1 / 33

4.2 (cont.) Standard Deviation of a Discrete Random Variable

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. Data Histogram measure of the center: sample mean x measure of spread:

dee
Download Presentation

4.2 (cont.) Standard Deviation of a Discrete Random Variable

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 4.2 (cont.) Standard Deviation of a Discrete Random Variable First center (expected value) Now - spread

  2. 4.2 (cont.) Standard Deviation of a Discrete Random Variable Measures how “spread out” the random variable is

  3. Data Histogram measure of the center: sample mean x measure of spread: sample standard deviation s Random variable Probability Histogram measure of the center: population mean m measure of spread: population standard deviation s Summarizing data and probability

  4. Example • x 0 100 p(x) 1/2 1/2 E(x) = 0(1/2) + 100(1/2) = 50 • y 49 51 p(y) 1/2 1/2 E(y) = 49(1/2) + 51(1/2) = 50

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

  6. Variation Variance of discrete random variable X

  7. Economic Scenario Profit ($ Millions) Probability X P Great 10 0.20 x1 P(X=x1) 5 Good 0.40 x2 P(X=x2) OK 1 0.25 x3 P(X=x3) Lousy -4 0.15 x4 P(X=x4) Variation Example 2 = (x1-µ)2 · P(X=x1) + (x2-µ)2 · P(X=x2) + (x3-µ)2 · P(X=x3) + (x4-µ)2 · P(X=x4) = (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 3.65 3.65 3.65 3.65 P. 207, Handout 4.1, P. 4

  8. Standard Deviation: of More Interest then the Variance

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

  10. Probability Histogram  = 4.40 µ=3.65

  11. Finance and Investment Interpretation • X = return on an investment (stock, portfolio, etc.) • E(x) = m = expected return on this investment • sis a measure of the risk of the investment

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

  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. 13 © 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. 14 © 2010 Pearson Education

  15. 68-95-99.7 Rule for Random Variables For random variables x whose probability histograms are approximately mound-shaped: • P(m - s  x  m + s)  .68 • P(m - 2s  x  m + 2s)  .95 • P(m -3s  x  m + 3s)  .997

  16. (m - 1s, m + 1s) (50-5, 50+5) (45, 55) P(m - s  X  m + s) = P(45  X  55) =.048+.057+.066+.073+.078+.08+.078+.073+ .066+.057+.048=.724

  17. Rules for E(X), Var(X) and SD(X):adding a constant a • If X is a rv and a is a constant: • E(X+a) = E(X)+a • Example: a = -1 • E(X+a)=E(X-1)=E(X)-1

  18. Rules for E(X), Var(X) and SD(X): adding constant a (cont.) • Var(X+a) = Var(X) SD(X+a) = SD(X) • Example: a = -1 • Var(X+a)=Var(X-1)=Var(X) • SD(X+a)=SD(X-1)=SD(X)

  19. Economic Scenario Profit ($ Millions) X Probability Economic Scenario Profit ($ Millions) X+2 Probability P P Great 10 0.20 Great 10+2 0.20 x1 x1+2 P(X=x1) P(X=x1) 5 5+2 Good 0.40 Good 0.40 x2 x2+2 P(X=x2) P(X=x2) OK 1 0.25 OK 1+2 0.25 x3 x3+2 P(X=x3) P(X=x3) Lousy -4 0.15 Lousy -4+2 0.15 x4 x4+2 P(X=x4) P(X=x4) E(x + a) = E(x) + a; SD(x + a)=SD(x); let a = 2  = 4.40  = 4.40 m=5.65 m=3.65

  20. 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

  21. 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

  22. Rules for E(X), Var(X) and SD(X): multiplying by constant b • E(bX)=b E(X) • Var(b X) = b2Var(X) • SD(bX)= |b|SD(X) • Example: b =-1 • E(bX)=E(-X)=-E(X) • Var(bX)=Var(-1X)= =(-1)2Var(X)=Var(X) • SD(bX)=SD(-1X)= =|-1|SD(X)=SD(X)

  23. 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

  24. Addition and Subtraction Rules for Random Variables • E(X+Y) = E(X) + E(Y); • E(X-Y) = E(X) - E(Y) • When 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)

  25. Motivation forVar(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) Has 2 components + Var(Y)

  26. Example: rv’s NOT independent • X=number of hours a randomly selected student from our class slept between noon yesterday and noon today. • Y=number of hours the same randomly selected student from our class was awake between noon yesterday and noon today. Y = 24 – X. • What are the expected value and variance of the total hours that a student is asleep and awake between noon yesterday and noon today? • Total hours that a student is asleep and awake between noon yesterday and noon today = X+Y • E(X+Y) = E(X+24-X) = E(24) = 24 • Var(X+Y) = Var(X+24-X) = Var(24) = 0. • We don't add Var(X) and Var(Y) since X and Y are not independent.

  27. Pythagorean Theorem of Statistics for Independent X and Y a2+b2=c2 Var(X+Y) =Var(X+Y) c2 +Var(Y) Var(X) Var(X) c a2 a SD(X+Y) SD(X) a + b ≠ c SD(X)+SD(Y) ≠SD(X+Y) b SD(Y) b2 Var(Y)

  28. Pythagorean Theorem of Statistics for Independent X and Y 32 + 42 = 52 Var(X)+Var(Y)=Var(X+Y) 25=9+16 Var(X) Var(X+Y) 5 9 3 SD(X+Y) SD(X) 3 + 4 ≠ 5 SD(X)+SD(Y) ≠SD(X+Y) 4 SD(Y) 16 Var(Y)

  29. Example: meal plans • Regular plan: X = daily amount spent • E(X) = $13.50, SD(X) = $7 • Expected value and stan. dev. of total spent in 2 consecutive days? • E(X1+X2)=E(X1)+E(X2)=$13.50+$13.50=$27 SD(X1 + X2) ≠ SD(X1)+SD(X2) = $7+$7=$14

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

  31. For random variables, X+X≠2X • Let 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 • Suppose insurance policies are sold to 2 people. The annual payouts are X1 and X2. Assume the 2 people behave independently. What are the expected value and standard deviation of the total payout? • E(X1 + X2)=E(X1) + E(X2) = $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.

More Related