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 using z - score(s), what is the probability that a passenger will take longer than 40 minutes to get through the security check? (all problems in this edla were done using the formulas and the \by hand\ method. if you are using a calculator then round your answers to 3 decimal places and you should be able to match the correct answer.)
0.0105
0.0033
0.9895
0.0000

Explanation:

Step1: Calculate the difference between the value and the mean

The mean \(\mu = 20.6\) and the value \(x = 40\). The difference \(x-\mu=40 - 20.6=19.4\)

Step2: Compare with standard deviation

The standard deviation \(\sigma = 8.4\). Now, \(\frac{x-\mu}{\sigma}=\frac{19.4}{8.4}\approx2.31\)
In a normal distribution, about \(99.7\%\) of the data lies within \(3\sigma\) of the mean (\(\mu - 3\sigma\) to \(\mu+3\sigma\)). Also, about \(95\%\) lies within \(2\sigma\) (\(\mu - 2\sigma\) to \(\mu + 2\sigma\)) and about \(68\%\) lies within \(1\sigma\) (\(\mu-\sigma\) to \(\mu+\sigma\)).
Since \(x = 40\) is more than \(2\sigma\) ( \(2\sigma=2\times8.4 = 16.8\)) and \(3\sigma=3\times8.4=25.2\)
We know that the total area under the normal curve is \(1\). The area within \(2.31\sigma\) above the mean can be approximated.
Using the empirical rule - based approximations (for non - z - score method as per problem requirement), we know that for a value more than \(2\sigma\) above the mean, the tail area is small.
The area within \(2\sigma\) of the mean is about \(0.95\), so the area outside \(2\sigma\) is \(1 - 0.95=0.05\) (split equally in two tails). For a value more than \(2.3\sigma\) above the mean, the area is much less than \(0.025\) (the area in one tail for \(2\sigma\)).
If we use a calculator for the normal distribution \(P(X>x)=P(X > 40)\) with \(X\sim N(\mu = 20.6,\sigma = 8.4)\)
\(P(X>40)=1 - P(X\leq40)\)
Using the formula for the normal distribution \(P(X\leq x)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{(t - \mu)^2}{2\sigma^2}}dt\)
\(P(X>40)\approx0.0105\)

Answer:

0.0105