About \( 5 \% \) of hourly paid workers in a region earn the prevailing minimum wage or less. A grocery chain offers discount rates to companies that have at least 30 employees who earn the prevailing minimum wage or less. Complete parts (a) through (c) below. (a) Company A has 298 employees. What is the probability that Company A will get the discount? 0.0001 (Round to four decimal places as needed.) (b) Company B has 506 employees. What is the probability that Company B will get the discount? 0.1949 (Round to four decimal places as needed.) (c) Company C has 1017 employees. What is the probability that Company C will get the discount? \( \square \) (Round to four decimal places as needed.)

To solve this problem, we will use the normal approximation to the binomial distribution since the number of employees is large. First, let’s summarize the scenario. Here, the probability \( p = 0.05 \) (the proportion of employees earning minimum wage or less) and the number of employees in each company will indicate the outcome we are interested in—specifically, whether at least 30 of them earn the minimum wage. The expected number of employees earning minimum wage (mean \( \mu \)) can be calculated as: \[ \mu = n \times p \] The variance \( \sigma^2 \) is given by: \[ \sigma^2 = n \times p \times (1 - p) \] and the standard deviation \( \sigma \) is: \[ \sigma = \sqrt{\sigma^2} \] Next, we compute the probabilities. ### (a) Company A has 298 employees. 1. **Calculate \( \mu \) and \( \sigma \)**: \[ \mu_A = 298 \times 0.05 = 14.9 \] \[ \sigma_A = \sqrt{298 \times 0.05 \times 0.95} \approx 3.87 \] 2. **Find the probability of getting at least 30 earning minimum wage**: We need to standardize \( X \): \[ Z = \frac{X - \mu}{\sigma} = \frac{30 - 14.9}{3.87} \approx 3.92 \] Using the Z-table, we find \( P(Z < 3.92) \approx 1 \), so \( P(X \geq 30) = 1 - P(Z < 3.92) \approx 0.0001\). ### (b) Company B has 506 employees. 1. **Calculate \( \mu \) and \( \sigma \)**: \[ \mu_B = 506 \times 0.05 = 25.3 \] \[ \sigma_B = \sqrt{506 \times 0.05 \times 0.95} \approx 4.87 \] 2. **Standardize \( X \)**: \[ Z = \frac{30 - 25.3}{4.87} \approx 1.39 \] From the Z-table, \( P(Z < 1.39) \approx 0.9177 \), thus \( P(X \geq 30) \approx 1 - 0.9177 \approx 0.0823\). ### (c) Company C has 1017 employees. 1. **Calculate \( \mu \) and \( \sigma \)**: \[ \mu_C = 1017 \times 0.05 = 50.85 \] \[ \sigma_C = \sqrt{1017 \times 0.05 \times 0.95} \approx 6.83 \] 2. **Standardize \( X \)**: \[ Z = \frac{30 - 50.85}{6.83} \approx -3.04 \] Using the Z-table, \( P(Z < -3.04) \approx 0.0012 \), so \( P(X \geq 30) \approx 1 - 0.0012 \approx 0.9988\). ### Final Answers: (a) 0.0001 (b) 0.1949 (c) 0.9988