Question

The mean preparation fee h&r block charged retail customers last year was $183 (the wall street journal, march 7, 2012). use this price as the population mean and assume the population standard deviation of preparation fees is $50.
a. what is the probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean?
b. what is the probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean?
c. what is the probability that the mean price for a sample of 100

Answer 1

Given that the population mean, $\mu=\ 183$ and the population standard deviation, $\sigma=\ 50$

Part A:

The probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean is evaluated as follows:

$P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{30} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{30} } } \right) \\ \\ =P(z\leq0.8764)-P(z\leq-0.8764) \\ \\ =P(z\leq0.8764)-[1-P(z\leq0.8764)] \\ \\ =2P(z\leq0.8764)-1=2(0.80958)-1 \\ \\ =1.61916-1=0.61916\approx62\%$



Part B:

The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:

$P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{50} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{50} } } \right) \\ \\ =P(z\leq1.131)-P(z\leq-1.131) \\ \\ =P(z\leq1.131)-[1-P(z\leq1.131)] \\ \\ =2P(z\leq1.131)-1=2(0.87105)-1 \\ \\ =1.7421-1=0.7421\approx74\%$



Part C:

The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:

$P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{100} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{100} } } \right) \\ \\ =P(z\leq1.6)-P(z\leq-1.6) \\ \\ =P(z\leq1.6)-[1-P(z\leq1.6)] \\ \\ =2P(z\leq1.6)-1=2(0.9452)-1 \\ \\ =1.8904-1=0.8904\approx89\%$