To solve this, we will use the properties of the normal distribution.

Given:

• Mean $\mu = 1.1 \, \text{kg}$
• Standard deviation $\sigma = 4.6 \, \text{kg}$

We are asked to find the probability that a randomly selected male college student gains between 0 kg and 3 kg during their freshman year. This is equivalent to finding the probability that $X$ (the weight gain) is between 0 and 3, i.e., $P(0 \leq X \leq 3)$.

We can standardize this using the z-score formula:

$z = \frac{x - \mu}{\sigma}$

Step 1: Find the z-scores for 0 kg and 3 kg.

• For $X = 0 \, \text{kg}$: $z_0 = \frac{0 - 1.1}{4.6} = \frac{-1.1}{4.6} \approx -0.2391$

• For $X = 3 \, \text{kg}$: $z_3 = \frac{3 - 1.1}{4.6} = \frac{1.9}{4.6} \approx 0.4130$

Step 2: Use the standard normal distribution table or a calculator to find the probabilities corresponding to these z-scores.

• $P(z_0) \approx 0.4052$ (probability corresponding to $z_0 = -0.2391$)
• $P(z_3) \approx 0.6604$ (probability corresponding to $z_3 = 0.4130$)

Step 3: Calculate the probability.

The probability that the student gains between 0 kg and 3 kg is:

$P(0 \leq X \leq 3) = P(z_3) - P(z_0) = 0.6604 - 0.4052 = 0.2552$

So, the probability is approximately 0.2552.

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Would you like further clarification, or any details on this process?

Here are 5 additional questions based on this problem:

1. How would the probability change if the standard deviation was larger?
2. What is the probability that a student gains more than 3 kg during their freshman year?
3. What z-score corresponds to a weight gain of 5 kg?
4. If we randomly select 5 students, how would the probability of their average weight gain being between 0 and 3 kg change?
5. How does the assumption of normal distribution affect the interpretation of this result?

Tip: The z-score standardizes values, allowing us to compare different normal distributions using the standard normal (mean 0, standard deviation 1).