A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is \( 0: 05 \). If 220 are sampled, what is the probability that the sample proportion will differ from the population proportion by greater than 0.04 ? Round your answer to four decimal places.

To find the probability that the sample proportion will differ from the population proportion by greater than 0.04, we can use the normal approximation to the binomial distribution. Given: - True proportion \( p = 0.05 \) - Sample size \( n = 220 \) First, we calculate the mean and standard deviation of the sample proportion \( \hat{p} \): 1. **Mean of the sample proportion**: \[ \mu_{\hat{p}} = p = 0.05 \] 2. **Standard deviation of the sample proportion**: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.05(1-0.05)}{220}} = \sqrt{\frac{0.05 \times 0.95}{220}} = \sqrt{\frac{0.0475}{220}} \approx \sqrt{0.000215909} \approx 0.0147 \] Next, we want to find the probability that the sample proportion \( \hat{p} \) differs from the population proportion \( p \) by more than 0.04. This means we are looking for: \[ P(|\hat{p} - p| > 0.04) \] This can be rewritten as: \[ P(\hat{p} < 0.01 \text{ or } \hat{p} > 0.09) \] Now we convert these sample proportions to z-scores: 1. For \( \hat{p} < 0.01 \): \[ z_1 = \frac{0.01 - 0.05}{0.0147} \approx \frac{-0.04}{0.0147} \approx -2.7211 \] 2. For \( \hat{p} > 0.09 \): \[ z_2 = \frac{0.09 - 0.05}{0.0147} \approx \frac{0.04}{0.0147} \approx 2.7211 \] Now we can find the probabilities corresponding to these z-scores using the standard normal distribution: - Using a standard normal distribution table or calculator: - \( P(Z < -2.7211) \) gives us the left tail probability. - \( P(Z > 2.7211) = 1 - P(Z < 2.7211) \) gives us the right tail probability. Using a standard normal distribution calculator: - \( P(Z < -2.7211) \approx 0.0032 \) - \( P(Z < 2.7211) \approx 0.9968 \) Thus, \[ P(Z > 2.7211) = 1 - P(Z < 2.7211) \approx 1 - 0.9968 = 0.0032 \] Now, we combine these probabilities: \[ P(|\hat{p} - p| > 0.04) = P(Z < -2.7211) + P(Z > 2.7211) \approx 0.0032 + 0.0032 = 0.0064 \] Finally, rounding to four decimal places, the probability that the sample proportion will differ from the population proportion by greater than 0.04 is: \[ \boxed{0.0064} \]