An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. The airline collected survey data from 1,000 passengers on their time from arrival at the airport to reaching the boarding gate. The sample mean was 1 hour and 20 minutes, with a sample standard deviation of 30 minutes. Based on this sample, how long prior to a flight should a passenger arrive at the airport to have a 95% probability of making it to the gate on time? A) Two hours, ten minutes. B) Two hours, thirty minutes. C) One hour, fifty minutes.
is it C
is it A ?? I used the z dist = (x- sample mean)/s.d the question says 95% prob … if its option A …then (130-80)/30 …(converted into mins) = 1.66 whats the right answer ?
Answer A/B… Explaination: ->> Sample mean = 1Hr 20 min = 80 min 80 +/- 1.96(30) = 138 min…
A. 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. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.)