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