Q: quants

Hello, What would be an interpretation of the below? thanks An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. Based on a survey of 1,000 passengers, the mean time from arrival at the airport to reaching the boarding gate was 1 hour, 20 minutes, with a standard deviation of 30 minutes. If the airline wants to make sure at the 95% confidence level that passengers have sufficient time to catch their flight, how much time ahead of their flight should passengers be advised to arrive at the airport? A) Two hours, ten minutes. B) One hour, fifty minutes. C) Two hours, thirty minutes. D) Two hours, forty-five minutes. The correct answer was A. We can use standard distribution tables because the sample is so large. From a table of area under a normally distributed curve, the z value corresponding to a 95%, one-tail test is: 1.65. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.) Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight. The answer is one hour, twenty minutes + 1.65(30 minutes) = 2 hours, 10 minutes.