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?
d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within $8 of the population mean?
A. .6212
this is correct
To solve these probability questions, we will use the Central Limit Theorem (CLT) because we have a large enough sample size (n ≥ 30).
The CLT states that for a large sample size, the sampling distribution of the sample mean follows a normal distribution, regardless of the shape of the population distribution.
Let's denote:
μ = population mean = $183
σ = population standard deviation = $50
n_a = sample size for part (a) = 30
n_b = sample size for part (b) = 50
n_c = sample size for part (c) = 100
a. To find the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean, we need to calculate the z-score for the upper and lower bounds of the range.
z_lower = (183 - 8 - μ) / (σ / sqrt(n_a))
z_upper = (183 + 8 - μ) / (σ / sqrt(n_a))
Now, we can use the z-scores to find the probabilities using a standard normal distribution table or a calculator. The probability we need is the area between these two z-scores.
P(a) = P(z_lower < Z < z_upper)
b. For part (b), we will follow the same steps as part (a), but using the sample size for part (b).
z_lower = (183 - 8 - μ) / (σ / sqrt(n_b))
z_upper = (183 + 8 - μ) / (σ / sqrt(n_b))
P(b) = P(z_lower < Z < z_upper)
c. Similarly for part (c), we will follow the same steps but using the sample size for part (c).
z_lower = (183 - 8 - μ) / (σ / sqrt(n_c))
z_upper = (183 + 8 - μ) / (σ / sqrt(n_c))
P(c) = P(z_lower < Z < z_upper)
d. To find the sample size that gives us at least a .95 probability that the sample mean is within $8 of the population mean, we can repeat the calculations for different sample sizes until we find the one that satisfies the condition.
We can start checking with n = 30, and then increment it to 50, 100, etc., until we find the sample size that satisfies the condition.