Download STAT 155 Introductory Statistics Lecture 14: Means and Variances of

The UNIVERSITY of NORTH CAROLINA
at CHAPEL HILL
STAT 155 Introductory Statistics
Lecture 14: Means and Variances of
Random Variables
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Lecture 14
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Means & Variances of Random Variables
• Data
– Sample mean: x
– Sample variance:
s
2
• Random variable X
– Mean: E(X) =
µX
2
σ
– Variance: Var (X) = X
E(X) and Var (X) are summaries of the probability
distribution of X.
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Mean of a Discrete Random Variable (definition)
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Free 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).
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Free-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)
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Rules for Expected Value
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Variance of a Discrete Random Variable (definition)
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Car Sales
• The total number X of cars to be sold next week is
described by the following probability distribution
x
p(x)
0
1
.05 .15
2
.35
3
.25
4
.20
• Find E(X) and Var (X).
5
µ X = ∑ xi p ( xi ) = 0(0.05) + 1(0.15) + 2(0.35) + 3(0.25) + 4(0.20) = 2.40
i =1
5
σ X = ∑ ( xi − 2.4) 2 p( xi ) = (0 − 2.4) 2 (.05) + (1 − 2.4) 2 (.15)
2
i =1
+ (2 − 2.4) 2 (.35) + (3 − 2.4) 2 (.25) + (4 − 2.4) 2 (.20) = 1.24
σ X = 1.24 = 1.11
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Car Salesman’s Commission (continued)
• 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) = 630
Var (Y) = (200)2 Var (X) = 40000 (1.24) = 49600
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Rules for Variances
• If X and Y are independent random variables and a
and b are constants, then
σ
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2
aX + bY
= a σ +b σ .
2
2
X
Lecture 14
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2
Y
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Rules for Variances
• If X and Y are random variables with correlation
ρ , then
2
2 2
2 2
σ aX
=
a
σ
+
b
σ Y + 2abρσ X σ Y .
X
+ bY
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Women’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?
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Women’s Height (continued)
• µX-Y = µX - µY = 64.3 – 64.3 = 0
• σ2X-Y = σ2X + σ2Y = 2.42 + 2.42 = 11.52, so σX-Y = 3.39.
• Need to find x1 & x2 so 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.
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Investment 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
µ X = 5 . 2 %, σ X = 2 . 9 %, µ Y = 13 . 3 %, σ Y = 17 . 0 %, ρ = − 0 . 1 .
• What are the mean and variance of the portfolio
return?
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Investment Portfolio (continued)
• Portfolio return R = 0.2 X + 0.8 Y
E(R) = 0.2 E(X) + 0.8 E(Y)
= 0.2 (5.2%) + 0.8 (13.3%) = …
Var (R) = (0.2)2 σ2X + (0.8)2 σ2Y
+ 2 (0.2) (0.8) corr (X,Y) σX σY
= 0.04 (2.9%)2 + 0.64 (17%)2
+ 2 (0.2) (0.8) (- 0.1) (2.9%) (17%)
=…
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Take 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 !
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