A population of values has a normal distribution with  
μ
=
115.2
  and  
σ
=
66.6
 . You intend to draw a random sample of size  
n
=
237
 .

Find the probability that a single randomly selected value is between 102.7 and 103.1.
P(102.7 < X < 103.1) = 


Find the probability that a sample of size  
n
=
237
  is randomly selected with a mean between 102.7 and 103.1.
P(102.7 < M < 103.1) = 


Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

### Given:
- Population mean \( \mu = 115.2 \)
- Population standard deviation \( \sigma = 66.6 \)
- Sample size \( n = 237 \)

We are tasked with finding the following probabilities:
1. \( P(102.7 < X < 103.1) \) where \( X \) is a single randomly selected value from the population.
2. \( P(102.7 < M < 103.1) \) where \( M \) is the sample mean of a sample of size \( n = 237 \).

### 1. Finding \( P(102.7 < X < 103.1) \) for a single value

This requires using the standard normal distribution. We first convert the values 102.7 and 103.1 into **z-scores** using the formula:

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

#### Calculating z-scores:
For \( X = 102.7 \):
\[
z_1 = \frac{102.7 - 115.2}{66.6} = \frac{-12.5}{66.6} \approx -0.1877
\]

For \( X = 103.1 \):
\[
z_2 = \frac{103.1 - 115.2}{66.6} = \frac{-12.1}{66.6} \approx -0.1817
\]

Now, we find the probability between these two z-scores by looking up the values in the standard normal distribution table (or using a calculator).

The cumulative probabilities for these z-scores are:
- \( P(Z < -0.1877) \approx 0.4255 \)
- \( P(Z < -0.1817) \approx 0.4280 \)

Thus, the probability \( P(102.7 < X < 103.1) \) is the difference between these two probabilities:
\[
P(102.7 < X < 103.1) = 0.4280 - 0.4255 = 0.0025
\]

### 2. Finding \( P(102.7 < M < 103.1) \) for a sample mean

When dealing with the sample mean, we use the **sampling distribution** of the mean. The standard deviation of the sampling distribution (standard error) is given by:

\[
\sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{66.6}{\sqrt{237}} \approx 4.3259
\]

Next, we convert the values 102.7 and 103.1 to z-scores using this standard error.

For \( M = 102.7 \):
\[
z_1 = \frac{102.7 - 115.2}{4.3259} \approx \frac{-12.5}{4.3259} \approx -2.8897
\]

For \( M = 103.1 \):
\[
z_2 = \frac{103.1 - 115.2}{4.3259} \approx \frac{-12.1}{4.3259} \approx -2.7967
\]

The cumulative probabilities for these z-scores are:
- \( P(Z < -2.8897) \approx 0.0019 \)
- \( P(Z < -2.7967) \approx 0.0026 \)

Thus, the probability \( P(102.7 < M < 103.1) \) is:
\[
P(102.7 < M < 103.1) = 0.0026 - 0.0019 = 0.0007
\]

### Final Answers:
- \( P(102.7 < X < 103.1) = 0.0025 \)
- \( P(102.7 < M < 103.1) = 0.0007 \)

Would you like further details on any part of the process? 

Here are some related questions:
1. How does increasing the sample size affect the probability for the sample mean?
2. What is the significance of the standard error in a sampling distribution?
3. How do cumulative probabilities relate to z-scores?
4. What assumptions are made when using the normal distribution for this problem?
5. How does the normal distribution apply to real-world situations?

**Tip:** When working with large samples, the distribution of the sample mean tends to be more concentrated around the population mean due to the smaller standard error.

A population of values has a normal distribution with μ = 115.2 and σ = 66.6. You intend to draw a random sample of size n = 237.

Find the probability that a single randomly selected value is between 102.7 and 103.1. P(102.7 < X < 103.1) = 

Find the probability that a sample of size n = 237 is randomly selected with a mean between 102.7 and 103.1. P(102.7 < M < 103.1) = 

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

This problem involves calculating the probability for a single randomly selected value and a sample mean within a normal distribution. Given a population with mean μ = 115.2 and standard deviation σ = 66.6, find the probability for values between 102.7 and 103.1. A step-by-step solution includes calculating z-scores and understanding sampling distribution for a sample size n = 237.

Math Problem Statement

A population of values has a normal distribution with
μ

115.2 and
σ

66.6 . You intend to draw a random sample of size
n

Find the probability that a sample of size
n

Solution

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