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BCOR 1020 Business Statistics

BCOR 1020 Business Statistics. Lecture 11 – February 21, 2008. Overview. Chapter 6 – Discrete Distributions Poisson Distribution Linear Transformations. Chapter 6 – Poisson Distribution. Poisson Processes:

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BCOR 1020 Business Statistics

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  1. BCOR 1020Business 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 l > 0 (occurrences per unit length/time) if the following conditions are satisfied… • The number of “occurrences” in non-overlapping intervals are independent. • The probability of exactly one “occurrences” in a sufficiently short interval of length h is lh. (i.e. If the interval is scaled by h, we also scale the parameter l by h.) • The probability of two or more “occurrences” in a sufficiently short interval is essentially zero. (i.e. there are no simultaneous “occurrences”.)

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

  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. 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. 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:

  8. Chapter 6 – Poisson Distribution Poisson Processes: • The Poisson model’s only parameter is l (Greek letter “lambda”). l 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 (l < 20). • To make l smaller, convert to a smaller time unit (e.g., convert hours to minutes).

  9. Chapter 6 – Poisson Distribution Poisson Processes: • The Poisson distribution is sometimes called the model of rare events. • 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.

  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 l: PMF of the Poisson distribution with parameter l: Mean of the Poisson distribution with parameter l: Variance and Standard Deviation of the Poisson distribution with parameter l:

  11. Chapter 6 – Poisson Distribution

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

  13. = 1.7 PDF = 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: Mean = l = 1.7 customers per minute. = 1.304 cust/min Standard deviation = s

  14. Chapter 6 – Poisson Distribution • Example: Credit Union Customers • Here is the Poisson probability distribution for l = 1.7 customers per minute on average. • 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:

  16. Poisson PDF for l = 1.7 Poisson CDF for l = 1.7 Chapter 6 – Poisson Distribution • Here are the graphs of the distributions: • 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 l = 0.2 calls per minute B = Poisson distribution with l = 0.8 calls per 15 minutes C = Poisson distribution with l = 3 calls per 15 min. D = Poisson distribution with l = 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 = m = 3 and s = 1.73 B = m = 3 and s = 3 C = m = 12 and s = 3.46 D = m = 12 and s = 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. P(X< 2) = P(0) + P(1) + P(2) PDF = 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

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

  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. • 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. Rule 1: maX+b = amX + b (mean of a transformed variable) Rule 2: saX+b = |a|sX (standard deviation of a transformed variable)

  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 mQ, sQ, 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 mQ = 500 units and standard deviation sQ = 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) sC = $35 B) sC = $40 C) sC = $1,400 D) sC = $25,400

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