Example 5.11 from Statistics, 5.3 The Exponential Distribution (Optional)

The Exponential Distribution (Optional): example 5.11

On average, how many minutes elapse between two successive arrivals?
When the store first opens, how long on average does it take for three customers to arrive?
After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
Seventy percent of the customers arrive within how many minutes of the previous customer?
Is an exponential distribution reasonable for this situation?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
2. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
3. Let X = the time between arrivals, in minutes. By Part a, μ = 2, so m = $\frac{1}{2}$ = 0.5.
  Therefore, X ~ Exp(0.5).
  The cumulative distribution function is P(X < x) = 1 – e^(–0.5)(x).
  Therefore P(X < 1) = 1 – e^(–0.5)(1) ≈ 0.3935.
  Using the TI-83, 83+, 84, 84+ Calculator
  
  1 - e^(–0.5) ≈ 0.3935
  
  Figure 5.28
4. P(X > 5) = 1 – P(X < 5) = 1 – (1 – e^(−0.5)(5)) = e^–2.5 ≈ 0.0821.
  
  Figure 5.29
  
  Using the TI-83, 83+, 84, 84+ Calculator
  
  $1 – (1 – e^(– 0.50) (5)) o r e^(– 0.50) (5)$
5. We want to solve 0.70 = P(X < x) for x.
  Substituting in the cumulative distribution function gives 0.70 = 1 – e^–0.5x, so that e^−0.5x = 0.30. Converting this to logarithmic form gives –0.5x = ln(0.30), or $x = \frac{l n (0.30)}{- 0.5} \approx 2.41$ minutes.
  Thus, 70 percent of customers arrive within 2.41 minutes of the previous customer.
  You are finding the 70^th percentile k so you can use the formula
  
  $k = \frac{l n (1 - A r e a_T o_T h e_L e f t_O f_k)}{(- m)}$
  
  $k = \frac{l n (1 - 0.70)}{(- 0.5)} \approx 2.41 minutes$
  5.1
  
  Figure 5.30
6. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.

How did it go?

Solution

Using the TI-83, 83+, 84, 84+ Calculator

Using the TI-83, 83+, 84, 84+ Calculator

Solution

Using the TI-83, 83+, 84, 84+ Calculator

Using the TI-83, 83+, 84, 84+ Calculator