Chapter 16 Random Variables

Streamlining Probability:Probability Distribution, Expected Value and Standard Deviation of

Random Variable

Graphically and Numerically Summarize a Random

ExperimentPrincipal vehicle by which we do this:random variables

Random Variables

Definition:A random variable is a numerical-

valued variable whose value is based on the outcome of a random event.

Denoted by upper-case letters X, Y, etc.

Examples

1. X = # of games played in a randomly selected World Series

Possible values of X are x=4, 5, 6, 7

2. Y=score on 13th hole (par 5) at Augusta National golf course for a randomly selected golfer on day 1 of 2011 Masters

y=3, 4, 5, 6, 7

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.

A probability distribution lists the possible values of a random variable and the probability that each value will occur.

Probability Distribution Of Number of Games Played in Randomly Selected World Series

Estimate based on results from 1946 to 2010.

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

Probability Histogram

4 5 6 70

0.1

0.2

0.3

0.4

0.185 0.1850.215

0.415

Number of Games in Randomly Selected World

Series

Probability Distribution Of Score on 13th hole (par 5) at Augusta National Golf Course on Day 1 of 2011 Masters

y 3 4 5 6 7

p(x) 0.040 0.414 0.465 0.051 0.030

3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.04

0.414

0.465

0.0510.03

Score on 13th Hole

Probability Histogram

Probability distributions: requirements

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

2. all x p(x) = 1

Expected Value of a Random Variable

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

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)

3 4 5 6 70

0.050.1

0.150.2

0.250.3

0.350.4

0.450.5

0.04

0.414

0.465

0.0510.03

Score on 13th Hole

4 5 6 70

0.1

0.2

0.3

0.4

0.5

0.185 0.185 0.215

0.415

Number of Games in Randomly Selected World

Series

k = the number of possible values of random variable

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

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

y 3 4 5 6 7

p(x) 0.040 0.414 0.465 0.051 0.030

k = the number of outcomes

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

GPA A=4, B=3, C=2, D=1, F=0

Baseball slugging percentage SLG (hr=4, 3b=3, 2b=2, 1b=1)

Baseball ticket prices Football ticket prices

Five 3-hour courses: 2 A's (6 hrs), 1 B (3 hrs), 2 C's (6 hrs)

4 * 6 3*3 2 * 6 45GPA: 3.0

15 15

4* 3*3 2*2 1*1

Babe Ruth 1920 (80 yrs): 458 AB; 54 hr, 9 3B, 36 2B, 73 1B

4*54 9*3 36*2 73*1 388.847

458 458

hr B B BSLG

AB

SLG

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414)=5.86 games

E(Y)= µ=3(.04)+4(0.414)+5(0.465)+6(0.051)+7(0.03)

=4.617 strokes

Mean orExpectedValue

k

i ii=1

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

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

y 3 4 5 6 7

p(x) 0.040 0.414 0.465 0.051 0.030

E(X)= µ =4(0.185)+5(0.185)+6(0.215)+7(0.414)

=5.86 games

Mean or Expected Value

µ=5.86

4 5 6 70

0.050.1

0.150.2

0.250.3

0.350.4

0.45

0.185 0.1850.215

0.415

Number of Games in Ran-domly Selected World Series

Interpretation

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

Interpretation of E(X)

E(X) is a “long run” average.The expected value of a random

variable is equal to the average value of the random variable if the chance process was repeated an infinite number of times. In reality, if the chance process is continually repeated, x 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)

Expected Value, Surprise Onside Kicks

http://www.advancednflstats.com/ The change in expected points for the kicking team: successful 1.9; fail -1.4.

X=change in expected points for kicking team when attempting surprise onside kick

What values of p make surprise onside kicks a good strategy?

X 1.9 -1.4

p(x) p 1-p

Expected change should be greater than 0

(1.9) ( 1.4)(1 ) 0

3.3 1.4 0

3.3 1.4

0.424

p p

p

p

p

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 Random Variable

First center (expected value)Now - spread

Standard Deviation of a Random Variable

Measures how “spread out” the random variable is

Summarizing data and probability

DataHistogrammeasure of the center: sample mean

xmeasure of spread:

sample standard deviation s

Random variableProbability Histogrammeasure of the center: population

mean mmeasure of spread: population

standard deviation s

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 random variable X

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)

= (4-5.86)2 · 0.185 + (5-5.86)2 · 0.185 +

(6-5.86)2 · 0.215 + (7-5.86)2 · 0.415 = 1.3204

Variation

5.86 5.86

5.86

5.86

2 2

1

= ( ) ( = )=

x P X xi ii

k

x 4 5 6 7

p(x) 12/65=0.185

12/65=0.185

14/65=0.215

27/65=0.415

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) = 1.3204 1.1491 games

2 = 1.3204

2 (or SD) =

© 2010 Pearson Education

33

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

34

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( - m s x + ) .68msP( - m s x + ) .95m sP( -3 m s x + 3 ) .997m s

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

073+ .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)

Carolina Panthers Next Season’s Profit

Probability

Great 0.20

Good 0.40

OK 0.25

Economy Profit X($ Millions)

5

1

-4Lousy 0.15

10

E(X)=10(0.20) + 5(0.40) + 1(0.25) – 4(0.15)

=3.65SD(X)=4.4

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5

1

-4Lousy 0.15

10

X

x1

x2

x3

x4

Probability

Great 0.20

Good 0.40

OK 0.25

EconomicScenario

Profit($ Millions)

5+2

1+2

-4+2Lousy 0.15

10+2

X+2

x1+2

x2+2

x3+2

x4+2

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

Profit=m 5.65

= 4.40Probability

0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4 6 8 10 12 14

Profit=m 3.65

= 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)=bE(X)

Var(bX) = b2Var(X)

SD(bX)= |b|SD(X) |b| denotes the

absolute value of b

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 the random variable X= season field goal shooting percentage for an NBA team. Suppose E(X)= 45.31 and SD(X)=1.67

The relationship between X and points scored per game for an NBA team can be described by 14.49 + 1.85X.

What are the mean and standard deviation of the points scored per game?

Points per game (ppg) = 14.49 + 1.85XE(ppg) = E(14.49+1.85X)=14.49+1.85E(X)=14.49+1.85*45.31=

= 14.49+83.82=98.31SD(ppg)=SD(14.49+1.85X)=SD(1.85X)=1.85*SD(X)=1.85*1.67=

=3.09

Note that the shift of 14.49 does NOT affect the standard deviation.

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 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.

a2

b2

Pythagorean Theorem of Statistics for Independent X and Y

a

b

c

Var(X)

Var(Y)

Var(X+Y)

SD(X)

SD(Y)

SD(X+Y)

a + b ≠ cSD(X)+SD(Y) ≠SD(X+Y)

c2

a2+b2=c2

Var(X)+Var(Y)=Var(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-X

E(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.