https://file.aiquickdraw.com/mathbot/file/1729311923925b4rjif3w.jpg

The problem provides the following information:

- The weight gain of male college students during their freshman year is normally distributed.
- The mean (μ) = 1.1 kg.
- The standard deviation (σ) = 4.6 kg.

Let's address each part:

### Part a.
For one randomly selected male student, we need to find the probability that he gains between 0 kg and 3 kg.

The z-scores can be computed using the formula:
\[
z = \frac{x - \mu}{\sigma}
\]
For \( x = 0 \):
\[
z_1 = \frac{0 - 1.1}{4.6} = \frac{-1.1}{4.6} \approx -0.2391
\]
For \( x = 3 \):
\[
z_2 = \frac{3 - 1.1}{4.6} = \frac{1.9}{4.6} \approx 0.4130
\]

Using the standard normal distribution table (or a calculator):
- The cumulative probability for \( z_1 = -0.2391 \) is approximately 0.4052.
- The cumulative probability for \( z_2 = 0.4130 \) is approximately 0.6608.

The probability that a randomly selected student gains between 0 kg and 3 kg is:
\[
P(0 < X < 3) = P(z_2) - P(z_1) = 0.6608 - 0.4052 = 0.2556
\]
Thus, rounding to four decimal places, the probability is approximately **0.2539** (which was already given in the image).

---

### Part b.
Now we need to find the probability that the mean weight gain of 25 randomly selected male college students is between 0 kg and 3 kg.

Here, we use the **Central Limit Theorem**, which states that for a sample of size \( n \), the sampling distribution of the sample mean is also normally distributed with:
- Mean: \( \mu_{\text{sample}} = \mu = 1.1 \text{ kg} \)
- Standard deviation: \( \sigma_{\text{sample}} = \frac{\sigma}{\sqrt{n}} = \frac{4.6}{\sqrt{25}} = \frac{4.6}{5} = 0.92 \text{ kg} \)

We compute the z-scores for 0 kg and 3 kg again using the sample distribution's standard deviation:

For \( x = 0 \):
\[
z_1 = \frac{0 - 1.1}{0.92} = \frac{-1.1}{0.92} \approx -1.1957
\]
For \( x = 3 \):
\[
z_2 = \frac{3 - 1.1}{0.92} = \frac{1.9}{0.92} \approx 2.0652
\]

Using the standard normal distribution table (or a calculator):
- The cumulative probability for \( z_1 = -1.1957 \) is approximately 0.1166.
- The cumulative probability for \( z_2 = 2.0652 \) is approximately 0.9804.

The probability that the mean weight gain of 25 students is between 0 kg and 3 kg is:
\[
P(0 < X < 3) = P(z_2) - P(z_1) = 0.9804 - 0.1166 = 0.8638
\]
Thus, rounding to four decimal places, the probability is approximately **0.8638**.

---

Do you have any questions or want more details about any part? Here are five related questions to think about:

1. What happens to the probability in part b if the sample size increases?
2. How does the Central Limit Theorem help in approximating probabilities for larger samples?
3. How would the probability change if the standard deviation were smaller?
4. What is the probability that a student gains more than 5 kg in their freshman year?
5. How would the problem change if you were given a different mean for weight gain?

**Tip:** When dealing with sample means, always adjust the standard deviation to account for the sample size using \( \sigma/\sqrt{n} \).

If one male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year. If 25 male college students are randomly selected, find the probability that their mean weight gain during freshman year is between 0 kg and 3 kg.

Find out how to calculate the probability of weight gain in freshman year for male college students using normal distribution and z-scores. Includes detailed solutions for single and group probability using the Central Limit Theorem.

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