1. BCOR 1020 Business Statistics Lecture 11 – February 21, 2008

2. Overview Chapter 6 – Discrete Distributions –Poisson Distribution –Linear Transformations

3. Chapter 6 – Poisson Distribution Poisson Processes: If the number of “occurrences” of interest on a given continuous interval (of time, length, etc.) are being counted, we say we have an approximate Poisson Process with parameter > 0 (occurrences per unit length/time) if the following conditions are satisfied… 1)The number of “occurrences” in non-overlapping intervals are independent. 2)The probability of exactly one “occurrences” in a sufficiently short interval of length h is h. (i.e. If the interval is scaled by h, we also scale the parameter by h.) 3)The probability of two or more “occurrences” in a sufficiently short interval is essentially zero. (i.e. there are no simultaneous “occurrences”.)

4. Chapter 6 – Poisson Distribution Poisson Distribution: The Poisson distribution describes the number of occurrences within a randomly chosen unit of time or space. For example, within a minute, hour, day, square foot, or linear mile. If X denotes the number of “occurrences” of interest observed on a given interval of length 1 unit of a Poisson Process with parameter > 0, then we say that X has the Poisson distribution with parameter.

5. Chapter 6 – Poisson Distribution Poisson Distribution: Called the model of arrivals, most Poisson applications model arrivals per unit of time. The events occur randomly and independently over a continuum of time or space: One Unit One Unit One Unit of Time of Time of Time |  ---  | |  ---  | |  ---  | Flow of Time  Each dot () is an occurrence of the event of interest.

6. Chapter 6 – Poisson Distribution Let X = the number of events per unit of time. X is a random variable that depends on when the unit of time is observed. For example, we could get X = 3 or X = 1 or X = 5 events, depending on where the randomly chosen unit of time happens to fall. One Unit One Unit One Unit of Time of Time of Time |  ---  | |  ---  | |  ---  | Flow of Time 

7. Chapter 6 – Poisson Distribution Arrivals (e.g., customers, defects, accidents) must be independent of each other. Some examples of Poisson models in which assumptions are sufficiently met are: X = number of customers arriving at a bank ATM in a given minute. X = number of file server virus infections at a data center during a 24-hour period. X = number of blemishes per sheet of white bond paper.

8. Chapter 6 – Poisson Distribution Poisson Processes: represents the mean number of events (occurrences) per unit of time or space. The unit of time should be short enough that the mean arrival rate is not large (  < 20). To make smaller, convert to a smaller time unit (e.g., convert hours to minutes). The Poisson model’s only parameter is (Greek letter “lambda”).

9. Chapter 6 – Poisson Distribution Poisson Processes: The number of events that can occur in a given unit of time is not bounded, therefore X has no obvious limit. However, Poisson probabilities taper off toward zero as X increases. The Poisson distribution is sometimes called the model of rare events.

10. Chapter 6 – Poisson Distribution Poisson Distribution: We can formulate the PMF, mean and variance (or standard deviation) of the Poisson distribution in terms of the parameter : PMF of the Poisson distribution with parameter : Mean of the Poisson distribution with parameter : Variance and Standard Deviation of the Poisson distribution with parameter :

11. Chapter 6 – Poisson Distribution Parameters  = mean arrivals per unit of time or space PDF RangeX = 0, 1, 2,... (no obvious upper limit) Mean  St. Dev. Random dataUse Excel’s Tools | Data Analysis | Random Number Generation Comments Always right-skewed, but less so for larger.

12. Chapter 6 – Poisson Distribution Poisson Processes: Poisson distributions are always right-skewed but become less skewed and more bell-shaped as increases. = 0.8 = 1.6 = 6.4

13. Chapter 6 – Poisson Distribution Example: Credit Union Customers On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the Oxnard University Credit Union at a mean rate of 102 customers per hour (or 1.7 customers per minute). Why would we consider this a Poisson distribution? Which units should we use? Why? Find the PDF, mean and standard deviation: PDF = Mean = = 1.7 customers per minute. Standard deviation =  = 1.7 = 1.304 cust/min

14. Chapter 6 – Poisson Distribution Example: Credit Union Customers Here is the Poisson probability distribution for = 1.7 customers per minute on average. x PDF P(X = x) CDF P(X  x) 0.1827 1.3106.4932 2.2640.7572 3.1496.9068 4.0636.9704 5.0216.9920 6.0061.9981 7.0015.9996 8.0003.9999 9.00011.0000 Note that x represents the number of customers. For example, P(X=4) is the probability that there are exactly 4 customers in the bank.

15. Chapter 6 – Poisson Distribution Using the Poisson Formula: These probabilities can be calculated using a calculator or Excel: Formula:Excel function: =POISSON(0,1.7,0) =POISSON(1,1.7,0) =POISSON(2,1.7,0) =POISSON(3,1.7,0) =POISSON(4,1.7,0)

16. Chapter 6 – Poisson Distribution Here are the graphs of the distributions: Poisson PDF for = 1.7Poisson CDF for = 1.7 The most likely event is 1 arrival (P(1)=.3106 or 31.1% chance). This will help the credit union schedule tellers.

17. Clickers Orders arrive at a pizza delivery franchise at an average rate of 12 calls per hour. If we want to model the number of calls arriving during a randomly-selected 15 minute interval, which distribution should we use? A = Poisson distribution with = 0.2 calls per minute B = Poisson distribution with = 0.8 calls per 15 minutes C = Poisson distribution with = 3 calls per 15 min. D = Poisson distribution with = 12 calls per hour

18. Clickers Orders arrive at a pizza delivery franchise at an average rate of 12 calls per hour. What are the mean and standard deviation of the number of calls arriving during a randomly-selected 15 minute interval? A =  = 3 and  = 1.73 B =  = 3 and  = 3 C =  = 12 and  = 3.46 D =  = 12 and  = 12

19. Clickers Orders arrive at a pizza delivery franchise at an average rate of 12 calls per hour. What is the probability of exactly two calls arriving during a randomly-selected 15 minute interval? A = 0.0004 B = 0.1494 C = 0.2240 D = 0.4481

20. Chapter 6 – Poisson Distribution Compound Events: Recall our earlier credit union example: On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the Oxnard University Credit Union at a mean rate of 102 customers per hour (or 1.7 customers per minute). Cumulative probabilities can be evaluated by summing individual X probabilities. What is the probability that two or fewer customers will arrive in a given minute? =.1827 +.3106 +.2640 =.7573 P(X < 2) = P(0) + P(1) + P(2) PDF =

21. Chapter 6 – Poisson Distribution Compound Events: What is the probability of at least three customers (the complimentary event)? = 1 -.7573 =.2427P(X > 3) = 1 - P(X < 2) P(X > 3) = P(3) + P(4) + P(5) + … Since X has no limit, this sum never ends. So, we will use the compliment.

22. Clickers Orders arrive at a pizza delivery franchise at an average rate of 12 calls per hour. What is the probability that more than two calls arrive during a randomly-selected 15 minute interval? A = 0.0498 B = 0.1494 C = 0.2240 D = 0.4232 E = 0.5768

23. Chapter 6 – Poisson Distribution Recognizing Poisson Applications: Can you recognize a Poisson situation? Look for arrivals of “rare” independent events with no obvious upper limit. In the last week, how many credit card applications did you receive by mail? In the last week, how many checks did you write? In the last week, how many e-mail viruses did your firewall detect?

24. Chapter 6 – Linear Transformations Linear Transformations: A linear transformation of a random variable X is performed by adding a constant or multiplying by a constant. Rule 1:  aX+b = a  X + b (mean of a transformed variable) Rule 2:  aX+b = |a|  X (standard deviation of a transformed variable) For example, consider defining a random variable Y in terms of the random variable X as follows: Where a and b are any two constants.

25. Chapter 6 – Linear Transformations Example: Total Cost The total cost of many goods is often modeled as a function of the good produced, Q (a random variable). Specifically, if there is a variable cost per unit v and a fixed cost F, then the total cost of the good, C, is given by … where v and F are constant values. For given values of  Q,  Q, v, and F, we can determine the mean and standard deviation of the total cost…

26. Clickers If Q is a random variable with mean  Q = 500 units and standard deviation  Q = 40 units, the variable cost is v = $35 per unit, and the fixed cost is F = $24,000, the mean of the total cost is Determine the standard deviation of the total cost. A)  C = $35 B)  C = $40 C)  C = $1,400 D)  C = $25,400