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We have a normal distribution with:
- Mean (\(\mu\)) = 80.2
- Standard deviation (\(\sigma\)) = 93.6
- Sample size (\(n\)) = 50

The problem requires us to find two probabilities:

1. **Probability that a single randomly selected value \(X\) is between 67 and 73.6**:
   \[
   P(67 < X < 73.6)
   \]

2. **Probability that the mean of a sample of size \(n = 50\) is between 67 and 73.6**:
   \[
   P(67 < M < 73.6)
   \]

### Step-by-Step Solution

#### 1. Probability for a Single Value \(X\) (Question 1)

For a single value \(X\) drawn from a normal distribution with mean \(\mu = 80.2\) and standard deviation \(\sigma = 93.6\), we standardize using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]

For \(X = 67\):
\[
Z_{67} = \frac{67 - 80.2}{93.6}
\]

For \(X = 73.6\):
\[
Z_{73.6} = \frac{73.6 - 80.2}{93.6}
\]

Calculate these Z-scores and then find the probability \(P(67 < X < 73.6)\) by finding the area between these two Z-scores.

#### 2. Probability for the Sample Mean \(M\) (Question 2)

When considering the sample mean \(M\) of size \(n = 50\), the distribution of the sample mean has a mean \(\mu_M = \mu = 80.2\) and a standard deviation \(\sigma_M\) given by:
\[
\sigma_M = \frac{\sigma}{\sqrt{n}}
\]

Calculate \(\sigma_M\):
\[
\sigma_M = \frac{93.6}{\sqrt{50}}
\]

Now, we standardize for the sample mean:

For \(M = 67\):
\[
Z_{67} = \frac{67 - 80.2}{\sigma_M}
\]

For \(M = 73.6\):
\[
Z_{73.6} = \frac{73.6 - 80.2}{\sigma_M}
\]

After calculating these Z-scores, find \(P(67 < M < 73.6)\) by looking up the area between the two Z-scores.

Let me perform these calculations for precise results.The calculated probabilities are:

1. **Probability that a single randomly selected value \(X\) is between 67 and 73.6**:
   \[
   P(67 < X < 73.6) \approx 0.0280
   \]

2. **Probability that the mean of a sample of size \(n = 50\) is between 67 and 73.6**:
   \[
   P(67 < M < 73.6) \approx 0.1497
   \]

Would you like further explanation on any part of this solution?

---

Here are some follow-up questions for practice:

1. What is the probability that a single randomly selected value is less than 67?
2. What is the probability that a single randomly selected value is greater than 73.6?
3. How would the probabilities change if the sample size increased to 100?
4. What would be the probability if the range changed to \(70 < X < 80\)?
5. Can you explain why the sample mean probability differs from the single value probability?

**Tip:** When working with normal distributions, remember that the standard deviation of the sample mean decreases as the sample size increases, which results in a narrower distribution for the sample mean.

A population of values has a normal distribution with μ = 80.2 and σ = 93.6. You intend to draw a random sample of size n = 50. Find the probability that a single randomly selected value is between 67 and 73.6. Find the probability that a sample of size n = 50 is randomly selected with a mean between 67 and 73.6.

This problem explores the calculation of probabilities for a normal distribution with μ = 80.2 and σ = 93.6. Learn how to find the probability that a single value lies between 67 and 73.6, and how to calculate the probability that a sample mean of n = 50 is between 67 and 73.6. Suitable for college-level statistics students.

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