Let X be the waiting time in minutes of a customer in a post

Let X be the waiting time (in minutes) of a customer in a post office. Suppose that X follows an exponential distribution with a probability density function

a) Find the probability that the waiting time is between 3 and 5 minutes.

b) Find the 50th percentile (median) of the waiting time.

Please include any calculator instructions.

Solution

A)

The mean of the distirbution is also the standard deviation, and is equal to 1/lambda:          
          
mean = standard deviation = 1/lambda =    0.5      
          
The area between two values in an exponential distribution ois          
          
Area = e^(-lambda*x2) - e^(-lambda*x1)          
          
As          
          
x1 = lower bound =   3      
x2 = upper bound =    5      
          
Then          
          
Area =    0.141045162   [ANSWER]  

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B)

As the left tailed area is given by

P(X<x) = 1 - exp(-x/2)

Then for the 50th percentile,

P(X<x) = 1 - exp(-x/2) = 0.5

exp(-x/2) = 0.5

-x/2 = ln(0.5)

x = -2*ln(0.5) = 1.386294361 [ANSWER]