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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