Poisson Distributions

Adjusting the Time Frame (Changing Lambda)

Learning Objectives

Understand and calculate lambda ([latex]\lambda[/latex]) for different length time periods.

Changing Lambda ([latex]\lambda[/latex])

  • If the duration of time given in the question doesn’t match the duration given in the average,
  • We need to adjust lambda ([latex]\lambda[/latex]) to match the duration given in the question
  • To do this, we can use: [latex]\lambda_{NEW}=\lambda_{OLD}\times \frac{t_{NEW}}{t_{OLD}}[/latex].

Changing Lambda ([latex]\lambda[/latex]) Example (VIDEO)

  • It is important to stress that we must always match the time units.
  • In the example from the previous section, the average arrival rate was given per minute.
  • The question’s time interval was also one minute: “What is the probability of 5 customers arriving in any given minute?”
  • So, no adjustment of lambda ([latex]\lambda[/latex]) was needed since the time intervals matched.
  • See the next example to see where an adjustment of [latex]\lambda[/latex] is needed.

Example 31.1

Problem Setup: The average number of shoppers arriving at a supermarket check-out counter waiting to be served by a cashier is 6 per minute. The arrivals follow a Poisson distribution.

Question: In any 10-second interval, what is the probability that 2 shoppers will arrive?

Changing Lambda ([latex]\lambda[/latex]):

  • There are 60 seconds in one minute. So, we will rewrite our ‘old’ time ([latex]t_{OLD}[/latex]) as 60 seconds.
  • Our new time interval is 10 seconds ([latex]t_{NEW}[/latex]).
  • This gives a new average of: [latex]\lambda_{NEW}=\lambda_{OLD}\times \frac{t_{NEW}}{t_{OLD}}=6\times \frac{10}{60}=1[/latex] arrivals per second.

Solving for the Probability:

  • Since we want the probability that two shoppers arrive, we use [latex]x=2[/latex].
  • Input [latex]x=2[/latex] and [latex]\lambda=1[/latex] into the formula [latex]P(x) =\frac{{\lambda}^x e^{-\lambda}}{x!}[/latex]
  • This gives: [latex]P(x=2) =\frac{1^2 e^{-1}}{2!}  = \frac{1\times 0.367879441}{2} = 0.1839[/latex]

Video Solution: Click here to download the written solutions. Also, see the video below:

Conclusion: There is an 18.39% chance that 2 shoppers will arrive in any 10-second interval.

Changing Lambda ([latex]\lambda[/latex]) Second Example (EXERCISE)

Now, let’s practice what you just learnt in the exercise in the next example

Example 31.2

Problem Setup: There is an average of 146 crashes per year at Willingdon Avenue and the  Highway#1 off-ramp (near BCIT). Let’s assume, for now, that the number of accidents is similar on all days of the week and they follow a Poisson distribution.

Question: What is the probability that there are no accidents at this location on any given day that you are at BCIT?

You Try: First, solve for the new average by placing the values in the correct positions below:

Click here to reveal the solutions:

  • There are 365 days in the year. So, we will write our ‘old’ time ([latex]t_{OLD}[/latex]) of one year minute as 365 days.
  • Our new time interval is 1 day (([latex]t_{NEW}[/latex])).
  • This gives a new average of: [latex]\lambda_{NEW}=\lambda_{OLD}\times \frac{t_{NEW}}{t_{OLD}}=146\times \frac{1}{365}=0.4[/latex] accidents per day.

You Try: Next, solve for probability by placing values in the correct positions below:

Click here to reveal the solutions:

  • Since we want the probability that no accidents occur, we use [latex]x=0[/latex].
  • Input [latex]x=0[/latex] and [latex]\lambda=0.4[/latex] into the formula [latex]P(x) =\frac{{\lambda}^x e^{-\lambda}}{x!}[/latex]
  • This gives: [latex]P(x=0) =\frac{0.4^0 e^{-0.4}}{0!}  = \frac{1\times 0.67032}{1} = 0.67032[/latex]

Changing Lambda in Excel Example (Video)

  • Let us practice changing lambda in Excel
  • There is no special Excel call – so it’s just like doing it by hand
  • None-the-less, let’s finish this section with a quick video on how to do this Excel

Example 31.3

Problem Setup: There are an average of 240 accidents per year at 264th Street and Highway #1. Let’s assume, for now, that the number of accidents is similar week-over-week and that they follow a Poisson distribution.

Question: What is the probability that there are 10 accidents in any given week at 264th and Highway 1?

Solution: Use = POISSON.DIST(10, 240/52, 0). Click here to download the Excel solutions below:

Key Takeaways (EXERCISE)

Key Takeaways: Adjusting the Time Frame (Changing Lambda)

Your Own Notes (EXERCISE)

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An Introduction to Business Statistics for Analytics (1st Edition) Copyright © 2024 by Amy Goldlist; Charles Chan; Leslie Major; Michael Johnson is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.

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