e Central Limit Theorem Question 5, 6.4.7-T HW Score: \( 33.33 \%, 4 \) of 12 points Part 1 of 3 Points: of 1 Save 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 \( \sigma=5.1 \mathrm{~kg} \). 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. The probability is \( \square \) (Round to four decimal places as needed.)

The weight gained by a male college student is modeled by a normal random variable \( X \sim N(1.1, 5.1^2) \). We are asked to compute: \[ P(0 \leq X \leq 3) \] **Step 1: Standardize the variable.** The standardization formula is: \[ Z = \frac{X - \mu}{\sigma} \] For the lower limit \( X = 0 \): \[ z_1 = \frac{0 - 1.1}{5.1} \approx \frac{-1.1}{5.1} \approx -0.2157 \] For the upper limit \( X = 3 \): \[ z_2 = \frac{3 - 1.1}{5.1} \approx \frac{1.9}{5.1} \approx 0.3725 \] **Step 2: Express the probability in terms of \( Z \).** \[ P(0 \leq X \leq 3) = P(z_1 \leq Z \leq z_2) = P(-0.2157 \leq Z \leq 0.3725) \] **Step 3: Use the standard normal distribution table or calculator.** The probability is computed as: \[ P(-0.2157 \leq Z \leq 0.3725) = \Phi(0.3725) - \Phi(-0.2157), \] where \(\Phi(z)\) is the standard normal cumulative distribution function. Using approximate values: - \(\Phi(0.3725) \approx 0.6457\) - \(\Phi(-0.2157) \approx 1 - \Phi(0.2157) \approx 1 - 0.5852 = 0.4148\) Thus: \[ P(0 \leq X \leq 3) \approx 0.6457 - 0.4148 = 0.2309 \] **Final Answer:** \[ P(0 \leq X \leq 3) \approx 0.2309 \]