The Central Limit Theorem Question 6, 6.4.9-T HW Score: \( 41.67 \%, 5 \) of 12 points Part 2 of 3 Points: of 1 Save An elevator has a placard stating that the maximum capacity is \( 4000 \mathrm{lb}-27 \) passengers. So, 27 adult male passengers can have a mean weight of up to \( 4000 / 27=148 \) pounds. Assume that weights of males are normally distributed with a mean of 182 lb and a standard deviation of 34 lb . a. Find the probability that 1 randomly selected adult male has a weight greater than 148 lb , b. Find the probability that a sample of 27 randomly selected adult males has a mean weight greater than 148 lb . c. What do you conclude about the safety of this elevator? a. The probability that 1 randomly selected adult male has a weight greater than 148 lb is \( \square \) 0.8413 (Round to four decimal places as needed.) b. The probability that a sample of 27 randomly selected adult males has a mean weight greater than 148 lb is \( \square \) (Round to four decimal places as needed.)

**Step 1. Identify the given information** - Mean weight of adult males: \(\mu = 182\) lb - Standard deviation for individual weight: \(\sigma = 34\) lb - Maximum allowable average weight per person per placard: \(\bar{x}_0 = 148\) lb - Sample size for part (b): \(n = 27\) --- **Step 2. Part (a): Probability that one randomly selected adult male weighs more than 148 lb** For a single individual, the weight \(X\) is normally distributed as: \[ X \sim N(182, 34) \] Convert \(X = 148\) lb to a \(z\)-score using: \[ z = \frac{X - \mu}{\sigma} \] Substitute the values: \[ z = \frac{148 - 182}{34} = \frac{-34}{34} = -1 \] The probability that an individual weighs more than 148 lb is: \[ P(X > 148) = P(z > -1) \] Since the standard normal distribution is symmetric, \[ P(z > -1) = 1 - P(z \leq -1) \] From standard normal tables: \[ P(z \leq -1) \approx 0.1587 \] Thus, \[ P(X > 148) \approx 1 - 0.1587 = 0.8413 \] --- **Step 3. Part (b): Probability that the sample mean weight of 27 adult males is greater than 148 lb** The sample mean \(\bar{X}\) is also normally distributed with: \[ \mu_{\bar{X}} = \mu = 182 \quad \text{and} \quad \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{34}{\sqrt{27}} \] Compute \(\sigma_{\bar{X}}\): \[ \sqrt{27} \approx 5.196, \quad \sigma_{\bar{X}} \approx \frac{34}{5.196} \approx 6.542 \] Convert \(\bar{X} = 148\) lb to a \(z\)-score for the sample mean: \[ z = \frac{148 - 182}{6.542} = \frac{-34}{6.542} \approx -5.196 \] Find the probability: \[ P(\bar{X} > 148) = P(z > -5.196) \] Since \(-5.196\) is extremely low on the standard normal distribution, almost the entire distribution lies above \(-5.196\). Therefore, \[ P(z > -5.196) \approx 1.0000 \] --- **Step 4. Part (c): Conclusion about the safety of the elevator** - For a single male, the probability of weighing more than 148 lb is approximately \(0.8413\). - For a group of 27 males, the probability that the mean weight exceeds 148 lb is virtually \(1\). This indicates that in almost every case, 27 adult males would have an average weight far exceeding the placard’s limit of 148 lb per person. In fact, the expected total weight for 27 males is: \[ 27 \times 182 = 4914 \text{ lb} \] which is significantly higher than the maximum capacity of 4000 lb. **Conclusion:** The elevator is not safe for 27 adult male passengers as it is very likely to exceed its safe operating weight limit.