1. 4 - 1 © 2001 prentice-Hall, Inc. Behavioral Statistics Discrete Random Variables Chapter 4

2. 4 - 2 © 2001 prentice-Hall, Inc. Learning Objectives 1.Distinguish Between the Two Types of Random Variables 2.Compute the Expected Value & Variance of Discrete Random Variables 3.Describe the Binomial and Poisson 4.Calculate Probabilities for Discrete Random Variables

3. 4 - 3 © 2001 prentice-Hall, Inc. Thinking Challenge You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right? If you guessed on all 33 questions, what would be your grade? pass?

4. 4 - 4 © 2001 prentice-Hall, Inc. Data Types

5. 4 - 5 © 2001 prentice-Hall, Inc. Discrete Random Variables

6. 4 - 6 © 2001 prentice-Hall, Inc. Discrete Random Variable 1. Random Variable A Numerical Outcome of an Experiment A Numerical Outcome of an Experiment Example: Number of Tails in 2 Coin Tosses Example: Number of Tails in 2 Coin Tosses 2. Discrete Random Variable Whole Number (0, 1, 2, 3 etc.) Whole Number (0, 1, 2, 3 etc.) Obtained by Counting Obtained by Counting Usually Finite Number of Values Usually Finite Number of Values Poisson Random Variable Is Exception (  ) Poisson Random Variable Is Exception (  )

7. 4 - 7 © 2001 prentice-Hall, Inc. Discrete Random Variable Examples ExperimentRandom Variable Possible Values Make 100 Sales Calls # Sales 0, 1, 2,..., 100 Inspect 70 Radios # Defective 0, 1, 2,..., 70 Answer 33 Questions # Correct 0, 1, 2,..., 33 Count Cars at Toll Between 11:00 & 1:00 # Cars Arriving 0, 1, 2,..., 

8. 4 - 8 © 2001 prentice-Hall, Inc. Discrete Probability Distribution 1.List of All possible [x, p(x)] pairs x = Value of Random Variable (Outcome) x = Value of Random Variable (Outcome) p(x) = Probability Associated with Value p(x) = Probability Associated with Value 2.Mutually Exclusive (No Overlap) 3.Collectively Exhaustive (Nothing Left Out) 4. 0  p(x)  1 5.  p(x) = 1

9. 4 - 9 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Example Probability Distribution Values, x Probabilities, p(x) 01/4 =.25 12/4 =.50 21/4 =.25 Experiment: Toss 2 Coins. Count # Tails. © 1984-1994 T/Maker Co.

10. 4 - 10 © 2001 prentice-Hall, Inc. Visualizing Discrete Probability Distributions ListingTable GraphEquation # Tails f(x) Count p(x) 01.25 12.50 21.25 px n xnx pp xnx () ! !()! ()    1.00.25.50 012 x p(x)

11. 4 - 11 © 2001 prentice-Hall, Inc. Summary Measures 1.Expected Value Mean of Probability Distribution Mean of Probability Distribution Weighted Average of All Possible Values Weighted Average of All Possible Values  = E(X) =  x p(x)  = E(X) =  x p(x) 2.Variance Weighted Average Squared Deviation about Mean Weighted Average Squared Deviation about Mean  2 = E[ (x    (x    p(x)  2 = E[ (x    (x    p(x)

12. 4 - 12 © 2001 prentice-Hall, Inc. Summary Measures Calculation Table xp(x)xp(x)x -  (x(x(x(x-  ) 2 ( x -  ) 2 p( p( x ) Total  xp(x)  ( x -  ) 2 x )

13. 4 - 13 © 2001 prentice-Hall, Inc. Thinking Challenge You toss 2 coins. You’re interested in the number of tails. What are the expected value & standard deviation of this random variable, number of tails? © 1984-1994 T/Maker Co.

14. 4 - 14 © 2001 prentice-Hall, Inc. Expected Value & Variance Solution* 0.2501.00.25 1.50.50000 2.25.501.001.00.25  = 1.0 = 1.0  2 =.50 =.50 xp(x)xp(x)x -  (x(x(x(x-  ) 2 ( x -  ) 2 p( p( x )

15. 4 - 15 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Function

16. 4 - 16 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Function 1.Type of Model Representation of Some Underlying phenomenon Representation of Some Underlying phenomenon 2.Mathematical Formula 3.Represents Discrete Random Variable 4.Used to Get Exact Probabilities PXx x () !  x e -

17. 4 - 17 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Models

18. 4 - 18 © 2001 prentice-Hall, Inc. Binomial Distribution

19. 4 - 19 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Models

20. 4 - 20 © 2001 prentice-Hall, Inc. Binomial Distribution 1.Number of ‘Successes’ in a Sample of n Observations (Trials) # Reds in 15 Spins of Roulette Wheel # Reds in 15 Spins of Roulette Wheel # Defective Items in a Batch of 5 Items # Defective Items in a Batch of 5 Items # Correct on a 33 Question Exam # Correct on a 33 Question Exam # Customers Who Purchase Out of 100 Customers Who Enter Store # Customers Who Purchase Out of 100 Customers Who Enter Store

21. 4 - 21 © 2001 prentice-Hall, Inc. Binomial Distribution Properties 1.Two Different Sampling Methods Infinite Population Without Replacement Infinite Population Without Replacement Finite Population With Replacement Finite Population With Replacement 2.Sequence of n Identical Trials 3.Each Trial Has 2 Outcomes ‘Success’ (Desired Outcome) or ‘Failure’ ‘Success’ (Desired Outcome) or ‘Failure’ 4.Constant Trial Probability 5.Trials Are Independent

22. 4 - 22 © 2001 prentice-Hall, Inc. Binomial Probability Distribution Function p(x) = Probability of x ‘Successes’ n=Sample Size p=Probability of ‘Success’ x=Number of ‘Successes’ in Sample (x = 0, 1, 2,..., n)

23. 4 - 23 © 2001 prentice-Hall, Inc. Binomial Probability Distribution Example Experiment: Toss 1 Coin 5 Times in a Row. Note # Tails. What’s the Probability of 3 Tails?

24. 4 - 24 © 2001 prentice-Hall, Inc. Binomial Probability Table (Portion) Cumulative Probabilities

25. 4 - 25 © 2001 prentice-Hall, Inc. Binomial Distribution Characteristics n = 5 p = 0.1 n = 5 p = 0.5 Mean Standard Deviation

26. 4 - 26 © 2001 prentice-Hall, Inc. Binomial Distribution Thinking Challenge You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p =.20). If you call 12 people tonight, what’s the probability of A. No sales? B. Exactly 2 sales? C. At most 2 sales? D. At least 2 sales?

27. 4 - 27 © 2001 prentice-Hall, Inc. Binomial Distribution Solution* Using the Binomial Tables: A. p(0) =.0687 B. p(2) =.2835 C. p(at most 2)= p(0) + p(1) + p(2) =.0687 +.2062 +.2835 =.5584 D. p(at least 2)= p(2) + p(3)...+ p(12) = 1 - [p(0) + p(1)] = 1 -.0687 -.2062 =.7251

28. 4 - 28 © 2001 prentice-Hall, Inc. Poisson Distribution

29. 4 - 29 © 2001 prentice-Hall, Inc. Discrete Probability Distribution Models

30. 4 - 30 © 2001 prentice-Hall, Inc. Poisson Distribution 1.Number of Events that Occur in an Interval Events Per Unit Events Per Unit Time, Length, Area, Space Time, Length, Area, Space 2.Examples # Customers Arriving in 20 minutes # Customers Arriving in 20 minutes # Strikes Per Year in the U.S. # Strikes Per Year in the U.S. # Defects Per Lot (Group) of VCR’s # Defects Per Lot (Group) of VCR’s

31. 4 - 31 © 2001 prentice-Hall, Inc. Poisson Process 1.Constant Event Probability Average of 60/Hr Is 1/Min for 60 1-Minute Intervals Average of 60/Hr Is 1/Min for 60 1-Minute Intervals 2.One Event Per Interval Don’t Arrive Together Don’t Arrive Together 3.Independent Events Arrival of 1 Person Does Not Affect Another’s Arrival Arrival of 1 Person Does Not Affect Another’s Arrival © 1984-1994 T/Maker Co.

32. 4 - 32 © 2001 prentice-Hall, Inc. Poisson Probability Distribution Function p(x) = Probability of x Given p(x) = Probability of x Given =Expected (Mean) Number of ‘Successes’ =Expected (Mean) Number of ‘Successes’ e=2.71828 (Base of Natural Logs) x=Number of ‘Successes’ Per Unit px x () ! xe-

33. 4 - 33 © 2001 prentice-Hall, Inc. Poisson Distribution Characteristics  = 0.5  = 6 Mean Standard Deviation

34. 4 - 34 © 2001 prentice-Hall, Inc. Poisson Distribution Example Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes? © 1995 Corel Corp.

35. 4 - 35 © 2001 prentice-Hall, Inc. Poisson Distribution Solution 72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

36. 4 - 36 © 2001 prentice-Hall, Inc. Poisson Probability Table (Portion) Cumulative Probabilities

37. 4 - 37 © 2001 prentice-Hall, Inc. Thinking Challenge You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction? © 1984-1994 T/Maker Co.

38. 4 - 38 © 2001 prentice-Hall, Inc. Poisson Distribution Solution: Finding * 75 words/min = (75 words/min)(60 min/hr) = 4500 words/hr 6 errors/hr= 6 errors/4500 words 6 errors/hr= 6 errors/4500 words =.00133 errors/word In a 255-word transaction (interval): = (.00133 errors/word )(255 words) = (.00133 errors/word )(255 words) =.34 errors/255-word transaction =.34 errors/255-word transaction

39. 4 - 39 © 2001 prentice-Hall, Inc. Poisson Distribution Solution: Finding p(0)*

40. 4 - 40 © 2001 prentice-Hall, Inc. Conclusion 1.Distinguished Between the Two Types of Random Variables 2.Computed the Expected Value & Variance of Discrete Random Variables 3.Described the Binomial and Poisson Distributions 4.Calculated Probabilities for Discrete Random Variables

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