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CHAPTER 16Standard Deviation of a
Discrete Random Variable
First center (expected value)Now - spread
Summarizing data and probability
DataHistogrammeasure of the
center: sample mean x
measure of spread:sample standard deviation s
Random variableProbability
Histogrammeasure of the
center: population mean
measure of spread: population standard deviation
Example
x 0 100p(x) 1/2 1/2
E(x) = 0(1/2) + 100(1/2) = 50
y 49 51p(y) 1/2 1/2
E(y) = 49(1/2) + 51(1/2) = 50
s =
(X X)
n - 1 =
1805.703
34 = 53.10892
i2
i=1
n
VarianceVariance
The deviations of the outcomes from the mean of the probability distribution xi - µ
2 (sigma squared) is the variance of the probability distribution
Variation
X - Xi
s =
(X X)
n - 1 =
1805.703
34 = 53.10892
i2
i=1
n
VarianceVariance
Variation
2 2
1
= ( ) ( = )=
x P X xi ii
k
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5
1
-4Lousy 0.15
10
P(X=x4)
X
x1
x2
x3
x4
P
P(X=x1)
P(X=x2)
P(X=x3)
P. 207, Handout 4.1, P. 4
Example2 = (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
Variation
3.65 3.65
3.65
3.65
2 2
1
= ( ) ( = )=
x P X xi ii
k
Standard Deviation: of More Interest then the Variance
variancepopulation theof
root square theisdeviation standard population The
Standard Deviation (s) =
Positive Square Root of the Variance
Standard DeviationStandard Deviation
s = s2
or SD, is the standard deviation of the probability distribution
Standard Deviation
(or SD) = 19.3275 4.40 ($ mil.)
2 = 19.3275
2 (or SD) =
Probability Histogram
-4 -2 0 2 4 6 8 10 12
Profit
Probability
Lousy
OK
Good
Great
.05
.10
.15
.40
.20
.25
.30
.35
µ=3.65
= 4.40
Finance and Investment Interpretation
X = return on an investment (stock, portfolio, etc.)
E(x) = expected return on this investment
is a measure of the risk of the investment
Example
866.75.
.75.
25.2 25. 25. 25.2
)5.13()5.12()5.11()5.10(
: variance theCompute8
1
8
3
8
3
8
1)(
3210
81
83
83
81
812
832
832
8122
xp
x
2 2
1
= ( ) ( = )=
x P X xi ii
k
Example (cont.)Specify the interval ()(1.5 - 1(.866), 1.5 + 1(.866))
(1.5 - .866, 1.5 + .866)
(.634, 2.366)P(.634 x 2.366) = p(1)+p(2)=3/8 + 3/8 = 3/4
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
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
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)
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5
1
-4Lousy 0.15
10
P(X=x4)
X
x1
x2
x3
x4
P
P(X=x1)
P(X=x2)
P(X=x3)
Probability
Great 0.20
Good 0.40
OK 0.25
EconomicScenario
Profit($ Millions)
5+2
1+2
-4+2Lousy 0.15
10+2
P(X=x4)
X+2
x1+2
x2+2
x3+2
x4+2
P
P(X=x1)
P(X=x2)
P(X=x3)
E(x + a) = E(x) + a; Var(x + a)=Var (x); let a = 2
Probability
0
0.1
0.2
0.3
0.4
0.5
-4 -2 0 2 4 6 8 10 12 14
Profit5.65
= 4.40Probability
0
0.1
0.2
0.3
0.4
0.5
-4 -2 0 2 4 6 8 10 12 14
Profit3.65
= 4.40
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
New Variance and SDLong (UNC-CH) way: (compute from
“scratch”)Var(X+2)=(12-5.65)2(0.20)+…
+(-2+5.65)2(0.15) = 19.3275SD(X+2) = √19.3275 = 4.40
Smart (NCSU) way:Var(X+2) = Var(X) = 19.3275SD(X+2) = SD(X) = 4.40
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)
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 + $25XE(cost) = E($100+$25X)=$100+$25E(X)=$100+$25*0.20== $100+$5=$105SD(cost)=SD($100+
$25X)=SD($25X)=$25*SD(X)=$25*0.55==$13.75
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)
( ) ( )Var X Var Y
( ) ( )Var X Var Y
In 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)
a2
c2
b2
Pythagorean Theorem of Statistics: variances add; standard deviations do not add
a
b
c
a2 + b2 = c2
Var(X)
Var(Y)
Var(X+Y)
SD(X)
SD(Y)
SD(X+Y)
Var(X)+Var(Y)=Var(X+Y)
a + b ≠ cSD(X)+SD(Y) ≠SD(X+Y)
9
25
16
Pythagorean Theorem of Statistics: variances add; standard deviations do not add
3
4
5
32 + 42 = 52
Var(X)
Var(Y)
Var(X+Y)
SD(X)
SD(Y)
SD(X+Y)
Var(X)+Var(Y)=Var(X+Y)
3 + 4 ≠ 5SD(X)+SD(Y) ≠SD(X+Y)
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-YVar(Z) = Var(X-Y) =
Var(X) + Var(Y)
Has 2 components
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=$400SD(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
1 2 1 2 1 2
2 2
SD(X + X )= ( ) ( ) ( )
(3867) (3867) 14,953,689 14,953,689
29,907,378
Var X X Var X Var X
$5,468.76
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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