Z < -2.0812] = 0.5914 – 0.0187 = 0.5727 standard deviation