Government workers: The Bureau of Labor Statistics reported that \( 15 \% \) of U.S. nonfarm workers are government employees. A randorn sample of 35 workers is drawn. Use the excel spreadsheet as needed. Part: \( 0 / 5 \) Part 1 of 5 (a) Is it appropriate to use the normal approximation to find the probability that less than \( 25 \% \) of the individuals in the sample are government employees? If so, find the probability. If not, explain why not. It (Choose one) \( \mathbf{v} \) appropriate to use the normal curve, since \( n p= \) is not is not

To determine if it's appropriate to use the normal approximation for a binomial distribution in this case, we need to check the conditions for the Central Limit Theorem, specifically if both \(np\) and \(n(1-p)\) are greater than 5. Here, \(n = 35\) and \(p = 0.15\). Calculating: \[ np = 35 \times 0.15 = 5.25 \] \[ n(1-p) = 35 \times (1 - 0.15) = 35 \times 0.85 = 29.75 \] Since both \(np\) and \(n(1-p)\) are greater than 5, it is **appropriate** to use the normal approximation. Now, we need the standard deviation and mean: - Mean \( \mu = np = 5.25 \) - Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{35 \times 0.15 \times 0.85} \approx 2.635 \) To find the probability that less than 25% of the individuals in the sample are government employees: \( 25\% \) of \( 35 \) is \( 8.75 \) workers. Since we can only have whole workers, we look at \( 8 \) workers. Now we'll apply the normal approximation \( X \sim N(5.25, 2.635^2) \): Using the z-score formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{8 - 5.25}{2.635} \approx 1.042 \] Using a z-table or calculator, we find \( P(Z < 1.042) \approx 0.8508 \). Therefore, the probability that less than \( 25\% \) of the sample are government employees is approximately \( 0.8508 \) or \( 85.08\% \).