To find the probability $P(M > 166)$, where $M$ is the sample mean, we will follow these steps:

1. Identify the given values:

• Population mean $\mu = 183.3$
• Population standard deviation $\sigma = 48.1$
• Sample size $n = 25$
• Sample mean $M = 166$

2. Calculate the standard error of the mean (SEM):

The standard error of the mean is given by the formula:

$\text{SEM} = \frac{\sigma}{\sqrt{n}}$

Substitute the known values:

$\text{SEM} = \frac{48.1}{\sqrt{25}} = \frac{48.1}{5} = 9.62$

3. Find the z-score:

To find the probability, we need to calculate the z-score for $M = 166$. The z-score is given by:

$z = \frac{M - \mu}{\text{SEM}}$

Substitute the values:

$z = \frac{166 - 183.3}{9.62} = \frac{-17.3}{9.62} \approx -1.798$

4. Find the corresponding probability:

Now, we need to find the probability corresponding to this z-score. From the z-table, or using a standard normal distribution calculator:

• The probability for $z = -1.798$ is approximately 0.0361 (this is the area to the left of $z = -1.798$).

However, we are looking for the probability that the sample mean is greater than 166, which is the area to the right of this z-score. Thus:

$P(M > 166) = 1 - 0.0361 = 0.9639$

Final Answer:

$P(M > 166) \approx 0.9639$

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:

1. How does the sample size affect the standard error of the mean?
2. What is the significance of the z-score in normal distribution problems?
3. How would the probability change if the sample size increased?
4. Can you explain why we subtract the z-score probability from 1?
5. What would happen if the population standard deviation was smaller?

Tip: The standard error decreases as the sample size increases, making the sample mean more likely to be closer to the population mean.