A director of reservations believes that $8 \%$ of the ticketed passengers are no-shows. If the director is right, what is the probability that the proportion of no-shows in a sample of 596 ticketed passengers would differ from the population proportion more than $3 \%$ ? Round your answer to four decimal places. Answer How to enter your answer (opens in new window) Tables Keyboard Sh

Step 1 ：Calculate the standard deviation of the sampling distribution (σp̂) using the formula: \(σp̂ = \sqrt{\frac{p(1 - p)}{n}}\)

Step 2 ：Substitute the given values into the formula: \(σp̂ = \sqrt{\frac{0.08(1 - 0.08)}{596}}\)

Step 3 ：Solve the equation to get: \(σp̂ = 0.011\)

Step 4 ：We are looking for the probability that the sample proportion differs from the population proportion by more than 0.03, which can be written as: \(P(|p̂ - p| > 0.03)\)

Step 5 ：This can be rewritten as: \(P(p̂ < p - 0.03 \text{ or } p̂ > p + 0.03)\)

Step 6 ：Standardize these values using the Z-score formula: \(Z = \frac{p̂ - p}{σp̂}\)

Step 7 ：For \(p̂ < p - 0.03\), the Z-score is: \(Z1 = \frac{0.08 - 0.03 - 0.08}{0.011} = -2.73\)

Step 8 ：For \(p̂ > p + 0.03\), the Z-score is: \(Z2 = \frac{0.08 + 0.03 - 0.08}{0.011} = 2.73\)

Step 9 ：Using the standard normal distribution table, the probability corresponding to Z1 is 0.0032 and the probability corresponding to Z2 is 0.9968

Step 10 ：The probability that the sample proportion differs from the population proportion by more than 0.03 is therefore \(P(Z < -2.73 \text{ or } Z > 2.73) = 2 * (1 - 0.9968)\)

Step 11 ：\(\boxed{0.0064}\) or \(\boxed{0.64\%}\) when rounded to four decimal places is the final answer