Question

the length of time it takes passengers to get through the security checks at a large urban airport follows a normal distribution with a mean of 20.6 minutes and a standard deviation of 8.4 minutes. using standardized values, determine the 75th percentile for the time it takes to get through security check. 26.9 minutes 15.5 minutes 26.2 minutes -5.4 minutes

Explanation:

Step1: Find the z - score for the 75th percentile

Using the standard normal distribution table (or a calculator with a normal - distribution function), the z - score \(z\) corresponding to a cumulative probability of \(P(Z\leq z)=0.75\). Looking up in the standard normal table, \(z\approx0.674\).

Step2: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\) to solve for \(x\)

We know that \(\mu = 20.6\) (mean), \(\sigma=8.4\) (standard deviation), and \(z = 0.674\).
Substitute the values into the formula \(0.674=\frac{x - 20.6}{8.4}\).
Multiply both sides by \(8.4\): \(0.674\times8.4=x - 20.6\).
\(0.674\times8.4=5.6616\).
Then \(x=5.6616 + 20.6\).
\(x=26.2616\approx26.2\)

Answer:

26.2 minutes