To solve this problem, we can use the Central Limit Theorem. Since the population distribution is normal, the sample mean $M$ will also follow a normal distribution. Here are the steps:

1. Problem Setup

• Population Mean ( $\mu$ ): 12.7
• Population Standard Deviation ( $\sigma$ ): 75
• Sample Size ( $n$ ): 48
• We need to find $P(M > 12.7)$, the probability that the sample mean $M$ is greater than 12.7.

2. Standard Error of the Mean

The standard error of the mean (SEM) is calculated by: $\text{SEM} = \frac{\sigma}{\sqrt{n}}$ Substitute the values: $\text{SEM} = \frac{75}{\sqrt{48}} \approx 10.8253$

3. Z-Score Calculation

The probability $P(M > 12.7)$ is the same as finding the probability that the Z-score of the sample mean is greater than 0 (since the sample mean equals the population mean). The Z-score formula for the sample mean is: $Z = \frac{M - \mu}{\text{SEM}}$ Substitute the values: $Z = \frac{12.7 - 12.7}{10.8253} = 0$

4. Finding the Probability

The Z-score of 0 corresponds to a probability of 0.5, meaning: $P(M > 12.7) = 0.5$

Final Answer

$P(M > 12.7) = 0.5000$

Would you like more details on this process, or do you have any other questions?

Here are some related questions you might find helpful:

1. How does the Central Limit Theorem apply when the sample size is small?
2. What is the difference between population standard deviation and standard error?
3. How can we calculate probabilities for non-normal distributions?
4. How would the probability change if the sample size were different?
5. What is the significance of Z-scores in probability calculations?

Tip: When dealing with sample means, always use the standard error (SEM) instead of the population standard deviation in your calculations.