Question

once checked in at the ticket kiosk, the amount of time it takes a airline passenger at the calgary international airport to clear security is a random variable that can be modeled by the exponential distribution with a \\( \mu = 38 \\) minutes.
a statistician randomly selects \\( n = 40 \\) airline passengers and records how long, in minutes, it takes each to clear security once each has checked in at the ticket kiosk.
(a) complete the statement below enter your answer using all the decimals you can.
the distribution of \\( \overline { x } \\) with a mean \\( \mu _ { \overline { x } } = \\) minutes and a standard deviation \\( \sigma _ { \overline { x } } = \\) minutes.
(b) what is the probu this sample of \\( n = 40 \\) airline passengers to clear security is between 28 minutes and 52 minutes? enter your answer using all
(c) \\( 95 \\% \\) of the time, th airline passengers to clear security after ticket kiosk check - in is at most how many minutes? enter your answer using all the d minutes

Explanation:

Step1: Recall the properties of the sampling distribution of the sample mean

For a sample of size \(n\) from a population with mean \(\mu\) and standard deviation \(\sigma\), the mean of the sampling distribution of the sample mean \(\bar{X}\) is \(\mu_{\bar{X}}=\mu\), and the standard deviation is \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\).
For an exponential distribution, the mean \(\mu = 38\) minutes and the standard deviation \(\sigma=\mu = 38\) minutes (since for an exponential distribution \(X\sim\text{Exp}(\lambda)\), \(\mu=\frac{1}{\lambda}\) and \(\sigma=\frac{1}{\lambda}\)).

Step2: Calculate the mean of the sampling distribution of \(\bar{X}\)

By the formula \(\mu_{\bar{X}}=\mu\), substituting \(\mu = 38\), we get \(\mu_{\bar{X}}=38\)

Step3: Calculate the standard deviation of the sampling distribution of \(\bar{X}\)

Using the formula \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\), with \(\sigma = 38\) and \(n = 40\), we have \(\sigma_{\bar{X}}=\frac{38}{\sqrt{40}}\approx\frac{38}{6.324555}\approx6.01\)

Answer:

The distribution of \(\bar{X}\) is approximately Normal with a mean \(\mu_{\bar{X}} = 38\) minutes and a standard deviation \(\sigma_{\bar{X}}\approx6.01\) minutes.