A population of values has a normal distribution with u = 141.5
 and o = 95.2
. If a random sample of size n = 18
 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) = 


Find the probability that a sample of size n = 18
 is randomly selected with a mean greater than 208.8. Round your answer to four decimals. 
P(M > 208.8) = 

We are given a normally distributed population with:

- Population mean \( \mu = 141.5 \)
- Population standard deviation \( \sigma = 95.2 \)
- A random sample of size \( n = 18 \)

Now, we are asked to find two probabilities.

### 1. Probability that a single randomly selected value is greater than 208.8:
We need to calculate \( P(X > 208.8) \).

For a single value from a normal distribution, we first standardize the value using the **z-score** formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where:
- \( X = 208.8 \)
- \( \mu = 141.5 \)
- \( \sigma = 95.2 \)

Let's calculate this z-score.

\[
z = \frac{208.8 - 141.5}{95.2}
\] 

Next, we'll find the probability from the standard normal distribution table for this z-score, and subtract that from 1 to get \( P(X > 208.8) \).

### 2. Probability that a sample mean is greater than 208.8:

For the sample mean, we use the **sampling distribution** of the sample mean, which has a mean \( \mu = 141.5 \) and a standard deviation of \( \sigma_{\bar{X}} \), called the **standard error of the mean**:
\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}
\]

So, the standard error \( \sigma_{\bar{X}} \) is:
\[
\sigma_{\bar{X}} = \frac{95.2}{\sqrt{18}}
\]

Then, we calculate the z-score for the sample mean:
\[
z = \frac{M - \mu}{\sigma_{\bar{X}}}
\]
where \( M = 208.8 \), \( \mu = 141.5 \), and \( \sigma_{\bar{X}} \) is what we calculated above.

Finally, we find the probability from the z-score table, and subtract that from 1 to get \( P(M > 208.8) \).

I'll calculate both these z-scores and probabilities for you now.Here are the results:

1. The probability that a single randomly selected value is greater than 208.8 is:
\[
P(X > 208.8) = 0.2398
\]

2. The probability that a sample of size \( n = 18 \) is randomly selected with a mean greater than 208.8 is:
\[
P(M > 208.8) = 0.0014
\]

Would you like more details on these calculations or any additional clarifications?

Here are 5 related questions to further explore:

1. How do you calculate the probability for a value less than a given number in a normal distribution?
2. What happens to the probability if the sample size increases significantly?
3. What is the effect of standard deviation on the probabilities for normal distributions?
4. Can you explain why the probability for the sample mean is much smaller than for a single value?
5. How can you use z-scores to compare different normal distributions?

**Tip:** As the sample size increases, the sampling distribution of the sample mean becomes narrower, which affects probabilities.

A population of values has a normal distribution with μ = 141.5 and σ = 95.2. If a random sample of size n = 18 is selected, find the probability that a single randomly selected value is greater than 208.8. P(X > 208.8) = ? Find the probability that a sample of size n = 18 is randomly selected with a mean greater than 208.8. P(M > 208.8) = ?

Calculate the probability of selecting a single value greater than 208.8 and the probability that a sample mean is greater than 208.8 from a population with μ = 141.5 and σ = 95.2. Learn key statistical concepts like Z-scores, normal distribution, and sampling distribution. Suitable for advanced high school and college students.

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