Download Random variables - NCSU Statistics

Random Variables
Streamlining Probability:
Probability Distribution,
Expected Value and Standard
Deviation of Random Variable
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
variable whose value is based on the
outcome of a random event.
Denoted by upper-case letters X, Y, etc.
When the number of possible values of X is finite
(number of heads in 3 tosses of a coin) or
countably infinite (number of tosses until you get
3 heads in a row), the random variable is
discrete. (Will study continuous rv’s later).
Examples: Discrete rv’s
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 2015 Masters
y=3, 4, 5, 6, 7
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, . . .)
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.
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
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 Distribution Of Number of
Games Played in Randomly Selected
World Series
Estimate based on results from 1946 to
2014.
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
Number of Games in Randomly
Selected World Series
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0.415
0.185
0.185
4
5
0.215
6
7
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
Expected Value of a
Discrete Random Variable
A measure of the “middle”
of the values of a random
variable
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
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
1. 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
2. GPA A=4, B=3, C=2, D=1, F=0
Five 3-hour courses: 2 A's (6 hrs), 1 B (3 hrs), 2 C's (6 hrs)
GPA:
4 * 6  3*3  2 * 6
15

45
15
 3.0
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 (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
2
3/8
3
1/8
Let X = number of heads in 3 tosses of a
fair coin.
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
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 …
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
Standard Deviation of a
Discrete Random Variable
Measures how “spread out”
the random variable is
Variance
Variation
n
s2 =
 (X
i
 X) 2
i=1
n-1
=
1805.703
= 53.1089
34
The deviations of the individual x ‘s from
the mean (expected value) of their
probability distribution:
Xi - X
xi - µ
Var(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
Var(X) = s
k
2
=
 (x  )
i
i =1
2
 P( X = xi )
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.)
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 $200 per policy
per year.
37
© 2010 Pearson Education
Standard Deviation of a Random Variable
Example: The probability model for a particular life insurance
policy is shown. Find the standard deviation of the annual payout.
38
© 2010 Pearson Education
68-95-99.7 Rule for
Random Variables
For random variables x whose probability
histograms are approximately moundshaped:
P  s  x    s  68
P  s  x    s  9
P( 3s  x    3s  997
(  s,   s) (50-5, 50+5) (45, 55)
P  s  X    s  P(45  X  55)
=.048+.057+.066+.073+.078+.08+.078+.073+
.066+.057+.048=.724
Rules for E(X), Var(X) and SD(X):
adding a constant a
If X is a rv and a is Example: a = -1
a constant:
 E(X+a) = E(X)+a
 E(X+a)=E(X-1)=E(X)-1
Rules for E(X), Var(X) and SD(X):
adding constant a (cont.)
Var(X+a) = Var(X)
SD(X+a) = SD(X)
Example: a = -1
 Var(X+a)=Var(X-1)=Var(X)
 SD(X+a)=SD(X-1)=SD(X)
Economic Profit X
Scenario ($ 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 9 am yesterday and 9 am today.
 Y=number of hours a randomly selected student from our
class was awake between 9 am yesterday and 9 am today.
Y = 24 – X.
 What are the expected value and variance of the total hours
that a student is asleep and awake between 9 am yesterday
and 9 am today?
 Total hours that a student is asleep and awake between 9
am yesterday and 9 am 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