1. Chapter 16 Random Variables, Expected Value, Standard Deviation Random Variables and Probability Models: Binomial, Geometric and Poisson Distributions

2. 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

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

4. 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

5. 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

6. 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,...)

7. 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

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

9. 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)

10. 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)

11. 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

12. 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}

13. 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,...

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

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

16. 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.

17. 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.

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

19. 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.

20. 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.

21. 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

22. 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

23. What are the chances? What are the chances that profits will be less than $5 million in 2014? P(X < 5) = P(X = 1 or X = -4) = P(X = 1) + P(X = -4) =.25 +.15 =.40 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 X x1x1 x2x2 x3x3 x4x4

24. What are the chances? What are the chances that profits will be less than $5 million in 2014 and less than $5 million in 2015? P(X < 5 in 2011 and X < 5 in 2012) = P(X < 5)·P(X < 5) =.40·.40 =.16 P(X < 5) =.40 Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 p(x 4 ) X x1x1 x2x2 x3x3 x4x4 P p(x 1 ) p(x 2 ) p(x 3 )

25. .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 )

26. .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 )

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. Expected Value of a Discrete Random Variable A measure of the “middle” of the values of a random variable

35. Center The mean of the probability distribution is the expected value of X, denoted E(X) E(X) is also denoted by the Greek letter µ (mu)

36. k = the number of possible values (k=4) E(x)= µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) +... + x k ·p(x k ) Weighted mean Mean or Expected Value Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

37. k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) +... + x k ·p(x k ) Weighted mean Each outcome is weighted by its probability Mean or Expected Value

38. Other Weighted Means zStock Market: The Dow Jones Industrial Average yThe “Dow” consists of 30 companies (the 30 companies in the “Dow” change periodically) yTo compute the Dow Jones Industrial Average, a weight proportional to the company’s “size” is assigned to each company’s stock price

39. k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) +... + x k ·p(x k ) EXAMPLE Mean Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

40. k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) +... + x k ·p(x k ) EXAMPLE µ = 10*.20 + 5*.40 + 1*.25 – 4*.15 = 3.65 ($ mil) Mean Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 )

41. k = the number of outcomes (k=4) µ = x 1 ·p(x 1 ) + x 2 ·p(x 2 ) + x 3 ·p(x 3 ) +... + x k ·p(x k ) EXAMPLE µ = 10·.20 + 5·.40 + 1·.25 - 4·.15 = 3.65 ($ mil) Mean µ=3.65

42. Interpretation zE(x) is not the value of the random variable x that you “expect” to observe if you perform the experiment once

43. Interpretation zE(x) is a “long run” average; if you perform the experiment many times and observe the random variable X each time, then the average x of these observed X- values will get closer to E(X) as you observe more and more values of the random variable X.

44. Example: Green Mountain Lottery zState of Vermont zchoose 3 digits from 0 through 9; repeats allowed zwin $500 x$0$500 p(x).999.001 E(x)=$0(.999) + $500(.001) = $.50

45. Example (cont.) zE(X)=$.50 zOn average, each ticket wins $.50. zImportant for Vermont to know zE(X) is not necessarily a possible value of the random variable (values of X are $0 and $500)

46. Example: coin tossing  Suppose a fair coin is tossed 3 times and we let x=the number of heads. Find  (x). zFirst we must find the probability distribution of x.

47. Example (cont.) zPossible values of x: 0, 1, 2, 3. zp(1)? zAn outcome where x = 1: THT zP(THT)? (½)(½)(½)=1/8 zHow many ways can we get 1 head in 3 tosses? 3 C 1 =3

48. Example (cont.)

49. zSo the probability distribution of x is: x0123 p(x)1/83/83/81/8

50. Example zSo the probability distribution of x is: x0123 p(x)1/83/83/81/8

51. US Roulette Wheel and Table zThe roulette wheel has alternating black and red slots numbered 1 through 36. zThere are also 2 green slots numbered 0 and 00. zA bet on any one of the 38 numbers (1-36, 0, or 00) pays odds of 35:1; that is... zIf you bet $1 on the winning number, you receive $36, so your winnings are $35 American Roulette 0 - 00 (The European version has only one 0.)

52. US Roulette Wheel: Expected Value of a $1 bet on a single number zLet X be your winnings resulting from a $1 bet on a single number; X has 2 possible values X-135 p(x)37/381/38 zE(X)= -1(37/38)+35(1/38)= -.05 zSo on average the house wins 5 cents on every such bet. A “fair” game would have E(X)=0. zThe roulette wheels are spinning 24/7, winning big $$ for the house, resulting in …

54. Standard Deviation of a Discrete Random Variable First center (expected value) Now - spread

55. Standard Deviation of a Discrete Random Variable Measures how “spread out” the random variable is

56. Summarizing data and probability Data zHistogram zmeasure of the center: sample mean x zmeasure of spread: sample standard deviation s Random variable z Probability Histogram  measure of the center: population mean   measure of spread: population standard deviation 

57. Example z x0100 p(x)1/21/2 E(x) = 0(1/2) + 100(1/2) = 50 z y4951 p(y)1/21/2 E(y) = 49(1/2) + 51(1/2) = 50

58. The deviations of the outcomes from the mean of the probability distribution x i - µ  2 (sigma squared) is the variance of the probability distribution Variation

59. Variance of discrete random variable X

60. Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 ) P. 207, Handout 4.1, P. 4 Example  2 = (x 1 -µ) 2 · P(X=x 1 ) + (x 2 -µ) 2 · P(X=x 2 ) + (x 3 -µ) 2 · P(X=x 3 ) + (x 4 -µ) 2 · P(X=x 4 ) = (10-3.65) 2 · 0.20 + (5-3.65) 2 · 0.40 + (1-3.65) 2 · 0.25 + (-4-3.65) 2 · 0.15 = 19.3275 Variation 3.65

61. Standard Deviation: of More Interest then the Variance

62.  or SD, is the standard deviation of the probability distribution Standard Deviation  2 = 19.3275

63. Probability Histogram µ=3.65  = 4.40

64. Finance and Investment Interpretation zX = return on an investment (stock, portfolio, etc.)  E(X) =  expected return on this investment   is a measure of the risk of the investment

65. Example A basketball player shoots 3 free throws. P(make) =P(miss)=0.5. Let X = number of free throws made.

66. © 2010 Pearson Education 66 Expected Value of a Random Variable Example: The probability model for a particular life insurance policy is shown. Find the expected annual payout on a policy. We expect that the insurance company will pay out $200 per policy per year.

67. © 2010 Pearson Education 67 Standard Deviation of a Random Variable Example: The probability model for a particular life insurance policy is shown. Find the standard deviation of the annual payout.

68. 68-95-99.7 Rule for Random Variables For random variables X whose probability histograms are approximately mound- shaped:  P  X   P  X    P(  X  

69. (  ) (50-5, 50+5) (45, 55) P  X  P(45  X  55) =.048+.057+.066+.073+.078+.08+.078+.073+.066+.057+.048=.724

71. Rules for E(X), Var(X) and SD(X)  adding a constant a zIf X is a rv and a is a constant:  E(X+a) = E(X)+a z Example: a = -1  E(X+a)=E(X-1)=E(X)-1

72. Rules for E(X), Var(X) and SD(X): adding constant a (cont.) zVar(X+a) = Var(X) SD(X+a) = SD(X) z Example: a = -1  Var(X+a)=Var(X-1)=Var(X)  SD(X+a)=SD(X-1)=SD(X)

73. Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5 1 -4Lousy0.15 10 P(X=x 4 ) X x1x1 x2x2 x3x3 x4x4 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 ) Probability Great0.20 Good0.40 OK0.25 Economic Scenario Profit ($ Millions) 5+2 1+2 -4+2Lousy0.15 10+2 P(X=x 4 ) X+2 x1+2x1+2 x 2 +2 x 3 +2 x 4 +2 P P(X=x 1 ) P(X=x 2 ) P(X=x 3 ) E(x + a) = E(x) + a; SD(x + a)=SD(x); let a = 2  5.65  = 4.40  3.65  = 4.40

74. New Expected Value Long (UNC-CH) way: E(x+2)=12(.20)+7(.40)+3(.25)+(-2)(.15) = 5.65 Smart (NCSU) way: a=2; E(x+2) =E(x) + 2 = 3.65 + 2 = 5.65

75. New Variance and SD Long (UNC-CH) way: (compute from “scratch”) Var(X+2)=(12-5.65) 2 (0.20)+… +(-2+5.65) 2 (0.15) = 19.3275 SD(X+2) = √19.3275 = 4.40 Smart (NCSU) way: Var(X+2) = Var(X) = 19.3275 SD(X+2) = SD(X) = 4.40

76. Rules for E(X), Var(X) and SD(X): multiplying by constant b zE(bX)=b E(X) zVar(b X) = b 2 Var(X) zSD(bX)= |b|SD(X) z Example: b =-1  E(bX)=E(-X)=-E(X)  Var(bX)=Var(-1X)= =(-1) 2 Var(X)=Var(X)  SD(bX)=SD(-1X)= =|-1|SD(X)=SD(X)

77. Expected Value and SD of Linear Transformation a + bx Let X=number of repairs a new computer needs each year. Suppose E(X)= 0.20 and SD(X)=0.55 The service contract for the computer offers unlimited repairs for $100 per year plus a $25 service charge for each repair. What are the mean and standard deviation of the yearly cost of the service contract? Cost = $100 + $25X E(cost) = E($100+$25X)=$100+$25E(X)=$100+$25*0.20= = $100+$5=$105 SD(cost)=SD($100+$25X)=SD($25X)=$25*SD(X)=$25*0.55= =$13.75

78. Addition and Subtraction Rules for Random Variables zE(X+Y) = E(X) + E(Y); zE(X-Y) = E(X) - E(Y) zWhen X and Y are independent random variables: 1.Var(X+Y)=Var(X)+Var(Y) 2.SD(X+Y)= SD’s do not add: SD(X+Y)≠ SD(X)+SD(Y) 3.Var(X−Y)=Var(X)+Var(Y) 4.SD(X −Y)= SD’s do not subtract: SD(X−Y)≠ SD(X)−SD(Y) SD(X−Y)≠ SD(X)+SD(Y)

79. Motivation for Var(X-Y)=Var(X)+Var(Y)  Let X=amount automatic dispensing machine puts into your 16 oz drink (say at McD’s)  A thirsty, broke friend shows up. Let Y=amount you pour into friend’s 8 oz cup  Let Z = amount left in your cup; Z = ?  Z = X-Y  Var(Z) = Var(X-Y) = Var(X) + Var(Y) Has 2 components

80. Example: rv’s NOT independent zX=number of hours a randomly selected student from our class slept between noon yesterday and noon today. zY=number of hours a randomly selected student from our class was awake between noon yesterday and noon today. Y = 24 – X. zWhat are the expected value and variance of the total hours that a student is asleep and awake between noon yesterday and noon today? zTotal hours that a student is asleep and awake between noon yesterday and noon today = X+Y zE(X+Y) = E(X+24-X) = E(24) = 24 zVar(X+Y) = Var(X+24-X) = Var(24) = 0. zWe don't add Var(X) and Var(Y) since X and Y are not independent.

81. a2a2 c 2 =a 2 +b 2 b2b2 Pythagorean Theorem of Statistics for Independent X and Y a b c a 2 + b 2 = c 2 Var(X) Var(Y) Var(X+Y) SD(X) SD(Y) SD(X+Y) Var(X)+Var(Y)=Var(X+Y) a + b ≠ c SD(X)+SD(Y) ≠SD(X+Y)

82. 9 25=9+16 16 Pythagorean Theorem of Statistics for Independent X and Y 3 4 5 3 2 + 4 2 = 5 2 Var(X) Var(Y) Var(X+Y) SD(X) SD(Y) SD(X+Y) Var(X)+Var(Y)=Var(X+Y) 3 + 4 ≠ 5 SD(X)+SD(Y) ≠SD(X+Y)

83. Example: meal plans zRegular plan: X = daily amount spent zE(X) = $13.50, SD(X) = $7 zExpected value and stan. dev. of total spent in 2 consecutive days? zE(X 1 +X 2 )=E(X 1 )+E(X 2 )=$13.50+$13.50=$27 SD(X 1 + X 2 ) ≠ SD(X 1 )+SD(X 2 ) = $7+$7=$14

84. Example: meal plans (cont.) zJumbo plan for football players Y=daily amount spent zE(Y) = $24.75, SD(Y) = $9.50 zAmount by which football player’s spending exceeds regular student spending is Y-X zE(Y-X)=E(Y)–E(X)=$24.75-$13.50=$11.25 SD(Y ̶ X) ≠ SD(Y) ̶ SD(X) = $9.50 ̶ $7=$2.50

85. For random variables, X+X≠2X zLet X be the annual payout on a life insurance policy. From mortality tables E(X)=$200 and SD(X)=$3,867. 1)If the payout amounts are doubled, what are the new expected value and standard deviation?  Double payout is 2X. E(2X)=2E(X)=2*$200=$400  SD(2X)=2SD(X)=2*$3,867=$7,734 2)Suppose insurance policies are sold to 2 people. The annual payouts are X 1 and X 2. Assume the 2 people behave independently. What are the expected value and standard deviation of the total payout?  E(X 1 + X 2 )=E(X 1 ) + E(X 2 ) = $200 + $200 = $400 The risk to the insurance co. when doubling the payout (2X) is not the same as the risk when selling policies to 2 people.