One of the nation's biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. If a sample of size n = 16 turn times was selected at random from the population, the chances of the mean of this sample exceeding 20 minutes is 0.2327.

Question

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Given the population parameters (mean = 17.23, standard deviation = 3.79) and the sample size (n = 16), we can calculate the standard error of the mean (SEM) as the population standard deviation divided by the square root of the sample size. The SEM would be 3.79 / sqrt(16) = 3.79 / 4 = 0.9475. To find the probability of the sample mean exceeding 20 minutes, we would use the Z-score formula: (X - μ) / SEM, where X is the sample mean (20 minutes), and μ is the population mean (17.23 minutes). The Z-score would be (20 - 17.23) / 0.9475 = 2.92. Looking up this Z-score in a standard normal distribution table or using a calculator would give a probability much lower than 0.2327, indicating the statement is false.

One of the Nation's Biggest Regional Airlines Has Tracked 4,000