5/28/10 Lecture 13 1STOR 155 Introductory StatisticsLecture 13: Means and Variances of Random VariablesThe UNIVERSITY of NORTH CAROLINAat CHAPEL HILL5/28/10 Lecture 13 2Means & Variances of Random Variables• Data– Sample mean: – Sample variance:• Random variable X– Mean: E(X) =– Variance: Var (X) =E(X) and Var (X) are summaries of the probability distribution of X.x2s2XX5/28/10 Lecture 13 3Mean of a Discrete Random Variable (definition)5/28/10 Lecture 13 4Free Throws• A Tar Heel basketball player is a 80% free throw shooter.• Suppose he shoots 10 free throws during each practice.• X: number of free throws he makes in a practice.• Find E(X).5/28/10 Lecture 13 5Free-throws (continued)• E(X) = 0 P(X=0) + 1 P(X=1) + 2 P(X=2) +… + 10 P(X=10) = … (by definition)• Need to compute P(X=x) for x=0,1,…, 10.• Any better method ?• Intuition: E(X) = 10 (0.8) = 8• In fact, it’s correct! Why?• Some useful rules create shortcuts …(detail later)5/28/10 Lecture 13 6Rules for Expected Value5/28/10 Lecture 13 7Variance of a Discrete Random Variable (definition)5/28/10 Lecture 13 8• The total number X of cars to be sold next week is described by the following probability distribution• Find E(X) and Var (X).x 0 1 2 3 4p(x) .05 .15 .35 .25 .20Car Sales11.124.124.1)20(.)4.24()25(.)4.23()35(.)4.22( )15(.)4.21()05(.)4.20()()4.2(40.2)20.0(4)25.0(3)35.0(2)15.0(1)05.0(0)(22222512251XiiiXiiiXxpxxpx5/28/10 Lecture 13 95/28/10 Lecture 13 10• Suppose a salesman earns a fixed weekly wage of $150 plus $200 commission for each car sold. • Let Y be his weekly earnings = wage + commission• Find E(Y) and Var (Y).• Note: Y = 150 + 200 X• E(Y) = 150 + 200 E(X) = 150 + 200 (2.4) = 630Var (Y) = (200)2 Var (X) = 40000 (1.24) = 49600Car Salesman’s Commission (continued)5/28/10 Lecture 13 11• If X and Y are independent random variables and aand b are constants, then--- This is a special case!Rules for Variances.22222YXbYaXba5/28/10 Lecture 13 12• If X and Y are random variables with correlation , then--- This is a general rule!Rules for Variances.222222YXYXbYaXabba5/28/10 Lecture 13 13Women’s Height• The height of American women aged 18 – 24 is approximately normally distributed with mean 64.3 inches and s.d. 2.4 inches. Two women in the age group are randomly selected. • What is the interquartile range of the heights of women in the age group? (done before)• What are the mean and the s.d. of the height difference between the two women?5/28/10 Lecture 13 14Women’s Height (continued)•X-Y= X-Y = 64.3 – 64.3 = 0•2X-Y = 2X + 2Y = 2.42+ 2.42= 11.52, so X-Y = 3.39.• Need to find x1& x2so that P (X < x1) = .25, P(X > x2) = .25.• .25 = P(X < x1) = P(Z < (x1– 64.3)/2.4). From the table,P(Z < -0.67) = .25. So, (x1– 64.3)/2.4 = -0.67, i.e., x1= 62.69.• Similarly, .25 = P(X > x2) = P(Z > (x2– 64.3)/2.4). P(Z > 0.67) = .25. So, (x2– 64.3)/2.4 = 0.67, i.e., x2= 65.91.So, IQR = 65.91 – 62.69 = 3.22 in.5/28/10 Lecture 13 15Investment Portfolio• Zadie has invested 20% of her funds in Treasury Bills and 80% in an “index fund” that consists of many U.S. common stocks. • Let X be the annual return on T-bills and Y the annual return on the index fund• Suppose• What are the mean and variance of the portfolio return? .1.0%,0.17%,3.13%,9.2%,2.5 YYXX5/28/10 Lecture 13 16Investment Portfolio (continued)• Portfolio return R = 0.2 X + 0.8 YE(R) = 0.2 E(X) + 0.8 E(Y)= 0.2 (5.2%) + 0.8 (13.3%) = …Var (R) = (0.2)22X+ (0.8)22Y+ 2 (0.2) (0.8) corr (X,Y)XY= 0.04 (2.9%)2 + 0.64 (17%)2+ 2 (0.2) (0.8) (- 0.1) (2.9%) (17%)= …5/28/10 Lecture 13 17Take Home Message• Definition of mean E(X) and variance Var (X) for random variable X• How to compute E(X) and Var (X) ?(1) Use definition (for discrete RV’s)(2) Use appropriate rules (for discrete and continuous RV’s) --- a more flexible and extensively used method
View Full Document
Unlocking...