Question

1. 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? a. the newspaper article is correct because the actual percentage of flights delayed by more than 30 minutes is about 10%. b. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 25%. c. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 15% or 16%. d. the newspaper article is incorrect because the actual percentage of flights delayed by more than 30 minutes is about 4% or 5%. e. 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 formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 30$ (the value we are interested in), $\mu=18$ (the mean), and $\sigma = 7$ (the standard deviation). So, $z=\frac{30 - 18}{7}=\frac{12}{7}\approx1.71$.

Step2: Find the probability using the standard normal table

We want to find $P(X>30)$, which is equivalent to $P(Z > 1.71)$ in the standard - normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq1.71)$ in the standard - normal table, we find that $P(Z\leq1.71)\approx0.9564$. So, $P(Z > 1.71)=1 - 0.9564 = 0.0436\approx4\%$ or $5\%$.

Answer:

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