Question

departure delays for flights leaving a major airport follow a normal distribution, with a mean of 18 minutes and a standard deviation of 7 minutes. based on this information, a newspaper article claims that about one in every ten flights leaving this airport is delayed by more than 30 minutes. is the newspaper article correct? the newspaper article is correct because the actual percentage of flights delayed by more than 30 minutes is about 10%. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 25%. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 15% or 16%. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 4% or 5%. unless the newspaper was given access to a full data set containing delay information for hundreds of flights, they would have no way of making such a claim.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 30$ (the value we are interested in), $\mu=18$ (mean) and $\sigma = 7$ (standard deviation). So $z=\frac{30 - 18}{7}=\frac{12}{7}\approx1.71$.

Step2: Find the proportion of values above the z - score

We use the standard normal distribution table. The table gives the proportion of values to the left of a z - score. The proportion of values to the left of $z = 1.71$ is approximately $0.9564$. So the proportion of values to the right of $z=1.71$ (flights delayed more than 30 minutes) is $1 - 0.9564=0.0436\approx4\%-5\%$.

Answer:

The newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 4% or 5%.