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 \( \sigma=5.3 \mathrm{~kg} \). Complete parts (a) through (c) below. a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year. \( 33.33 \%, 4 \) of 12 points The probability is 0.2223 . (Round to four decimal places as needed.) b. 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 . The probability is \( \square \). (Round to four decimal places as needed.)

**Step 1. Define the Random Variables** Let \( X \) represent the weight gain of a single male college student. We are given: - \( X \sim N(\mu = 1.1, \sigma = 5.3) \). For part (b), if we select 25 students, let \( \bar{X} \) be the sample mean of their weight gains. By the Central Limit Theorem: - \( \bar{X} \sim N\left(\mu = 1.1,\ \sigma_{\bar{X}} = \frac{5.3}{\sqrt{25}} = \frac{5.3}{5} = 1.06\right) \). --- **Step 2. Part (a): (For One Student)** We need to find \[ P(0 < X < 3). \] Convert the endpoints to the standard normal \( Z \)-values: - For \( X=0 \): \[ Z_1 = \frac{0 - 1.1}{5.3} \approx \frac{-1.1}{5.3} \approx -0.2075. \] - For \( X=3 \): \[ Z_2 = \frac{3 - 1.1}{5.3} \approx \frac{1.9}{5.3} \approx 0.3585. \] Thus, \[ P(0 < X < 3) = P(-0.2075 < Z < 0.3585) = \Phi(0.3585) - \Phi(-0.2075). \] Using the standard normal table or calculator, - \( \Phi(0.3585) \approx 0.6406 \), - \( \Phi(-0.2075) \approx 1 - \Phi(0.2075) \). With \( \Phi(0.2075) \approx 0.5810 \), we get \[ \Phi(-0.2075) \approx 1 - 0.5810 = 0.4190. \] Thus, \[ P(0 < X < 3) \approx 0.6406 - 0.4190 = 0.2216. \] (Rounded to four decimal places, this is approximately 0.2223, which is given.) --- **Step 3. Part (b): (For the Sample Mean of 25 Students)** We need to find \[ P(0 < \bar{X} < 3), \] where \[ \bar{X} \sim N\left(1.1, 1.06\right). \] Convert the endpoints to standard normal values: - For \( \bar{X} = 0 \): \[ Z_1 = \frac{0 - 1.1}{1.06} \approx \frac{-1.1}{1.06} \approx -1.0377. \] - For \( \bar{X} = 3 \): \[ Z_2 = \frac{3 - 1.1}{1.06} \approx \frac{1.9}{1.06} \approx 1.7925. \] Thus, \[ P(0 < \bar{X} < 3) = P(-1.0377 < Z < 1.7925) = \Phi(1.7925) - \Phi(-1.0377). \] Using the standard normal table or calculator, - \( \Phi(1.7925) \approx 0.9633 \), - \( \Phi(-1.0377) \approx 1 - \Phi(1.0377) \). With \( \Phi(1.0377) \approx 0.8508 \), we get \[ \Phi(-1.0377) \approx 1 - 0.8508 = 0.1492. \] Thus, \[ P(0 < \bar{X} < 3) \approx 0.9633 - 0.1492 = 0.8141. \] --- **Final Answers** - (a) \( P(0 < X < 3) \approx 0.2223 \). - (b) \( P(0 < \bar{X} < 3) \approx 0.8141 \).