# Solved: The time required to pass through security screening at the airport can be annoying to travelers

#### ByDr. Raju Chaudhari

Feb 24, 2021

The time required to pass through security screening at the airport can be annoying to travelers. The mean wait time during peak periods at Cincinnati/Northern Kentucky International Airport is 12.1 minutes (The Cincinnati Enquirer, February 2, 2006). Assume the time to pass through security screening follows an exponential distribution.

a. What is the probability it will take less than 10 minutes to pass through security screening during a peak period?
b. What is the probability it will take more than 20 minutes to pass through security screening during a peak period?
c. What is the probability it will take between 10 and 20 minutes to pass through security screening during a peak period?
d. It is 8:00 A.M. (a peak period) and you just entered the security line. To catch your plane you must be at the gate within 30 minutes. If it takes 12 minutes from the time you clear security until you reach your gate, what is the probability you will miss your flight?

### Solution

Let $X$ denote the time required to pass through security screening. $X\sim \exp(12.1)$.

The pdf of $X$ is

 \begin{aligned} f(x) & = \frac{1}{12.1}e^{-x/12.1}, x>0 \end{aligned}

The distribution function of $X$ is

 \begin{aligned} F(x) & = 1-e^{-x/12.1} \end{aligned}

(a) The probability it will take less than 10 minutes to pass through security screening during a peak period is

 \begin{aligned} P(X < 10) &= F(10)\\ &= 1 - e^{-10/12.1}\\ & = e^{-10/12.1} \\ &= 0.4376 \end{aligned}

(b) The probability it will take more than 20 minutes to pass through security screening during a peak period is

 \begin{aligned} P(X > 20) &= 1- P(X\leq 20)\\ & = 1- F(20)\\ &= 1 - (1 - e^{-20/12.1})\\ & = e^{-20/12.1} \\ &= 0.1915 \end{aligned}

(c) The probability it will take between 10 and 20 minutes to pass through security screening during a peak period is

 \begin{aligned} P(10 < X < 20) &= F(20)-F(10)\\ & = (1 - e^{-20/12.1}) - (1 - e^{-10/12.1})\\ & = e^{-10/12.1}- e^{-20/12.1} \\ &= 0.4376 - 0.1915\\ &= 0.2461 \end{aligned}

(d) To catch your plane you must be at the gate within 30 minutes. If it takes 12 minutes from the time you clear security, then the probability you will miss your flight is

 \begin{aligned} P(X > 18) &= 1-F(18)\\ &= 1- (1- e^{-18/12.1}\\ &= e^{-18/12.1}\\ &= 0.2259 \end{aligned}