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Let's break down both parts of this problem step by step.

### 1. Probability for a single randomly selected value:
We are asked to find the probability that a single randomly selected value \( X \) falls between 8.7 and 28.1, given that the population has a normal distribution with mean \( \mu = 27.4 \) and standard deviation \( \sigma = 96.8 \).

We will use the **z-score formula** to transform these values into standard normal distribution values:

\[
z = \frac{X - \mu}{\sigma}
\]

- For \( X = 8.7 \):
  \[
  z_1 = \frac{8.7 - 27.4}{96.8} = \frac{-18.7}{96.8} \approx -0.1932
  \]

- For \( X = 28.1 \):
  \[
  z_2 = \frac{28.1 - 27.4}{96.8} = \frac{0.7}{96.8} \approx 0.0072
  \]

Now, using a **standard normal table** or a calculator:
- The cumulative probability corresponding to \( z_1 = -0.1932 \) is approximately **0.4235**.
- The cumulative probability corresponding to \( z_2 = 0.0072 \) is approximately **0.5029**.

Thus, the probability that \( X \) is between 8.7 and 28.1 is:
\[
P(8.7 < X < 28.1) = 0.5029 - 0.4235 = 0.0794
\]

### 2. Probability for a sample mean:
Now, we are finding the probability that a sample of size \( n = 168 \) has a mean between 8.7 and 28.1. For this, we use the **sampling distribution of the sample mean**. The standard deviation of the sample mean (standard error) is given by:

\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{96.8}{\sqrt{168}} \approx 7.467
\]

We again use the z-score formula, but this time for the sample mean:

\[
z = \frac{M - \mu}{\sigma_{\bar{X}}}
\]

- For \( M = 8.7 \):
  \[
  z_1 = \frac{8.7 - 27.4}{7.467} = \frac{-18.7}{7.467} \approx -2.504
  \]

- For \( M = 28.1 \):
  \[
  z_2 = \frac{28.1 - 27.4}{7.467} = \frac{0.7}{7.467} \approx 0.094
  \]

Now, using a **standard normal table** or a calculator:
- The cumulative probability corresponding to \( z_1 = -2.504 \) is approximately **0.0062**.
- The cumulative probability corresponding to \( z_2 = 0.094 \) is approximately **0.5374**.

Thus, the probability that the sample mean is between 8.7 and 28.1 is:
\[
P(8.7 < M < 28.1) = 0.5374 - 0.0062 = 0.5312
\]

### Final Answers:
1. \( P(8.7 < X < 28.1) = 0.0794 \)
2. \( P(8.7 < M < 28.1) = 0.5312 \)

Would you like more clarification on any of the steps?

Here are 5 related questions:
1. How would the probability change if the sample size increased to 200?
2. What is the probability that a single randomly selected value is less than 27.4?
3. How do you find the z-scores for values far outside the mean, like 50 or 100?
4. What assumptions must hold for the sample mean to follow a normal distribution?
5. How would the probabilities change if the standard deviation were 50 instead of 96.8?

**Tip:** When dealing with large sample sizes, the sampling distribution of the sample mean has less variability than the population, which is why the sample mean often results in higher probabilities compared to individual values.

A population of values has a normal distribution with mean (μ) = 27.4 and standard deviation (σ) = 96.8. You intend to draw a random sample of size n = 168. Find the probability that a single randomly selected value is between 8.7 and 28.1. Also, find the probability that a sample of size n = 168 is randomly selected with a mean between 8.7 and 28.1.

Solve a statistics problem involving a normal distribution with mean 27.4 and standard deviation 96.8. Learn how to calculate the probability that a single value or sample mean falls within a specified range using z-scores and the Central Limit Theorem.

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