Question

1. A researcher tells you that flights between two cities in the United States are uniformly distributed with arrival times that vary between 4 hours and 4 hours30 minutes, with a mean of 4 hours 15 minutes.
   1. Sketch out a related graph, being sure to label the base and the height appropriately.
   2. What is the probability of a randomly chosen flight between these two cities arriving in less than4 hours 10 minutes?
   3. What is the probability of a randomly chosen flight between these two cities arriving in more than4 hours 20 minutes?
   4. You share this test problem with a pilot friend of yours.  S/he is very surprised at the idea that such flights would actually be distributed uniformly.  Please explain why we might expect such an objection.

Answer 1

Solution:

We are given that arrival times follows uniform distribution with a = 4 hours = 4*60 = 240 minutes and b = 4 hours 30 minutes = 4*60 + 30 = 270 minutes.

a = 240, b = 270

Part a

The required graph for the above uniform distribution is given as below:

P(x) 0.33 Arrival times 0.00 (in minutes) 250

Part b

Here, we have to find P(X<4 hours 10 minutes) = P(X < 4*60 + 10 ) = P(X<250)

P(X<m) = (m – a)/(b – a)

P(X<250) = (250 – 240) / (270 – 240) = 10/30 = 1/3 = 0.3333

Required probability = 0.3333

Part c

Here, we have to find P(X>4 hours 20 minutes) = P(X>4*60+20) = P(X>260)

P(X>m) = (b – m)/(b – a)

P(X>260) = (270 – 260)/(270 – 240) = 10/30 = 1/3 = 0.3333

Required probability = 0.3333

Part d

We might expect such an objection because usually arrival times do not follows uniform distribution most of the times. In most of the cases, arrival times follow Poisson distribution.