1. CHAPTER 16 Standard Deviation of a Discrete Random Variable First center (expected value) Now - spread

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

3. 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 

4. 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

5. 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

7. 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

8. Standard Deviation: of More Interest then the Variance

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

10. Probability Histogram µ=3.65  = 4.40

11. 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

12. Example

13. Example (cont.)  Specify the interval (  ) z(1.5 - 1(.866), 1.5 + 1(.866)) (1.5 -.866, 1.5 +.866) (.634, 2.366) zP(.634  x  2.366) = zp(1)+p(2)=3/8 + 3/8 = 3/4 

14. 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  

16. 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

17. 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)

18. 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; Var(x + a)=Var (x); let a = 2  5.65  = 4.40  3.65  = 4.40

19. 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

20. 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

21. 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)

22. 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

23. 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:  Var(X+Y)=Var(X)+Var(Y)  SD(X+Y)= SD’s do not add: SD(X+Y)≠ SD(X)+SD(Y)  Var(X−Y)=Var(X)+Var(Y)  SD(X −Y)= SD’s do not subtract: SD(X−Y)≠ SD(X)−SD(Y) SD(X−Y)≠ SD(X)+SD(Y)

24. zIn general, if X and Y are RVs, Var(X+Y) ≠ Var(X)+Var(Y) SD(X+Y) ≠ SD(X)+SD(Y) Special case (previous slide): If RVs are Independent: Variances add: Var(X+Y)=Var(X)+Var(Y) but… Standard Deviations still DO NOT add: SD(X+Y)≠SD(X)+SD(Y)

25. a2a2 c2c2 b2b2 Pythagorean Theorem of Statistics: variances add; standard deviations do not add 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)

26. 9 25 16 Pythagorean Theorem of Statistics: variances add; standard deviations do not add 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)

27. 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

28. 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. 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 zSuppose 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.