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. without standardizing the value(s), determine the 75th percentile for the time it takes to get through security check. 26.9 minutes 15.5 minutes 26.3 minutes 14.9 minutes

Explanation:

Step1: Recall the property of normal distribution

For a normal distribution \(X\sim N(\mu,\sigma^{2})\), the \(p\) - th percentile \(x_p\) satisfies \(P(X\leq x_p)=p\). The 75th percentile means \(p = 0.75\).

Step2: Use the fact that for a normal distribution \(X\sim N(\mu,\sigma^{2})\), the 75th percentile \(x_{0.75}\) is greater than the mean \(\mu\)

Since the mean \(\mu=20.6\) and the standard deviation \(\sigma = 8.4\). We know that values less than the mean correspond to percentiles less than 50%. So we can eliminate the values \(15.5\) and \(14.9\) (because \(15.5<20.6\) and \(14.9<20.6\))

Step3: Further analysis between the remaining two values

We know that the 75th percentile \(x_{0.75}\) is not too far from the mean in a normal distribution. The formula for the \(z\) - score is \(z=\frac{x-\mu}{\sigma}\). For a normal distribution, the \(z\) - score corresponding to the 75th percentile (using standard normal table or properties of normal distribution) is approximately \(z = 0.674\). Then \(x=\mu+z\sigma\approx20.6+0.674\times8.4\approx20.6 + 5.66=26.26\approx26.3\)

Answer:

26.3 minutes