To find the probabilities requested, we'll use the properties of the normal distribution.
Part 1: Probability that a single randomly selected value is greater than 33.2
The given population parameters are:
- Mean (μ) = 21
- Standard deviation (σ) = 33.7
We want to find: \[ P(X > 33.2) \]
First, we need to calculate the Z-score for \( x = 33.2 \):
\[ Z = \frac{X - \mu}{\sigma} = \frac{33.2 - 21}{33.7} = \frac{12.2}{33.7} \approx 0.3629 \]
Now, we can look up the Z-score in the standard normal distribution table or use a calculator to find \( P(Z < 0.3629) \).
Using a Z-table or calculator, we find:
\[ P(Z < 0.3629) \approx 0.6436 \]
Thus, to find \( P(X > 33.2) \):
\[ P(X > 33.2) = 1 - P(Z < 0.3629) \approx 1 - 0.6436 = 0.3564 \]
Part 2: Probability that a sample of size n=15 has a mean greater than 33.2
For the sample mean, we will use the Central Limit Theorem. The mean of the sample means (\( \mu_M \)) remains \( \mu = 21 \), but the standard deviation of the sample means (standard error, \( \sigma_M \)) is calculated as:
\[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{33.7}{\sqrt{15}} \approx \frac{33.7}{3.8729} \approx 8.693 \]
Now, we want to find: \[ P(M > 33.2) \]
We calculate the Z-score for the sample mean:
\[ Z = \frac{M - \mu_M}{\sigma_M} = \frac{33.2 - 21}{8.693} = \frac{12.2}{8.693} \approx 1.402 \]
Now we can look up the Z-score in the standard normal distribution table or use a calculator to find \( P(Z < 1.402) \).
Using a Z-table or calculator, we find:
\[ P(Z < 1.402) \approx 0.9191 \]
Thus, to find \( P(M > 33.2) \):
\[ P(M > 33.2) = 1 - P(Z < 1.402) \approx 1 - 0.9191 = 0.0809 \]
Final Answers:
- \( P(X > 33.2) \approx 0.3564 \)
- \( P(M > 33.2) \approx 0.0809 \)
Therefore:
- \( P(X > 33.2) = 0.3564 \)
- \( P(M > 33.2) = 0.0809 \)