The Central Limit Theorem Question 5, 6.4.7-T Part 4 of 3 HW score: \( 33.38 \%, 4 \) of 12 points Pointss of 1 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. The probability is \( \square \) (Round to four decimal places as needed.)

Let \(X\) be the weight gain of a male college student during his freshman year. We know that \[ X \sim N(\mu=1.1, \sigma=5.3). \] We wish to find \[ P(0 \leq X \leq 3). \] **Step 1. Standardize the variable.** For a general value \(x\), the standard \(z\)-score is given by \[ z = \frac{x-\mu}{\sigma}. \] For \(x=0\): \[ z_1 = \frac{0-1.1}{5.3} = \frac{-1.1}{5.3} \approx -0.2075. \] For \(x=3\): \[ z_2 = \frac{3-1.1}{5.3} = \frac{1.9}{5.3} \approx 0.3585. \] **Step 2. Express the probability in terms of \(z\)-scores.** We have \[ P(0 \leq X \leq 3)=P\left(-0.2075 \leq z \leq 0.3585\right)=\Phi(0.3585)-\Phi(-0.2075), \] where \(\Phi(z)\) is the cumulative distribution function (CDF) of the standard normal distribution. **Step 3. Use the symmetry property of the normal distribution.** Recall that \[ \Phi(-z)=1-\Phi(z). \] Hence, \[ \Phi(-0.2075)=1-\Phi(0.2075). \] **Step 4. Look up (or compute) the CDF values.** Using a standard normal table or calculator: \[ \Phi(0.3585) \approx 0.6406, \] \[ \Phi(0.2075) \approx 0.5817. \] Thus, \[ \Phi(-0.2075) \approx 1-0.5817 = 0.4183. \] **Step 5. Compute the probability.** Subtract: \[ P(0 \leq X \leq 3) \approx 0.6406 - 0.4183 = 0.2223. \] The probability is \( \boxed{0.2223} \).