Question

2. The uniform distribution

The Transportation Security Administration (TSA) collects data on wait time at each of its airport security checkpoints. For flights departing from Terminal 9 at John F. Kennedy International Airport (JFK) between 3:00 and 4:00 PM on Wednesday, the mean wait time is 10 minutes, and the maximum wait time is 18 minutes. [Source: Transportation Security Administration, summary statistics based on historical data collected between January 8, 2008, and February 5, 2008.]

Assume that x, the wait time at the Terminal 9 checkpoint at JFK for flights departing between 3:00 and 4:00 PM on Wednesday, is uniformly distributed between 2 and 18 minutes.

The height of the graph of the probability density function f(x) varies with x as follows (round to four decimal places):

x	Height of the Graph of the Probability Density Function
x < 2
2 ≤ x ≤ 18
x > 18

You are flying out of Terminal 9 at JFK on a Wednesday afternoon between 3:00 and 4:00 PM. You get stuck in a traffic jam on the way to the airport, and if it takes you longer than 12 minutes to clear security, you’ll miss your flight. The probability that you'll miss your flight is____.

• 0.6667
• 0.375
• 0.3333
• 0.625

The mean wait time is_____ minutes,   

• 10
• 9
• 16
• 3

and the standard deviation is____.

• 3.46
• 3.16
• 4.62
• 4.24

Answer 1

x	Height of the Graph of the Probability Density Function
x < 2	0
2 ≤ x ≤ 18	1/16
x > 18

X> 18 , 0

====================

P(miss flight)=P(X > 12 ) =   (b-x)/(b-a) = (18-12)/(18-2) = 0.375

==========================

mean =    (a+b)/2 = (2+18)/2 = 10
======================

variance =    (b-a)²/12 = (18-2)²/12 =  21.33333333
      
std dev =   √ variance =    4.62