To solve the problem, let's first establish the parameters of the normal distribution:
- Mean (μ) = 58.1
- Standard deviation (σ) = 33.9
- Sample size (n) = 16
1. Find the probability that a single randomly selected value is greater than 66.6.
We need to compute \( P(X > 66.6) \). For a normal distribution, we can convert a single value to a z-score using the formula:
\[ z = \frac{X - \mu}{\sigma} \]
Substituting the values:
\[ z = \frac{66.6 - 58.1}{33.9} = \frac{8.5}{33.9} \approx 0.250735 \]
Now, we can look up this z-score in the standard normal distribution table or use a calculator:
\[ P(Z > 0.250735) = 1 - P(Z \leq 0.250735) \]
From a standard normal distribution table or calculator:
\[ P(Z \leq 0.250735) \approx 0.5987 \]
Thus:
\[ P(Z > 0.250735) \approx 1 - 0.5987 = 0.4013 \]
So,
\[ P(X > 66.6) \approx 0.4013 \]
2. Find the probability that a sample of size n=16 is randomly selected with a mean greater than 66.6.
For a sample mean, the standard deviation (standard error) is given by:
\[ \sigma_M = \frac{\sigma}{\sqrt{n}} \]
Calculating the standard error for the sample size \( n = 16 \):
\[ \sigma_M = \frac{33.9}{\sqrt{16}} = \frac{33.9}{4} = 8.475 \]
Now, we need to calculate \( P(M > 66.6) \) using the z-score formula for the sample mean:
\[ z = \frac{M - \mu}{\sigma_M} \]
Substituting the values:
\[ z = \frac{66.6 - 58.1}{8.475} = \frac{8.5}{8.475} \approx 1.00295 \]
Now again, we can look up this z-score:
\[ P(Z > 1.00295) = 1 - P(Z \leq 1.00295) \]
From a standard normal distribution table or calculator:
\[ P(Z \leq 1.00295) \approx 0.8413 \]
Thus:
\[ P(Z > 1.00295) \approx 1 - 0.8413 = 0.1587 \]
So,
\[ P(M > 66.6) \approx 0.1587 \]
Summary of results:
- \( P(X > 66.6) \approx 0.4013 \)
- \( P(M > 66.6) \approx 0.1587 \)
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