Download Lecture Unit 5

Chapter 16 Random
Variables, Expected Value,
Standard Deviation
Random Variables and
Probability Models:
Binomial, Geometric and
Poisson Distributions
Graphically and
Numerically Summarize a
Random Experiment
Principal vehicle by which we do this:
random variables
A 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
Random variable
Number line
Examples
S = {HH, TH, HT, TT}
the random variable:
x = # of heads in 2 tosses of a coin
Possible 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 family
Number 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’s
The 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 customer
b 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>0
d x=the diameter of precision-engineered “5
inch diameter” ball bearings coming off an
assembly line
d continuous; range could be {4.5<x<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, ...
Data Variables and Data Distributions
CUSIP
60855410
40262810
81180410
46489010
69318010
26157010
90249410
4886910
87183910
62475210
36473510
00755P10
23935910
68555910
16278010
51460610
4523710
74555310
80819410
19770920
23790310
11457710
00431L10
29605610
23303110
64124610
59492810
22821010
190710
46978310
531320
49766010
30205210
46065P10
19247910
IND
4
5
4
9
9
7
4
5
9
4
7
9
2
4
4
4
4
4
4
9
4
4
9
4
4
4
6
7
4
6
4
4
4
5
4
CONAME
MOLEX INC
GULFMARK INTL INC
SEAGATE TECHNOLOGY
ISOMEDIX INC
PCA INTERNATIONAL INC
DRESS BARN INC
TYSON FOODS INC
ATLANTIC SOUTHEAST AIRLINES
SYSTEM SOFTWARE ASSOC INC
MUELLER (PAUL) CO
GANTOS INC
ADVANTAGE HEALTH CORP
DAWSON GEOPHYSICAL CO
ORBIT INTERNATIONAL CP
CHECK TECHNOLOGY CORP
LANCE INC
ASPECT TELECOMMUNICATIONS
PULASKI FURNITURE CORP
SCHULMAN (A.) INC
COLUMBIA HOSPITAL CORP
DATA MEASUREMENT CORP
BROOKTREE CORP
ACCESS HEALTH MARKETING INC
ESCALADE INC
DBA SYSTEMS INC
NEUTROGENA CORP
MICROAGE INC
CROWN BOOKS CORP
AST RESEARCH INC
JACO ELECTRONICS INC
ADAC LABORATORIES
KIRSCHNER MEDICAL CORP
EXIDE ELECTRS GROUP INC
INTERPROVINCIAL PIPE LN
COHERENT INC
PE
24.7
21.4
21.3
25.2
21.4
24.5
20.9
20.1
23.7
14.5
15.7
23.3
14.9
15.0
17.1
19.0
25.7
22.0
19.4
18.3
11.3
13.8
22.4
10.8
6.3
27.2
9.0
24.4
9.7
31.9
18.5
33.0
29.0
11.9
40.2
NPM
8.7
8.1
2.2
21.1
4.7
4.5
3.9
15.7
11.6
3.9
1.8
5.3
9.3
3.0
3.2
8.5
8.2
2.1
6.0
3.1
2.6
13.6
11.0
2.0
5.0
9.0
0.5
1.8
7.3
0.4
10.6
0.8
2.4
19.2
1.2
CUSIP IND CONAME
60855410 4 MOLEX INC
40262810 5 GULFMARK INTL INC
81180410 4 SEAGATE TECHNOLOGY
46489010 9 ISOMEDIX INC
69318010 9 PCA INTERNATIONAL INC
26157010 7 DRESS BARN INC
PE NPM
24.7 8.7
21.4 8.1
21.3 2.2
25.2 21.1
21.4 4.7
24.5 4.5
Data variables are
known outcomes.
Data Variables and Data Distributons
CUSIP
60855410
40262810
81180410
46489010
69318010
26157010
90249410
4886910
87183910
62475210
36473510
00755P10
23935910
68555910
16278010
51460610
4523710
74555310
80819410
19770920
23790310
11457710
00431L10
29605610
23303110
Class
64124610
(bin)
59492810
22821010
1
190710
46978310
2
531320
49766010
3
30205210
46065P10
4
19247910
IND
CONAME
4
MOLEX INC
5
GULFMARK INTL INC
4
SEAGATE TECHNOLOGY
9
ISOMEDIX INC
9
PCA INTERNATIONAL INC
7
DRESS BARN INC
4
TYSON FOODS INC
5
ATLANTIC SOUTHEAST AIRLINES
9
SYSTEM SOFTWARE ASSOC INC
4
MUELLER (PAUL) CO
7
GANTOS INC
9
ADVANTAGE HEALTH CORP
2
DAWSON GEOPHYSICAL CO
4
ORBIT INTERNATIONAL CP
4
CHECK TECHNOLOGY CORP
4
LANCE INC
4
ASPECT TELECOMMUNICATIONS
4
PULASKI FURNITURE CORP
4
SCHULMAN (A.) INC
9
COLUMBIA HOSPITAL CORP
4
DATA MEASUREMENT CORP
4
BROOKTREE CORP
9
ACCESS HEALTH MARKETING INC
4
ESCALADE INC
4
DBA SYSTEMS INC
Class
4
NEUTROGENA
TallyCORPFrequency
Boundary
6
MICROAGE INC
76.00-12.99
CROWN BOOKS
|||| | CORP 6
4
AST RESEARCH INC
6
JACO ELECTRONICS
INC 10
13.00-19.99
|||| ||||
4
ADAC LABORATORIES
4
KIRSCHNER
CORP
20.00-26.99
|||| ||||MEDICAL
||||
14
4
EXIDE ELECTRS GROUP INC
5
INTERPROVINCIAL
PIPE LN4
27.00-33.99
||||
4
COHERENT INC
PE
NPM
24.7
8.7
21.4
8.1
21.3
2.2
25.2
21.1
21.4
4.7
24.5
4.5
20.9
3.9
20.1
15.7
23.7
11.6
14.5
3.9
15.7
1.8
23.3
5.3
14.9
9.3
15.0
3.0
17.1
3.2
19.0
8.5
25.7
8.2
22.0
2.1
19.4
6.0
18.3
3.1
11.3
2.6
13.8
13.6
22.4
11.0
10.8
2.0
6.3
5.0
Relative
27.2
9.0
Frequency
9.0
0.5
24.4= 0.1711.8
6/35
9.7
7.3
31.9= 0.2860.4
10/35
18.5
10.6
33.0
14/35 = 0.4000.8
29.0
2.4
11.9= 0.114
19.2
4/35
40.2
1.2
CUSIP IND CONAME
60855410 4 MOLEX INC
40262810 5 GULFMARK INTL INC
81180410 4 SEAGATE TECHNOLOGY
46489010 9 ISOMEDIX INC
69318010 9 PCA INTERNATIONAL INC
26157010 7 DRESS BARN INC
5
DATA DISTRIBUTION
Price-Earnings Ratios
34.00-40.99
|
1
1/35 = 0.029
PE NPM
24.7 8.7
21.4 8.1
21.3 2.2
25.2 21.1
21.4 4.7
24.5 4.5
Data variables are
known outcomes.
Data distributions
tell us what happened.
Handout 2.1, P. 10
Random Variables and
Probability Distributions
Random variables are
unknown chance
outcomes.
Probability distributions
tell us what is likely
to happen.
Data variables are
known outcomes.
Data distributions
tell us what happened.
Profit Scenarios
Economic
Scenario
Profit
($ Millions)
Great
10
Good
5
Random variables are
unknown chance
outcomes.
Probability distributions
tell us what is likely
to happen.
Handout 4.1, P. 3
Profit Scenarios
Economic
Scenario
Profit
($ Millions)
Great
10
Good
5
OK
1
Lousy
-4
Probability
Economic
Scenario
Profit
($ Millions)
Probability
Great
10
0.20
Good
5
0.40
OK
1
0.25
Lousy
-4
0.15
The proportion of the time an outcome is
expected to happen.
Probability Distribution
Economic
Scenario
Profit
($ Millions)
Probability
Great
10
0.20
Good
5
0.40
OK
1
0.25
Lousy
-4
0.15
Shows all possible values of a random
variable and the probability associated
with each outcome.
Notation
Economic
Scenario
Profit X
($ Millions)
Probability
Great
x1 10
0.20
Good
x2 5
0.40
OK
x3 1
0.25
Lousy
x4 -4
0.15
X = the random variable (profits)
xi = outcome i
x1 = 10
x2 = 5
x3 = 1
x4 = -4
Notation
Economic
Scenario
Profit X
($ Millions)
Probability
Great
x1 10
Pr(X=x1) 0.20
Good
x2 5
Pr(X=x2) 0.40
OK
x3 1
Pr(X=x3) 0.25
Lousy
Pr(X=x4) 0.15
x4 -4
P is the probability
p(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
What are the
chances?
Economic
Scenario
Profit X
($ Millions)
Probability
Great
x1 10
0.20
Good
x2 5
0.40
OK
x3 1
0.25
Lousy
x4 -4
0.15
What are the chances that profits will be
less than $5 million in 2014?
P(X < 5)
= P(X = 1 or X = -4)
= P(X = 1) + P(X = -4)
= .25 + .15 = .40
What are the
chances?
Economic
Scenario
Profit X
($ Millions)
Probability P
Great
x1 10
p(x1)
0.20
Good
x2 5
p(x2)
0.40
OK
x3 1
p(x3)
0.25
Lousy
x4 -4
p(x4)
0.15
P(X < 5) = .40
What are the chances that profits will be
less than $5 million in 2014 and less
than $5 million in 2015?
P(X < 5 in 2011 and X < 5 in 2012)
= P(X < 5)·P(X < 5) = .40·.40 = .16
Economic
Scenario
Probability
Histogram
Profit X
($ Millions)
Great
x1 10
p(x1)
0.20
Good
x2 5
p(x2)
0.40
OK
x3 1
p(x3)
0.25
Lousy
x4 -4
p(x4)
0.15
Probability
.40
.35
.30
.25
.20
.15
.10
.05
-4
-2
0
2
Probability
4
Profit
6
8
10
12
Economic
Scenario
Probability
Histogram
Probability
.40
.35
.30
.25
.20
.15
Lousy
Profit X
($ Millions)
Great
x1 10
p(x1)
0.20
Good
x2 5
p(x2)
0.40
OK
x3 1
p(x3)
0.25
Lousy
x4 -4
p(x4)
0.15
Good
OK
Great
.10
.05
-4
-2
0
2
Probability P
4
Profit
6
8
10
12
Probability distributions:
requirements
Notation: p(x)= Pr(X = x) is the probability that
the random variable X has value x
Requirements
1. 0  p(x)  1 for all values x of X
2. all x p(x) = 1
Example
x
0
1
2
p(x)
.20
.90
-.10
property 1) violated:
p(2) = -.10
x
-2
-1
1
2
p(x)
.3
.3
.3
.3
property 2) violated:
p(x) = 1.2
Example (cont.)
x
p(x)
-1
.25
0
.65
1
.10
OK 1) satisfied: 0  p(x)  1 for all x
2) satisfied: all x p(x) = .25+.65+.10 = 1
Example: light bulbs
20% 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 X
S: bulb lasts at least 800 hrs
F: bulb fails to last 800 hrs
P(S) = .2; P(F) = .8
S
F
S -
SS
F -
SF
S -
FS
F -
FF
Example (cont.)
Possible outcomes
(S,S)
(S,F)
(F,S)
(F,F)
probability
distribution of x:
P(outcome)
(.2)(.2)=.04
(.2)(.8)=.16
(.8)(.2)=.16
(.8)(.8)=.64
x
0
1
p(x) .64 .32
x
2
1
1
0
2
.04
Example: 3-child family
Outcomes P(outcome)
MMM
(1/2)3=1/8
3 child family;
MMF
1/8
X=#of boys
MFM
1/8
M: child is male
P(M)=1/2
FMM
1/8
(0.5121; from .5134)MFF
1/8
F: child is female
FMF
1/8
P(F)=1/2
FFM
1/8
(0.4879)
FFF
1/8
x
3
2
2
2
1
1
1
0
Probability Distribution of
x
x
0
1
2
3
p(x)
1/8 3/8 3/8 1/8
Probability of at least 1 boy:
P(x  1)= 3/8 + 3/8 +1/8 = 7/8
Probability 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
Center
Probability
.40
.35
.30
.25
.20
.15
Lousy
Good
OK
Great
.10
.05
-4
-2
0
2
4
6
8
10
12
Profit
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)
Economic
Scenario
Mean or
Expected
Value
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of possible values (k=4)
E ( x)   =
k
x
i
 P(X=x i )
i=1
E(x)= µ = x1·p(x1) + x2·p(x2) + x3·p(x3) +
... + xk·p(xk)
Weighted mean
Sample Mean
Mean or
Expected
Value
X
=
n
X

i
i = 1
n
x +x +x +...+x
n
X= 1 2 3
n
1
1
1
1
= x + x + x +...+ x
n 1 n 2 n 3
n n
k = the number of outcomes (k=4)
E ( x)   =
k
x
i
 P(X=x i )
i=1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
Weighted mean
Each outcome is weighted by its probability
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
Economic
Scenario
Mean
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of outcomes (k=4)
E ( x)   =
k
x
i
 P(X=x i )
i=1
µ = x1·p(x1) + x2·p(x2) + x3·p(x3) + ... +
xk·p(xk)
EXAMPLE
Economic
Scenario
Mean
Profit X
($ Millions)
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k = the number of outcomes (k=4)
E ( x)   =
k
x
i
Probability P
 P(X=x i )
i=1
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)
Probability
Mean
.40
.35
.30
.25
.20
.15
Lousy
Good
OK
Great
.10
.05
-4
-2
0
2
4
6
8
10
12
Profit
µ=3.65
k = the number of outcomes (k=4)
E ( x)   =
k
x
i
 P(X=x i )
i=1
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)
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 Xvalues will get closer to E(X) as you
observe more and more values of the
random variable X.
Example: Green Mountain
Lottery
State of Vermont
choose 3 digits from 0 through 9; repeats
allowed
win $500
x
$0
$500
p(x)
.999
.001
E(x)=$0(.999) + $500(.001) = $.50
Example (cont.)
E(X)=$.50
On average, each ticket wins $.50.
Important for Vermont to know
E(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
E(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: THT
P(THT)? (½)(½)(½)=1/8
How many ways can we get 1 head in 3
tosses? 3C1=3
Example (cont.)
p(0)  3 C0 
 
1 1 1 2
p(1)  3 C1  2   2   83
1 2 1 1
p(2)  3 C2  2   2   83
1 3 1 0
p(3)  3 C3  2   2   81
1 0
2
1 3
2
 81
Example (cont.)
So the probability distribution of x is:
x
p(x)
0
1/8
1
3/8
2
3/8
3
1/8
 So the probability distribution of x is:
Example
x
p(x)
0
1/8
1
3/8
E(x) (or μ ) is
E(x)
4
  x  p(x )
i
i
i 1
 (0  1 )  ( 1 3 )  (2  3 )  (3  1 )
8
8
8
8
 12  1.5
8
2
3/8
3
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
p(x)
-1
37/38
35
1/38
E(X)= -1(37/38)+35(1/38)= -.05
So 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
Data
Histogram
measure of the
center: sample mean
x
measure of spread:
sample standard
deviation s
Random variable
Probability Histogram
measure of the
center: population
mean 
measure of spread:
population standard
deviation s
Example
x
0
100
p(x)
1/2 1/2
E(x) = 0(1/2) + 100(1/2) = 50
y
49 51
p(y)
1/2 1/2
E(y) = 49(1/2) + 51(1/2) = 50
Variance
Variation
n
s2 =
 (X
i
 X) 2
i=1
n-1
=
1805.703
= 53.1089
34
The deviations of the outcomes from the
mean of the probability distribution
xi - µ
Xi - X
s2 (sigma squared) is the variance of the
probability distribution
Variance
Variation
n
s2 =
 (X
i
 X) 2
i=1
n-1
=
1805.703
= 53.1089
34
Variance of discrete random variable X
s
k
2
=
 (x
i =1
  )  P( X = x i )
2
i
Economic
Scenario
Variation
s2
Profit X
($ Millions)
Probability P
Great
x1 10
P(X=x1) 0.20
Good
x2 5
P(X=x2) 0.40
OK
x3 1
P(X=x3) 0.25
Lousy
x4 -4
P(X=x4) 0.15
k
=
2
(
x


)
 P( X = x i )
 i
i =1
Example 3.65
3.65
3.65
s2 = (x1-µ)2 · P(X=x1) + (x2-µ)2 · P(X=x2) +
3.65
(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
P. 207, Handout 4.1, P. 4
Standard Deviation: of
More Interest then the
Variance
The population standard deviation is the square root
of the population variance
s  s

Standard Deviation
Standard
Deviation
Standard Deviation (s) =
Positive Square Root of the Variance
s =
s2
s2 = 19.3275
s, or SD, is the standard deviation of the
probability distribution
s (or SD) = s
2
s (or SD) = 19.3275  4.40 ($ mil.)
Probability Histogram
Probability
.40
.35
.30
.25
.20
.15
Lousy
s = 4.40
Good
OK
Great
.10
.05
-4
-2
0
2
4
Profit
µ=3.65
6
8
10
12
Finance and Investment
Interpretation
X = return on an investment (stock,
portfolio, etc.)
E(X) =   expected return on this
investment
s is a measure of the risk of the
investment
s
k
2
Example
=
2
(
x

E
(
X
))
 P ( X = xi )
 i
i =1
A basketball player shoots 3 free throws. P(make)
=P(miss)=0.5. Let X = number of free throws made.
x
0
1
2
3
1
3
3
1
8
8
8
Compute the variance:
8
p( x)
E(X) 
s 2  (0  1.5) 2  18   (1  1.5) 2  83   (2  1.5) 2  83   (3  1.5) 2  18 
 2.25  18   .25  83   .25  83   2.25  81 
 .75.
s  s


.75  .866
Expected Value of a Random Variable
Example: 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 $200 per policy
per year.
66
© 2010 Pearson Education
Standard Deviation of a Random Variable
Example: The probability model for a particular life insurance
policy is shown. Find the standard deviation of the annual payout.
67
© 2010 Pearson Education
68-95-99.7 Rule for
Random Variables
For random variables X whose probability
histograms are approximately moundshaped:
P  s  X    s  68
P  s  X    s  9
P( 3s  X    3s  997
(  s,   s) (50-5, 50+5) (45, 55)
P  s  X    s  P(45  X  55)
=.048+.057+.066+.073+.078+.08+.078+.073+
.066+.057+.048=.724
Rules for E(X), Var(X) and SD(X):
adding a constant a
If X is a rv and a is Example: a = -1
a constant:
 E(X+a) = E(X)+a
 E(X+a)=E(X-1)=E(X)-1
Rules for E(X), Var(X) and SD(X):
adding constant a (cont.)
Var(X+a) = Var(X)
SD(X+a) = SD(X)
Example: a = -1
 Var(X+a)=Var(X-1)=Var(X)
 SD(X+a)=SD(X-1)=SD(X)
Economic Profit X
Scenario ($ Millions)
Probability P
Economic Profit X+2
Scenario ($ Millions)
Great
x1 10
P(X=x1) 0.20
Great
Good
x2 5
P(X=x2) 0.40
OK
x3 1
Lousy
x4 -4
Probability P
P(X=x1) 0.20
Good
x1+ 10+2
2
x2+2 5+2
P(X=x3) 0.25
OK
x3+2 1+2
P(X=x3) 0.25
P(X=x4) 0.15
Lousy
x4+2 -4+2
P(X=x4) 0.15
P(X=x2) 0.40
E(x + a) = E(x) + a; SD(x + a)=SD(x); let a = 2
s = 4.40
Probability
-4
-2
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0
2
4
Profit
3.65
6
8
10
12
14
s = 4.40
Probability
-4
-2
0
2
4
6
8
Profit
5.65
10
12
14
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 SD
Long (UNC-CH) way: (compute from
“scratch”)
Var(X+2)=(12-5.65)2(0.20)+…
+(-2+5.65)2(0.15) = 19.3275
SD(X+2) = √19.3275 = 4.40
Smart (NCSU) way:
Var(X+2) = Var(X) = 19.3275
SD(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 + $25X
E(cost) = E($100+$25X)=$100+$25E(X)=$100+$25*0.20=
= $100+$5=$105
SD(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)= Var ( X )  Var (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)= Var ( X )  Var (Y )
SD’s do not subtract:
SD(X−Y)≠ SD(X)−SD(Y)
SD(X−Y)≠ SD(X)+SD(Y)
Motivation for
Var(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-Y
Has 2 +
components
Var(Y)
 Var(Z) = Var(X-Y) = Var(X)
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.
Pythagorean Theorem of Statistics
for Independent X and Y
a2 + b2 = c 2
Var(X)+Var(Y)=Var(X+Y)
c2=a2+b2
Var(X)
a2
Var(X+Y)
a
c
SD(X+Y)
SD(X)
b
SD(Y)
b2
Var(Y)
a+b≠c
SD(X)+SD(Y) ≠SD(X+Y)
Pythagorean Theorem of Statistics
for Independent X and Y
32 + 42 = 52
Var(X)+Var(Y)=Var(X+Y)
25=9+16
Var(X)
9
Var(X+Y)
3
5
SD(X+Y)
SD(X)
4
SD(Y)
16
Var(Y)
3+4≠5
SD(X)+SD(Y) ≠SD(X+Y)
Example: meal plans
Regular plan: X = daily amount spent
E(X) = $13.50, SD(X) = $7
Expected value and stan. dev. of total spent in
2 consecutive days?
E(X
+X
)=E(X
)+E(X
)=$13.50+$13.50=$27
1
2
1
2
SD(X + X ) ≠ SD(X )+SD(X ) = $7+$7=$14
1
2
1
2
SD( X 1  X 2 )  Var ( X 1  X 2 )  Var ( X 1 )  Var ( X 2 )
 ($7)  ($7)  $ 49  $ 49  $ 98  $9.90
2
2
2
2
2
Example: meal plans (cont.)
Jumbo plan for football players Y=daily
amount spent
E(Y) = $24.75, SD(Y) = $9.50
Amount 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
SD(Y ̶ X) ≠ SD(Y) ̶ SD(X) = $9.50 ̶ $7=$2.50
SD(Y  X )  Var (Y  X )  Var (Y )  Var ( X )
 ($9.50)  ($7)  $ 90.25  $ 49  $ 139.25  $11.80
2
2
2
2
2
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?
The risk to the
 Double payout is 2X. E(2X)=2E(X)=2*$200=$400
insurance co. when
 SD(2X)=2SD(X)=2*$3,867=$7,734 doubling the payout
is notThe
the same
2) Suppose insurance policies are sold to 2 (2X)
people.
as 2
thepeople
risk when
annual payouts are X1 and X2. Assume the
selling policies
behave independently. What are the expected
value to 2
people.
and standard deviation of the total payout?
 E(X1 + X2)=E(X1) + E(X2) = $200 + $200 = $400
SD(X1 + X2 )= Var ( X1  X 2 )  Var ( X1 )  Var ( X 2 )
 (3867)2  (3867)2  14,953,689  14,953,689
 29,907,378  $5,468.76