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.