1. Chapter 16 Random Variables Streamlining Probability: Probability Distribution, Expected Value and Standard Deviation of Random Variable

2. Graphically and Numerically Summarize a Random Experiment zPrincipal vehicle by which we do this: random variables

3. Random Variables zDefinition: A random variable is a numerical-valued variable whose value is based on the outcome of a random event. Denoted by upper-case letters X, Y, etc.

4. Examples 1.X = # of games played in a randomly selected World Series Possible values of X are x=4, 5, 6, 7 2.Y=score on 13 th hole (par 5) at Augusta National golf course for a randomly selected golfer on day 1 of 2011 Masters y=3, 4, 5, 6, 7

5. 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. A probability distribution lists the possible values of a random variable and the probability that each value will occur.

6. Probability Distribution Of Number of Games Played in Randomly Selected World Series zEstimate based on results from 1946 to 2010. x4567 p(x)12/65=0.185 14/65=0.21527/65=0.415 Probability Histogram

7. Probability Distribution Of Score on 13 th hole (par 5) at Augusta National Golf Course on Day 1 of 2011 Masters y34567 p(x)0.0400.4140.4650.0510.030 Probability Histogram

8. Probability distributions: requirements zRequirements 1. 0  p(x)  1 for all values x of X 2.  all x p(x) = 1

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

10. 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)

11. k = the number of possible values of random variable 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 x4567 p(x)12/65=0.185 14/65=0.21527/65=0.415 y34567 p(x)0.0400.4140.4650.0510.030

12. k = the number of outcomes 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

13. Other Weighted Means zGPA A=4, B=3, C=2, D=1, F=0 zBaseball slugging percentage SLG (hr=4, 3b=3, 2b=2, 1b=1) zBaseball ticket prices Football ticket pricesBaseball ticket pricesFootball ticket prices

14. E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414) =5.86 games E(Y)= µ=3(.04)+4(0.414)+5(0.465)+6(0.051)+7(0.03) =4.617 strokes Mean or Expected Value x4567 p(x)12/65=0.185 14/65=0.21527/65=0.415 y34567 p(x)0.0400.4140.4650.0510.030

15. E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414) =5.86 games Mean or Expected Value µ=5.86

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

17. Interpretation of E(X) zE(X) is a “long run” average. zThe expected value of a random variable is equal to the average value of the random variable if the chance process was repeated an infinite number of times. In reality, if the chance process is continually repeated, x will get closer to E(x) as you observe more and more values of the random variable x.

18. 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

19. 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)

20. Expected Value, Surprise Onside Kicks zhttp://www.advancednflstats.com/ The change in expected points for the kicking team: successful 1.9; fail -1.4.http://www.advancednflstats.com/ zX=change in expected points for kicking team when attempting surprise onside kick zWhat values of p make surprise onside kicks a good strategy? X1.9-1.4 p(x)p1-p

21. 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.)

22. 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 …

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

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

26. 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 

27. 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

28. 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

29. Variance of random variable X

30. 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 ) = (4-5.86) 2 · 0.185 + (5-5.86) 2 · 0.185 + (6-5.86) 2 · 0.215 + (7-5.86) 2 · 0.415 = 1.3204 Variation 5.86 x4567 p(x)12/65=0.185 14/65=0.21527/65=0.415

31. Standard Deviation: of More Interest then the Variance

32.  or SD, is the standard deviation of the probability distribution Standard Deviation  2 = 1.3204

33. © 2010 Pearson Education 33 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.

34. © 2010 Pearson Education 34 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.

35. 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  

36. (  ) (50-5, 50+5) (45, 55) P  X  P(45  X  55) =.048+.057+.066+.073+.078+.08+.078+.073+.066+.057+.048=.724

38. 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

39. 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)

40. Carolina Panthers Next Season’s Profit Probability Great0.20 Good0.40 OK0.25 Economy Profit X ($ Millions) 5 1 -4Lousy0.15 10 zE(X)=10(0.20) + 5(0.40) + 1(0.25) – 4(0.15) z=3.65 zSD(X)=4.4

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

42. 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

43. 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

44. Rules for E(X), Var(X) and SD(X): multiplying by constant b zE(bX)=bE(X) zVar(bX) = b 2 Var(X) zSD(bX)= |b|SD(X) |b| denotes the absolute value of b 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)

45. Expected Value and SD of Linear Transformation a + bx Let the random variable X= season field goal shooting percentage for an NBA team. Suppose E(X)= 45.31 and SD(X)=1.67 The relationship between X and points scored per game for an NBA team can be described by 14.49 + 1.85X. What are the mean and standard deviation of the points scored per game? Points per game (ppg) = 14.49 + 1.85X E(ppg) = E(14.49+1.85X)=14.49+1.85E(X)=14.49+1.85*45.31= = 14.49+83.82=98.31 SD(ppg)=SD(14.49+1.85X)=SD(1.85X)=1.85*SD(X)=1.85*1.67= =3.09 Note that the shift of 14.49 does NOT affect the standard deviation.

46. 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)

47. 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

48. 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 the same 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.

49. a2a2 b2b2 Pythagorean Theorem of Statistics for Independent X and Y a b c Var(X) Var(Y) Var(X+Y) SD(X) SD(Y) SD(X+Y) a + b ≠ c SD(X)+SD(Y) ≠SD(X+Y) c2c2 a 2 +b 2 =c 2 Var(X) +Var(Y ) =Var(X+Y )

50. 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)

51. 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

52. 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

53. 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.