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eBook The mean preparation fee H&R Block charged retail customers in 2012 was  (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is . Use z-table. Round your answers to four decimal places. a. What is the probability that the mean price for a sample of  H&R Block retail customers is within  of the population mean? b. What is the probability that the mean price for a sample of  H&R Block retail customers is within  of the population mean? c. What is the probability that the mean price for a sample of  H&R Block retail customers is within  of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a  probability that the sample mean is within  of the population mean? - Select your answer -Sample of 30 H&R Block retail customersSample of 50 H&R Block retail customersSample of 100 H&R Block retail customersNone of the sample sizes in parts (a), (b), and (c) are large.All of the sample sizes in parts (a), (b), and (c) are large.

Answer 1

Given: Population mean, =$183 and Population standard deviation, =$50

a.

Margin of error, ME =$8 and sample size, n =30

Interval = =1838 =(175, 191)

Z =

P(175 191) =P(​​​​​​) =P(-0.8764 0.8764) =0.6192

b.

Margin of error, ME =$8 and sample size, n =50

Interval = =1838 =(175, 191)

Z =

P(175191)=P(​)

=P(-1.1314 1.1314) =0.7421

c.

Margin of error, ME =$8 and sample size, n =100

Interval = =1838 =(175, 191)

Z =

P(175191) =P(​​​​​​) =P(-1.6 1.6) =0.8904​​​​​​

d.

Neither of the sample sizes in parts (a), (b), and (c) I would recommend to have at least a 0.95 probability that the sample mean is within $8 of the population mean because the probabilities in (a), (b) and (c) above are less than 0.95