Assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of $\mu=1.1 \mathrm{~kg}$ and a standard deviation of $\sigma=5.7 \mathrm{~kg}$. Complete parts (a) through (c) below.
a. If 1 male college student is randomly selected, find the probability that he gains between $0 \mathrm{~kg}$ and $3 \mathrm{~kg}$ during freshman year.
The probability is
(Round to four decimal places as needed)
b. If 4 male college students are randomly selected, find the probability that their mean weight gain during freshman year is between $0 \mathrm{~kg}$ and $3 \mathrm{~kg}$.
The probability is
(Round to four decimal places as needed.)

Step 1 ：First, we need to standardize the weights by subtracting the mean and dividing by the standard deviation. This will give us a standard normal distribution with a mean of 0 and a standard deviation of 1.

Step 2 ：For part (a), we want to find the probability that a randomly selected student gains between 0 kg and 3 kg. We first convert these weights to the standard normal distribution: \(Z_1 = \frac{0 - 1.1}{5.7} = -0.193\) and \(Z_2 = \frac{3 - 1.1}{5.7} = 0.333\).

Step 3 ：We then look up these Z-scores in the standard normal distribution table or use a calculator to find the corresponding probabilities. The probability that Z is between -0.193 and 0.333 is \(P(-0.193 < Z < 0.333) = P(Z < 0.333) - P(Z < -0.193)\).

Step 4 ：Using the standard normal distribution table or a calculator, we find that \(P(Z < 0.333) = 0.6293\) and \(P(Z < -0.193) = 0.4236\).

Step 5 ：So, the probability that a randomly selected student gains between 0 kg and 3 kg is \(0.6293 - 0.4236 = 0.2057\).

Step 6 ：For part (b), we want to find the probability that the mean weight gain of 4 randomly selected students is between 0 kg and 3 kg. The standard deviation of the mean of 4 students is \(\frac{5.7}{\sqrt{4}} = 2.85\).

Step 7 ：We convert these weights to the standard normal distribution: \(Z_1 = \frac{0 - 1.1}{2.85} = -0.386\) and \(Z_2 = \frac{3 - 1.1}{2.85} = 0.667\).

Step 8 ：The probability that Z is between -0.386 and 0.667 is \(P(-0.386 < Z < 0.667) = P(Z < 0.667) - P(Z < -0.386)\).

Step 9 ：Using the standard normal distribution table or a calculator, we find that \(P(Z < 0.667) = 0.7522\) and \(P(Z < -0.386) = 0.3497\).

Step 10 ：So, the probability that the mean weight gain of 4 randomly selected students is between 0 kg and 3 kg is \(0.7522 - 0.3497 = 0.4025\).

Step 11 ：Thus, the probability that a randomly selected student gains between 0 kg and 3 kg is approximately \(\boxed{0.2057}\) and the probability that the mean weight gain of 4 randomly selected students is between 0 kg and 3 kg is approximately \(\boxed{0.4025}\).