To find the probability $P(36.8 < X < 39.3)$, where $X$ is normally distributed with a mean $\mu = 36.8$ and standard deviation $\sigma = 2.5$, we need to standardize the values using the Z-score formula:

$Z = \frac{X - \mu}{\sigma}$

Step 1: Find the Z-scores

1. For $X = 36.8$: $Z_1 = \frac{36.8 - 36.8}{2.5} = 0$
2. For $X = 39.3$: $Z_2 = \frac{39.3 - 36.8}{2.5} = \frac{2.5}{2.5} = 1$

Step 2: Use the standard normal distribution table

• The probability corresponding to $Z_1 = 0$ is 0.5000.
• The probability corresponding to $Z_2 = 1$ is 0.8413.

Step 3: Calculate the probability

The probability $P(36.8 < X < 39.3)$ is the difference between the two probabilities:

$P(36.8 < X < 39.3) = P(Z_2) - P(Z_1) = 0.8413 - 0.5000 = 0.3413$

Final Answer:

$P(36.8 < X < 39.3) = 0.3413$

Would you like further explanation or details?

Here are 5 related questions to expand on this:

1. How do you find probabilities for other intervals using Z-scores?
2. What is the meaning of a Z-score in statistics?
3. How does changing the standard deviation affect the probability in a normal distribution?
4. What if the population were not normally distributed—how would you approach this problem?
5. How do you calculate probabilities for values outside one standard deviation from the mean?

Tip: Always ensure to standardize values with the Z-score formula when dealing with normal distributions.