2. Lecture Unit 4 Section 4.2 Random Variables and Probability Models: Binomial, Geometric and Poisson Distributions

3. Streamline Treatment of Probability zSample spaces and events are good starting points for probability zSample spaces and events become quite cumbersome when applied to real-life business-related processes zRandom variables allow us to apply probability, risk and uncertainty to meaningful business-related situations

4. Bring Together Lecture Unit 2, and Section 4.1 zIn Lecture Unit 2 we saw that data could be graphically and numerically summarized in terms of midpoints, spreads, outliers, etc. zIn Section 4.1 we saw how probabilities could be assigned to outcomes of an experiment. Now we bring them together

5. First: Two Quick Examples 1. Hardee’s vs. The Colonel

6. Hardee’s vs The Colonel zOut of 100 taste-testers, 63 preferred Hardee’s fried chicken, 37 preferred KFC zEvidence that Hardee’s is better? A landslide? zWhat if there is no difference in the chicken? (p=1/2, flip a fair coin) zIs 63 heads out of 100 tosses that unusual?

7. Example 2. Mothers Identify Newborns

8. Mothers Identify Newborns zAfter spending 1 hour with their newborns, blindfolded and nose-covered mothers were asked to choose their child from 3 sleeping babies by feeling the backs of the babies’ hands z22 of 32 women (69%) selected their own newborn z“far better than 33% one would expect…” zIs it possible the mothers are guessing? zCan we quantify “far better”?

9. Graphically and Numerically Summarize a Random Experiment zPrincipal vehicle by which we do this: random variables zA random variable assigns a number to each outcome of an experiment

10. Random Variables zDefinition: A random variable is a numerical-valued function defined on the outcomes of an experiment S Number line Random variable

11. Examples zS = {HH, TH, HT, TT} the random variable: zx = # of heads in 2 tosses of a coin zPossible values of x = 0, 1, 2

12. Two Types of Random Variables zDiscrete: random variables that have a finite or countably infinite number of possible values zTest: for any given value of the random variable, you can designate the next largest or next smallest value of the random variable

13. Examples: Discrete rv’s zNumber of girls in a 5 child family zNumber of customers that use an ATM in a 1-hour period. zNumber of tosses of a fair coin that is required until you get 3 heads in a row (note that this discrete random variable has a countably infinite number of possible values: x=3, 4, 5, 6, 7,...)

14. Two types (cont.) zContinuous: a random variable that can take on all possible values in an interval of numbers zTest: given a particular value of the random variable, you cannot designate the next largest or next smallest value

15. Which is it, Discrete or Continuous? zDiscrete random variables “count” zContinuous random variables “measure” (length, width, height, area, volume, distance, time, etc.)

16. Examples: continuous rv’s zThe time it takes to run the 100 yard dash (measure) zThe time between arrivals at an ATM machine (measure) zTime spent waiting in line at the “express” checkout at the grocery store (the probability is 1 that the person in front of you is buying a loaf of bread with a third party check drawn on a Hungarian bank) (measure)

17. Examples: cont. rv’s (cont.) zThe length of a precision-engineered magnesium rod (measure) zThe area of a silicon wafer for a computer chip coming off a production line (measure)

18. Classify as discrete or continuous ax=the number of customers who enter a particular bank during the noon hour on a particular day adiscrete x={0, 1, 2, 3, …} bx=time (in seconds) required for a teller to serve a bank customer bcontinuous x>0

19. Classify (cont.) cx=the distance (in miles) between a randomly selected home in a community and the nearest pharmacy ccontinuous x>0 dx=the diameter of precision-engineered “5 inch diameter” ball bearings coming off an assembly line dcontinuous; range could be {4.5  x<5.5}

20. Classify (cont.) ex=the number of tosses of a fair coin required to observe at least 3 heads in succession ediscrete x=3, 4, 5,...

21. Data Variables and Data Distributions Data variables are known outcomes.

22. Handout 2.1, P. 10 Data Variables and Data Distributons Data variables are known outcomes. Data distributions tell us what happened.

23. Random Variables and Probability Distributions Random variables are unknown chance outcomes. Probability distributions tell us what is likely to happen. Data variables are known outcomes. Data distributions tell us what happened.

24. Great Good Economic Scenario Profit ($ Millions) 5 10 Profit Scenarios Handout 4.1, P. 3 Random variables are unknown chance outcomes. Probability distributions tell us what is likely to happen.

25. Great Good OK Economic Scenario Profit ($ Millions) 5 1 -4Lousy 10 Profit Scenarios

26. Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 The proportion of the time an outcome is expected to happen.

27. Probability Distribution Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 Shows all possible values of a random variable and the probability associated with each outcome.

28. X = the random variable (profits) x i = outcome i x 1 = 10 x 2 = 5 x 3 = 1 x 4 = -4 Notation Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 x4x4 X x1x1 x2x2 x3x3

29. P is the probability p(x i )= Pr(X = x i ) is the probability of X being outcome x i p(x 1 ) = Pr(X = 10) =.20 p(x 2 ) = Pr(X = 5) =.40 p(x 3 ) = Pr(X = 1) =.25 p(x 4 ) = Pr(X = -4) =.15 Notation Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 Pr(X=x 4 ) X Pr(X=x 1 ) Pr(X=x 2 ) Pr(X=x 3 ) x1x1 x2x2 x3x3 x4x4

30. What are the chances? What are the chances that profits will be less than $5 million in 2013? P(X < 5) = P(X = 1 or X = -4) = P(X = 1) + P(X = -4) =.25 +.15 =.40 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 X x1x1 x2x2 x3x3 x4x4

31. What are the chances? What are the chances that profits will be less than $5 million in 2013 and less than $5 million in 2014? P(X < 5 in 2013 and X < 5 in 2014) = P(X < 5)·P(X < 5) =.40·.40 =.16 P(X < 5) =.40 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 p(x 4 ) X x1x1 x2x2 x3x3 x4x4 P p(x 1 ) p(x 2 ) p(x 3 )

32. .05.10.15.40.20.25.30.35 Probability Histogram -4-2024681012 Profit Probability.05.10.15.40.20.25.30.35 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 p(x 4 ) X x1x1 x2x2 x3x3 x4x4 p(x 1 ) p(x 2 ) p(x 3 )

33. .05.10.15.40.20.25.30.35 Probability Histogram -4-2024681012 Profit Probability Lousy OK Good Great.05.10.15.40.20.25.30.35 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 p(x 4 ) X x1x1 x2x2 x3x3 x4x4 P p(x 1 ) p(x 2 ) p(x 3 )

34. Probability distributions: requirements zNotation: p(x)= Pr(X = x) is the probability that the random variable X has value x zRequirements 1. 0  p(x)  1 for all values x of X 2.  all x p(x) = 1

35. Example xp(x) 0.20 1.90 2-.10 zproperty 1) violated: p(2) = -.10 xp(x) -2.3 -1.3 1.3 2.3 z property 2) violated:  p(x) = 1.2

36. Example (cont.) xp(x) -1.25 0.65 1.10 OK1) satisfied: 0  p(x)  1 for all x 2) satisfied:  all x p(x) =.25+.65+.10 = 1

37. Example: light bulbs 20% of light bulbs last at least 800 hrs; you have just purchased 2 light bulbs. X=number of the 2 bulbs that last at least 800 hrs (possible values of x: 0, 1, 2) Find the probability distribution of X zS: bulb lasts at least 800 hrs zF: bulb fails to last 800 hrs zP(S) =.2; P(F) =.8

38. Example (cont.) Possible outcomesP(outcome)x (S,S)(.2)(.2)=.042 (S,F)(.2)(.8)=.161 (F,S)(.8)(.2)=.161 (F,F)(.8)(.8)=.640 probability x012 distribution of x:p(x).64.32.04 S S - SS F - SF S - FS F - FF F

39. Example: 3-child family 3 child family; X=#of boys M: child is male P(M)=1/2 (0.5121; from.5134) F: child is female P(F)=1/2 (0.4879) OutcomesP(outcome) x MMM(1/2) 3 =1/8 3 MMF1/8 2 MFM1/8 2 FMM1/8 2 MFF1/8 1 FMF1/8 1 FFM1/8 1 FFF1/8 0

40. Probability Distribution of x x0123 p(x)1/83/83/81/8 Probability of at least 1 boy: P(x  1)= 3/8 + 3/8 +1/8 = 7/8 Probability of no boys or 1 boy: p(0) + p(1)= 1/8 + 3/8 = 4/8 = 1/2