Question

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?

Answer 1

Given the question:

The mean tax-return 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.

Round your answers to four decimal places.

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 H&R Block retail customers is within $8 of the population mean?



Part A:

The probability that a sample of size, n, of a normally distributed data with a mean of μ, and standard deviation of σ, is within, a, units of the mean is given by

 $2P(X\ \textless \ a)-1=2P\left(z\ \textless \ \frac{a}{\sigma/\sqrt{n}} \right)-1$

Thus, the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean is given by:

$2P\left(z\ \textless \ \frac{8}{50/\sqrt{30}} \right)-1=2P(z\ \textless \ 0.8764)-1 \\ \\ =2(0.80958)-1=1.61916-1=\bold{0.6192}$



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 given by:

$2P\left(z\ \textless \ \frac{8}{50/\sqrt{50}} \right)-1=2P(z\ \textless \ 1.131)-1 \\ \\ =2(0.87105)-1=1.7421-1=\bold{0.7421}$



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 given by:

$2P\left(z\ \textless \ \frac{8}{50/\sqrt{100}} \right)-1=2P(z\ \textless \ 1.6)-1 \\ \\ =2(0.9452)-1=1.8904-1=\bold{0.8904}$