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Let's solve each part of the question step-by-step.

### Given Data
- Population mean (\(\mu\)) = 188.1
- Population standard deviation (\(\sigma\)) = 97.2
- Sample size (\(n\)) = 21

### Part (a)
We need to find the probability that a single randomly selected value is greater than 162.6.

1. **Calculate the Z-score for a single value \(X\):**
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Plugging in the values:
   \[
   Z = \frac{162.6 - 188.1}{97.2} = \frac{-25.5}{97.2} \approx -0.2623
   \]

2. **Find \(P(X > 162.6)\):**
   This is equivalent to finding \(P(Z > -0.2623)\).
   
   Using the standard normal distribution table, we find \(P(Z < -0.2623) \approx 0.3966\).

   Therefore, \(P(X > 162.6) = 1 - 0.3966 = 0.6034\).

   So, the answer to part (a) is approximately:
   \[
   P(X > 162.6) \approx 0.6034
   \]

### Part (b)
Now we need to find the probability that a sample mean (\(M\)) of size \(n = 21\) is greater than 162.6.

1. **Calculate the standard error of the mean (\(\sigma_M\)):**
   \[
   \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{97.2}{\sqrt{21}} \approx 21.22
   \]

2. **Calculate the Z-score for the sample mean \(M\):**
   \[
   Z = \frac{M - \mu}{\sigma_M} = \frac{162.6 - 188.1}{21.22} \approx -1.2024
   \]

3. **Find \(P(M > 162.6)\):**
   This is equivalent to finding \(P(Z > -1.2024)\).
   
   Using the standard normal distribution table, we find \(P(Z < -1.2024) \approx 0.1143\).

   Therefore, \(P(M > 162.6) = 1 - 0.1143 = 0.8857\).

   So, the answer to part (b) is approximately:
   \[
   P(M > 162.6) \approx 0.8857
   \]

### Summary of Answers
- **(a)** \(P(X > 162.6) \approx 0.6034\)
- **(b)** \(P(M > 162.6) \approx 0.8857\)

Would you like further details on any step or clarification on how to interpret these probabilities?

#### Additional Related Questions
1. What would happen to \(P(X > 162.6)\) if the mean \(\mu\) were lower?
2. How does the sample size affect the probability in part (b)?
3. What is the probability that a single randomly selected value is less than 162.6?
4. How would the answers change if the standard deviation were smaller?
5. What would be the Z-score if the sample mean \(M\) were exactly equal to 188.1?

#### Tip:
When calculating probabilities involving sample means, always adjust the standard deviation by dividing by the square root of the sample size, as this accounts for the Central Limit Theorem.

A population of values has a normal distribution with μ = 188.1 and σ = 97.2. If a random sample of size n = 21 is selected, a) Find the probability that a single randomly selected value is greater than 162.6. b) Find the probability that a sample of size n = 21 is randomly selected with a mean greater than 162.6.

Solve a normal distribution problem where the population has a mean of 188.1 and standard deviation of 97.2. Learn how to calculate the probability for a single value and the sample mean greater than 162.6, using Z-scores and the standard normal distribution.

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