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

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

4.2 (cont.) Standard Deviation of a Discrete Random Variable Measures how “spread out” the random variable is

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 

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

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

Variance of discrete random variable X

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

Standard Deviation: of More Interest then the Variance

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

Probability Histogram µ=3.65  = 4.40

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

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

© 2010 Pearson Education 13 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.

© 2010 Pearson Education 14 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.

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

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

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

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)

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

New Expected Value Long (UNC-CH) way: (compute from “scratch”) 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

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

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)

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

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)

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

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.

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 )

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)

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

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

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.

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