To solve these problems, we will use the properties of the normal distribution.
Part 1: Finding \( P(X > 66.6) \)
Given that the population mean (μ) is 58.1 and the population standard deviation (σ) is 33.9, we want to find the probability that a single randomly selected value \( X \) is greater than 66.6.
First, we will compute the z-score for \( X = 66.6 \):
\[ z = \frac{X - \mu}{\sigma} = \frac{66.6 - 58.1}{33.9} \]
Calculating the numerator:
\[ 66.6 - 58.1 = 8.5 \]
Now calculating the z-score:
\[ z = \frac{8.5}{33.9} \approx 0.2507 \]
Next, we look up this z-score in the standard normal distribution table, or we can use a calculator to find \( P(Z > 0.2507) \):
\[ P(Z > 0.2507) = 1 - P(Z \leq 0.2507) \]
Using the standard normal distribution table or computational tools, we find:
\[ P(Z \leq 0.2507) \approx 0.5987 \]
Therefore,
\[ P(Z > 0.2507) \approx 1 - 0.5987 = 0.4013 \]
Answer for Part 1:
\[ P(X > 66.6) \approx 0.4013 \]
Part 2: Finding \( P(M > 66.6) \)
For a sample mean \( M \), we need to adjust the standard deviation using the formula for the standard error of the mean (SEM):
\[ SEM = \frac{\sigma}{\sqrt{n}} = \frac{33.9}{\sqrt{16}} = \frac{33.9}{4} = 8.475 \]
Now we will find the z-score for the sample mean:
\[ z = \frac{M - \mu}{SEM} = \frac{66.6 - 58.1}{8.475} \]
Calculating the numerator:
\[ 66.6 - 58.1 = 8.5 \]
Calculating the z-score:
\[ z = \frac{8.5}{8.475} \approx 1.0029 \]
Now, we look up this z-score to find \( P(Z > 1.0029) \):
\[ P(Z > 1.0029) = 1 - P(Z \leq 1.0029) \]
Using the standard normal distribution table or computational tools, we find:
\[ P(Z \leq 1.0029) \approx 0.8413 \]
Thus,
\[ P(Z > 1.0029) \approx 1 - 0.8413 = 0.1587 \]
Answer for Part 2:
\[ P(M > 66.6) \approx 0.1587 \]
Summary of Results:
- \( P(X > 66.6) \approx 0.4013 \)
- \( P(M > 66.6) \approx 0.1587 \)