A population of values has a normal distribution with 
 and 
. If a random sample of size 
 is selected,

Find the probability that a single randomly selected value is greater than 208.8. Round your answer to four decimals. 
P(X > 208.8) = 
0.1894
Incorrect

Find the probability that a sample of size 
 is randomly selected with a mean greater than 208.8. Round your answer to four decimals. 
P(M > 208.8) = 
0


To solve this, we need to break it down into two separate parts:

### Part 1: Probability that a single randomly selected value \( X \) is greater than 208.8
The problem provides us with the following information:
- The population has a normal distribution, with a given mean \( \mu \) and standard deviation \( \sigma \).
- We need to find \( P(X > 208.8) \), which is the probability that a randomly selected value is greater than 208.8.

Unfortunately, you did not provide the mean \( \mu \) and standard deviation \( \sigma \) of the population. I’ll assume they are missing from the problem statement you copied, and you should provide these values.

#### The procedure for finding \( P(X > 208.8) \):
1. **Standardize the value** to a \( Z \)-score using the formula:

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

2. Use the standard normal distribution to find the probability associated with this \( Z \)-value.
3. Subtract the cumulative probability from 1, since you want \( P(X > 208.8) \).

---

### Part 2: Probability that a sample mean \( M \) is greater than 208.8
For the second part of the question, where you are asked to find the probability of the **sample mean** being greater than 208.8, you need the sample size \( n \). Again, the value for \( n \) is missing from the information provided.

In this case, we work with the **sampling distribution** of the sample mean, which has a mean \( \mu \) and a standard deviation known as the **standard error** of the mean, calculated as:

\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}}
\]

The steps to find \( P(M > 208.8) \):
1. **Standardize** the sample mean to a \( Z \)-score:

   \[
   Z = \frac{M - \mu}{\text{SE}} = \frac{208.8 - \mu}{\frac{\sigma}{\sqrt{n}}}
   \]

2. Use the standard normal distribution to find the corresponding probability.
3. As before, subtract the cumulative probability from 1 to find \( P(M > 208.8) \).

To complete these calculations, please provide the following:
- The population mean \( \mu \)
- The population standard deviation \( \sigma \)
- The sample size \( n \)

Would you like me to proceed with any assumptions, or can you provide the missing values?

---

Here are 5 related questions to expand understanding:
1. How do you calculate the standard error of the mean when the population standard deviation is known?
2. What is the difference between the population mean and sample mean?
3. How does the sample size affect the standard error of the mean?
4. Why do we subtract from 1 to find \( P(X > \text{value}) \) in a normal distribution?
5. How does the Central Limit Theorem justify using a normal distribution for sample means?

**Tip**: Always check if the problem is asking for probabilities of individual values or sample means; the approaches differ!

A population of values has a normal distribution with mean (μ) and standard deviation (σ). If a random sample of size n is selected, find the probability that a single randomly selected value is greater than 208.8. Find the probability that a sample of size n is randomly selected with a mean greater than 208.8.

This guide walks through solving a probability problem in a normal distribution for a single randomly selected value and a sample mean greater than 208.8. Learn to apply Z-scores and the Central Limit Theorem to find probabilities. Suitable for Grades 11-12 or undergraduate statistics students.

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