Chapter 16 Random Variables, Expected Value, Standard

Deviation

Random Variables and Probability Models:

Binomial, Geometric and Poisson Distributions

Graphically and Numerically Summarize a Random

ExperimentPrincipal vehicle by which we do this:random variablesA random variable assigns a number

to each outcome of an experiment

Random Variables

Definition:A random variable is a numerical-

valued function defined on the outcomes of an experiment

S Number lineRandom variable

Examples

S = {HH, TH, HT, TT}the random variable: x = # of heads in 2 tosses of a coinPossible values of x = 0, 1, 2

Two Types of Random Variables

Discrete: random variables that have a finite or countably infinite number of possible values

Test: for any given value of the random variable, you can designate the next largest or next smallest value of the random variable

Examples: Discrete rv’s

Number of girls in a 5 child familyNumber of customers that use an ATM

in a 1-hour period.Number of tosses of a fair coin that is

required until you get 3 heads in a row (note that this discrete random variable has a countably infinite number of possible values: x=3, 4, 5, 6, 7, . . .)

Two types (cont.)

Continuous: a random variable that can take on all possible values in an interval of numbers

Test: given a particular value of the random variable, you cannot designate the next largest or next smallest value

Which is it, Discrete or Continuous?

Discrete random variables “count”Continuous random variables

“measure” (length, width, height, area, volume, distance, time, etc.)

Examples: continuous rv’sThe time it takes to run the 100 yard

dash (measure)The time between arrivals at an ATM

machine (measure)Time spent waiting in line at the

“express” checkout at the grocery store (the probability is 1 that the person in front of you is buying a loaf of bread with a third party check drawn on a Hungarian bank) (measure)

Examples: cont. rv’s (cont.)

The length of a precision-engineered magnesium rod (measure)

The area of a silicon wafer for a computer chip coming off a production line (measure)

Classify as discrete or continuous

a x=the number of customers who enter a particular bank during the noon hour on a particular day

a discrete x={0, 1, 2, 3, …}b x=time (in seconds) required for a

teller to serve a bank customerb continuous x>0

Classify (cont.)

c x=the distance (in miles) between a randomly selected home in a community and the nearest pharmacy

c continuous x>0d x=the diameter of precision-

engineered “5 inch diameter” ball bearings coming off an assembly line

d continuous; range could be {4.5x<5.5}

Classify (cont.)

e x=the number of tosses of a fair coin required to observe at least 3 heads in succession

e discrete x=3, 4, 5, ...

CUSIP IND CONAME PE NPM60855410 4 MOLEX INC 24.7 8.740262810 5 GULFMARK INTL INC 21.4 8.181180410 4 SEAGATE TECHNOLOGY 21.3 2.246489010 9 ISOMEDIX INC 25.2 21.169318010 9 PCA INTERNATIONAL INC 21.4 4.726157010 7 DRESS BARN INC 24.5 4.590249410 4 TYSON FOODS INC 20.9 3.94886910 5 ATLANTIC SOUTHEAST AIRLINES 20.1 15.787183910 9 SYSTEM SOFTWARE ASSOC INC 23.7 11.662475210 4 MUELLER (PAUL) CO 14.5 3.936473510 7 GANTOS INC 15.7 1.800755P10 9 ADVANTAGE HEALTH CORP 23.3 5.323935910 2 DAWSON GEOPHYSICAL CO 14.9 9.368555910 4 ORBIT INTERNATIONAL CP 15.0 3.016278010 4 CHECK TECHNOLOGY CORP 17.1 3.251460610 4 LANCE INC 19.0 8.54523710 4 ASPECT TELECOMMUNICATIONS 25.7 8.274555310 4 PULASKI FURNITURE CORP 22.0 2.180819410 4 SCHULMAN (A.) INC 19.4 6.019770920 9 COLUMBIA HOSPITAL CORP 18.3 3.123790310 4 DATA MEASUREMENT CORP 11.3 2.611457710 4 BROOKTREE CORP 13.8 13.600431L10 9 ACCESS HEALTH MARKETING INC 22.4 11.029605610 4 ESCALADE INC 10.8 2.023303110 4 DBA SYSTEMS INC 6.3 5.064124610 4 NEUTROGENA CORP 27.2 9.059492810 6 MICROAGE INC 9.0 0.522821010 7 CROWN BOOKS CORP 24.4 1.8190710 4 AST RESEARCH INC 9.7 7.346978310 6 JACO ELECTRONICS INC 31.9 0.4531320 4 ADAC LABORATORIES 18.5 10.649766010 4 KIRSCHNER MEDICAL CORP 33.0 0.830205210 4 EXIDE ELECTRS GROUP INC 29.0 2.446065P10 5 INTERPROVINCIAL PIPE LN 11.9 19.219247910 4 COHERENT INC 40.2 1.2

CUSIP IND CONAME PE NPM60855410 4 MOLEX INC 24.7 8.740262810 5 GULFMARK INTL INC 21.4 8.181180410 4 SEAGATE TECHNOLOGY 21.3 2.246489010 9 ISOMEDIX INC 25.2 21.169318010 9 PCA INTERNATIONAL INC 21.4 4.726157010 7 DRESS BARN INC 24.5 4.5

Data Variables and Data DistributionsData Variables and Data Distributions

Data variables are

known outcomes.

CUSIP IND CONAME PE NPM60855410 4 MOLEX INC 24.7 8.740262810 5 GULFMARK INTL INC 21.4 8.181180410 4 SEAGATE TECHNOLOGY 21.3 2.246489010 9 ISOMEDIX INC 25.2 21.169318010 9 PCA INTERNATIONAL INC 21.4 4.726157010 7 DRESS BARN INC 24.5 4.590249410 4 TYSON FOODS INC 20.9 3.94886910 5 ATLANTIC SOUTHEAST AIRLINES 20.1 15.787183910 9 SYSTEM SOFTWARE ASSOC INC 23.7 11.662475210 4 MUELLER (PAUL) CO 14.5 3.936473510 7 GANTOS INC 15.7 1.800755P10 9 ADVANTAGE HEALTH CORP 23.3 5.323935910 2 DAWSON GEOPHYSICAL CO 14.9 9.368555910 4 ORBIT INTERNATIONAL CP 15.0 3.016278010 4 CHECK TECHNOLOGY CORP 17.1 3.251460610 4 LANCE INC 19.0 8.54523710 4 ASPECT TELECOMMUNICATIONS 25.7 8.274555310 4 PULASKI FURNITURE CORP 22.0 2.180819410 4 SCHULMAN (A.) INC 19.4 6.019770920 9 COLUMBIA HOSPITAL CORP 18.3 3.123790310 4 DATA MEASUREMENT CORP 11.3 2.611457710 4 BROOKTREE CORP 13.8 13.600431L10 9 ACCESS HEALTH MARKETING INC 22.4 11.029605610 4 ESCALADE INC 10.8 2.023303110 4 DBA SYSTEMS INC 6.3 5.064124610 4 NEUTROGENA CORP 27.2 9.059492810 6 MICROAGE INC 9.0 0.522821010 7 CROWN BOOKS CORP 24.4 1.8190710 4 AST RESEARCH INC 9.7 7.346978310 6 JACO ELECTRONICS INC 31.9 0.4531320 4 ADAC LABORATORIES 18.5 10.649766010 4 KIRSCHNER MEDICAL CORP 33.0 0.830205210 4 EXIDE ELECTRS GROUP INC 29.0 2.446065P10 5 INTERPROVINCIAL PIPE LN 11.9 19.219247910 4 COHERENT INC 40.2 1.2

DATA DISTRIBUTIONDATA DISTRIBUTIONPrice-Earnings RatiosPrice-Earnings Ratios

|||| ||||

Class(bin)

ClassBoundary Tally Frequency

1 6.00-12.99 |||| | 6 6/35 = 0.171

2 13.00-19.99 10

3 20.00-26.99 |||| |||| |||| 14

4 27.00-33.99 |||| 4

5 34.00-40.99 | 1 1/35 = 0.029

RelativeFrequency

4/35 = 0.114

14/35 = 0.400

10/35 = 0.286

Handout 2.1, P. 10

CUSIP IND CONAME PE NPM60855410 4 MOLEX INC 24.7 8.740262810 5 GULFMARK INTL INC 21.4 8.181180410 4 SEAGATE TECHNOLOGY 21.3 2.246489010 9 ISOMEDIX INC 25.2 21.169318010 9 PCA INTERNATIONAL INC 21.4 4.726157010 7 DRESS BARN INC 24.5 4.5

Data Variables and Data DistributonsData Variables and Data Distributons

Data variables are

known outcomes.

Data distributions

tell us what happened.

Random Variables and Probability Distributions

Random variables areunknown chance outcomes.

Probability distributionstell us what is likely

to happen.

Data variables are

known outcomes.

Data distributions

tell us what happened.

Great

Good

EconomicScenario

Profit($ Millions)

5

10

Profit Scenarios

Handout 4.1, P. 3

Random variables areunknown chance outcomes.

Probability distributionstell us what is likely

to happen.

Great

Good

OK

EconomicScenario

Profit($ Millions)

5

1

-4Lousy

10

Profit Scenarios

Probability

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

The proportion of the time an outcome is expected to happen.

Probability Distribution

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

Shows all possible values of a random variable and the probability associated with each outcome.

X = the random variable (profits)xi = outcome i

x1 = 10

x2 = 5

x3 = 1

x4 = -4

Notation

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

x4

X

x1

x2

x3

P is the probabilityp(xi)= Pr(X = xi) is the probability of X being

outcome xi

p(x1) = Pr(X = 10) = .20

p(x2) = Pr(X = 5) = .40

p(x3) = Pr(X = 1) = .25

p(x4) = Pr(X = -4) = .15

Notation

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

Pr(X=x4)

X

Pr(X=x1)

Pr(X=x2)

Pr(X=x3)

x1

x2

x3

x4

What are the chances?

What are the chances that profits will be less than $5 million in 2014?

P(X < 5) = P(X = 1 or X = -4)= P(X = 1) + P(X = -4)= .25 + .15 = .40

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

X

x1

x2

x3

x4

What are the chances?

What are the chances that profits will be less than $5 million in 2014 and less than $5 million in 2015?

P(X < 5 in 2011 and X < 5 in 2012)

= P(X < 5)·P(X < 5) = .40·.40 = .16

P(X < 5) = .40

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

p(x4)

X

x1

x2

x3

x4

P

p(x1)

p(x2)

p(x3)

.05

.10

.15

.40

.20

.25

.30

.35

Probability Probability HistogramHistogram

-4 -2 0 2 4 6 8 10 12

Profit

Probability

.05

.10

.15

.40

.20

.25

.30

.35

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

p(x4)

X

x1

x2

x3

x4

p(x1)

p(x2)

p(x3)

.05

.10

.15

.40

.20

.25

.30

.35

Probability Probability HistogramHistogram

-4 -2 0 2 4 6 8 10 12

Profit

Probability

Lousy

OK

Good

Great

.05

.10

.15

.40

.20

.25

.30

.35

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

p(x4)

X

x1

x2

x3

x4

P

p(x1)

p(x2)

p(x3)

Probability distributions: requirements

Notation: p(x)= Pr(X = x) is the probability that the random variable X has value x

Requirements1. 0 p(x) 1 for all values x of X

2. all x p(x) = 1

Example

x p(x)0 .201 .902 -.10

property 1) violated:p(2) = -.10

x p(x)-2 .3-1 .31 .32 .3

property 2) violated:p(x) = 1.2

Example (cont.)

x p(x)-1 .250 .651 .10

OK 1) satisfied: 0 p(x) 1 for all x

2) satisfied: all x p(x) = .25+.65+.10 = 1

Example: light bulbs20% of light bulbs last at least 800 hrs;

you have just purchased 2 light bulbs.X=number of the 2 bulbs that last at

least 800 hrs (possible values of x: 0, 1, 2)

Find the probability distribution of XS: bulb lasts at least 800 hrsF: bulb fails to last 800 hrsP(S) = .2; P(F) = .8

Example (cont.)

Possible outcomes P(outcome) x(S,S) (.2)(.2)=.04 2(S,F) (.2)(.8)=.16 1(F,S) (.8)(.2)=.16 1(F,F) (.8)(.8)=.64 0

probability x 0 1 2distribution of x: p(x) .64 .32 .04

S

S - SS

F - SF

S - FS

F - FF

F

Example: 3-child family

3 child family;X=#of boysM: child is male

P(M)=1/2(0.5121;

from .5134)F: child is female

P(F)=1/2(0.4879)

Outcomes P(outcome) xMMM (1/2)3=1/8 3MMF 1/8 2MFM 1/8 2FMM 1/8 2MFF 1/8 1FMF 1/8 1FFM 1/8 1FFF 1/8 0

Probability Distribution of x

x 0 1 2 3p(x) 1/8 3/8 3/8 1/8Probability of at least 1 boy:P(x 1)= 3/8 + 3/8 +1/8 = 7/8Probability of no boys or 1 boy:p(0) + p(1)= 1/8 + 3/8 = 4/8 = 1/2

Expected Value of a Discrete Random Variable

A measure of the “middle” of the values of a random variable

-4 -2 0 2 4 6 8 10 12

Profit

Probability

Lousy

OK

Good

Great

.05

.10

.15

.40

.20

.25

.30

.35Center

The mean of the probability distribution is the expected value of X, denoted E(X)

E(X) is also denoted by the Greek letter µ (mu)

k = the number of possible values (k=4)

E(x)= µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

Weighted mean

Mean orExpectedValue

k

i ii=1

( ) = x P(X=x )E 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)

k = the number of outcomes (k=4)

µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

Weighted meanEach outcome is weighted by its probability

Mean orExpectedValue

Sample MeanSample Mean

n

n

1=ii

X

= X

nx

n

1 + ... +

3x

n

1 +

2x

n

1 +

1x

n

1 =

nn

x + ... + 3

x + 2

x + 1

x = X

k

i ii=1

( ) = x P(X=x )E x

Other Weighted Means

Stock Market: The Dow Jones Industrial Average The “Dow” consists of 30 companies

(the 30 companies in the “Dow” change periodically)

To compute the Dow Jones Industrial Average, a weight proportional to the company’s “size” is assigned to each company’s stock price

k = the number of outcomes (k=4)

µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

EXAMPLE

Mean

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)

k

i ii=1

( ) = x P(X=x )E x

k = the number of outcomes (k=4)

µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

EXAMPLE

µ = 10*.20 + 5*.40 + 1*.25 – 4*.15 = 3.65 ($ mil)

Mean

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)

k

i ii=1

( ) = x P(X=x )E x

-4 -2 0 2 4 6 8 10 12

Profit

Probability

Lousy

OK

Good

Great

.05

.10

.15

.40

.20

.25

.30

.35

k = the number of outcomes (k=4)

µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... + xk·p(xk)

EXAMPLE

µ = 10·.20 + 5·.40 + 1·.25 - 4·.15 = 3.65 ($ mil)

Mean

µ=3.65

k

i ii=1

( ) = x P(X=x )E x

Interpretation

E(x) is not the value of the random variable x that you “expect” to observe if you perform the experiment once

Interpretation

E(x) is a “long run” average; if you perform the experiment many times and observe the random variable X each time, then the average x of these observed X-values will get closer to E(X) as you observe more and more values of the random variable X.

Example: Green Mountain Lottery

State of Vermontchoose 3 digits from 0 through 9;

repeats allowedwin $500

x $0 $500p(x) .999 .001

E(x)=$0(.999) + $500(.001) = $.50

Example (cont.)

E(X)=$.50On average, each ticket wins $.50.Important for Vermont to knowE(X) is not necessarily a possible

value of the random variable (values of X are $0 and $500)

Example: coin tossing

Suppose a fair coin is tossed 3 times and we let x=the number of heads. Find (x).

First we must find the probability distribution of x.

Example (cont.)

Possible values of x: 0, 1, 2, 3.p(1)?An outcome where x = 1: THTP(THT)? (½)(½)(½)=1/8How many ways can we get 1 head

in 3 tosses? 3C1=3

Example (cont.)

0 31 1 13 0 2 2 8

1 2 31 13 1 2 2 8

2 1 31 13 2 2 2 8

3 01 1 13 3 2 2 8

(0)

(1)

(2)

(3)

p C

p C

p C

p C

Example (cont.)

So the probability distribution of x is:

x 0 1 2 3p(x) 1/8 3/8 3/8 1/8

Example

1.58

12

)81(3)

83(2)

831()

81(0

4

1i)

ip(x

ixE(x)

is )μ (orE(x)

So the probability distribution of x is:

x 0 1 2 3p(x) 1/8 3/8 3/8 1/8

US Roulette Wheel and Table

The roulette wheel has alternating black and red slots numbered 1 through 36.

There are also 2 green slots numbered 0 and 00.

A bet on any one of the 38 numbers (1-36, 0, or 00) pays odds of 35:1; that is . . .

If you bet $1 on the winning number, you receive $36, so your winnings are $35

American Roulette 0 - 00(The European version has only one 0.)

US Roulette Wheel: Expected Value of a $1 bet on a single number

Let X be your winnings resulting from a $1 bet on a single number; X has 2 possible values

X -1 35p(x) 37/38 1/38

E(X)= -1(37/38)+35(1/38)= -.05So on average the house wins 5 cents on

every such bet. A “fair” game would have E(X)=0.

The roulette wheels are spinning 24/7, winning big $$ for the house, resulting in …

Standard Deviation of a Discrete Random Variable

First center (expected value)Now - spread

Standard Deviation of a Discrete Random Variable

Measures how “spread out” the random variable is

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

Variance of discrete random variable 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)

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

ExampleA basketball player shoots 3 free throws. P(make)

=P(miss)=0.5. Let X = number of free throws made.

2 2 2 2 23 31 18 8 8 8

3 31 18 8 8 8

0 1 2 3

1 3 3 1( ) E(X)

8 8 8 8

Compute the variance:

(0 1.5) (1 1.5) (2 1.5) (3 1.5)

2.25 .25 .25 2.25

.75.

.75 .866

x

p x

2 2

=1

= ( ( )) ( = )k

i ii

x E X P X x

© 2010 Pearson Education

66

Expected Value of a Random VariableExample: 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

67

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+.0

73+ .066+.057+.048=.724

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; SD(x + a)=SD(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: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)

( ) ( )Var X Var Y

( ) ( )Var X Var 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

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 a 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.

a2

c2=a2+b2

b2

Pythagorean Theorem of Statistics for Independent X and Y

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=9+16

16

Pythagorean Theorem of Statistics for Independent X and Y

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)

Example: meal plansRegular plan: X = daily amount spentE(X) = $13.50, SD(X) = $7Expected value and stan. dev. of total spent

in 2 consecutive days?E(X1+X2)=E(X1)+E(X2)=$13.50+$13.50=$27

1 2 1 2 1 2

2 2 2 2 2

( ) ( ) ( ) ( )

($7) ($7) $ 49 $ 49 $ 98 $9.90

SD X X Var X X Var X Var X

SD(X1 + X2) ≠ SD(X1)+SD(X2) = $7+$7=$14

Example: meal plans (cont.)Jumbo plan for football players Y=daily

amount spentE(Y) = $24.75, SD(Y) = $9.50Amount by which football player’s spending

exceeds regular student spending is Y-XE(Y-X)=E(Y)–E(X)=$24.75-$13.50=$11.25

2 2 2 2 2

( ) ( ) ( ) ( )

($9.50) ($7) $ 90.25 $ 49 $ 139.25 $11.80

SD Y X Var Y X Var Y Var X

SD(Y @ X) ≠ SD(Y) @ SD(X) = $9.50 @ $7=$2.50

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)=$3,867.1) 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=$400SD(2X)=2SD(X)=2*$3,867=$7,734

2)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.