HAVE YOU STARTED THE HOMEWORK
1.
2.
3.
Yes
33%
No
With a little help from a friend
33%
33%
Slide
1- 1
1
2
3
UPCOMING IN CLASS

Sunday Homework 5

Quiz 3 – Sept. 25th

Exam 1 – October 9th
CHAPTER 14
Random Variables
EVERYTHING
  E  X    x  P  x
  Var  X     x     P  x 
2
2
  SD  X   Var  X 
Slide
1- 4
ORANGE PROBLEM
A citrus farmer has observed the following
distribution for the number of oranges per tree.
 P(30 oranges )=0.10
 P(35 oranges) =0.40
 P(40 oranges) =0.30
 P(45 oranges) = 0.20
 The farmer expects 38 oranges per tree on
average

Slide
1- 5
WHAT IS THE STANDARD DEVIATION?
1.
2.
3.
4.
38
21
4.58
0
25%
25%
25%
25%
Slide
1- 6
1
2
3.
4
INSURANCE PROBLEM
An insurance policy costs $150 and will pay
policyholder $12,000 if they suffer a major injury
or $2,500 if they suffer a minor injury.
 The company estimates that 1 in every 1,984
policyholders will suffer major injuries and that 1
in every 469 will suffer minor injuries.

Slide
1- 7
WHAT IS THE RANDOM VARIABLE?
1.
2.
3.
4.
Profit
25%
25%
Number of policy holders
Number of injuries
Number of policy holders with injuries
25%
25%
Slide
1- 8
1
2
3
4
INSURANCE PROBLEM

Find the probability model for profit.
Slide
1- 9
FIND THE EXPECTED VALUE
1.
2.
3.
4.
138.22
138.62
160.59
160.99
25%
25%
25%
25%
Slide
1- 10
1.
2.
3.
4.
FIND THE STANDARD DEVIATION.
1.
2.
3.
4.
0.022
282.29
292.88
85,777
25%
25%
25%
25%
Slide
1- 11
1.
2.
3.
4
MORE ABOUT MEANS AND VARIANCES


Adding or subtracting a constant from data
E(X ± c) = E(X) ± c Var(X ± c) = Var(X)
Multiplying each value of a random variable by a
constant multiplies the mean by that constant
and the variance by the square of the constant:
E(aX) = aE(X)
Var(aX) = a2Var(X)
Slide
1- 12
MORE ABOUT MEANS AND VARIANCES
(CONT.)

In general,
The mean of the sum of two random variables is the
sum of the means.
 The mean of the difference of two random variables is
the difference of the means.

E(X ± Y) = E(X) ± E(Y)

If the random variables are independent, the
variance of their sum or difference is always the sum
of the variances.
Var(X ± Y) = Var(X) + Var(Y)
Slide
1- 13
FOR THE FOLLOWING PROBLEMS,
Let,
 Mean of X = 60
 SD of X = 10
 Mean of Y = 10
 SD of Y = 2

Slide
1- 14
FIND THE MEAN AND SD FOR THE
RANDOM VARIABLE 4X+2Y
1.
2.
3.
4.
4*60+2*10 and 4*10
4*60+2*10 and 2*2
4*60+2*10 and 4*10+2*2
4*60+2*10 and sqrt(16*100+4*4)
0%
0%
0%
0%Slide
1- 15
1
2
3
4
FIND THE MEAN AND SD FOR THE
RANDOM VARIABLE 2X-5Y
1.
2.
3.
4.
2*60-5*10 and sqrt(2*10-5*2)
2*60-5*10 and sqrt(4*10-25*2)
2*60-5*10 and sqrt(4*100-25*4)
2*60-5*10 and sqrt(4*100+25*4)
0%
0%
0%
0%Slide
1- 16
1
2
3
4
FIND THE MEAN AND SD FOR THE
RANDOM VARIABLE X1+X2
1.
2.
3.
4.
60+60 and sqrt(100+100)
60+60 and sqrt(10+10)
60+60 and 100+100
60+60 and 10+10
0%
0%
0%
0%Slide
1- 17
1
2
3
4
EGG EXAMPLE
A grocery supplier believes that in a dozen eggs,
the mean number of broken eggs is 0.5 with a SD
of 0.2 eggs.
 You buy 3 dozen eggs.

Slide
1- 18
HOW MANY BROKEN EGGS DO YOU EXPECT
IN THE THREE CARTONS?
1.
2.
3.
4.
0.5+0.5+0.5
0.5*0.5*0.5
0.5+0.2
I need a probability model to answer this
question
25%
25%
25%
25%
Slide
1- 19
1
2
3
4
WHAT’S THE SD OF THE NUMBER OF BROKEN
EGG WHEN BUYING THREE CARTONS?
1.
2.
3.
4.
0.2*0.2*0.2
0.2+0.2+0.2
Sqrt(0.2+0.2+0.2)
Sqrt(0.22+0.22+0.22)
25%
25%
25%
25%
Slide
1- 20
1
2
3
4
WHAT ASSUMPTION DID YOU MAKE ABOUT
THE EGGS?
1.
2.
3.
4.
The cartons are dependent
The number of broken eggs is a continuous
random variable.
The number of broken eggs is a discrete random
variable.
The cartons are independent of each other.
25%
25%
25%
25%
Slide
1- 21
1
2
3
4
*CORRELATION AND COVARIANCE

If X is a random variable with expected value E(X)=µ
and Y is a random variable with expected value E(Y)=ν,
then the covariance of X and Y is defined as
Cov( X , Y )  (( X   )(Y  ))

The covariance measures how X and Y vary together.
Slide
1- 22
*CORRELATION AND COVARIANCE (CONT.)
Covariance, unlike correlation, doesn’t have to be
between -1 and 1. If X and Y have large values,
the covariance will be large as well.
 To fix the “problem” we can divide the covariance
by each of the standard deviations to get the
correlation:

Corr ( X , Y ) 
Cov( X , Y )
 XY
Slide
1- 23
INSURANCE POLICIES

An insurance company estimates that it
should make an annual profit of $130 on
each homeowner’s policy, with a standard
deviation of $4,000.
Slide
1- 24
WHAT IS THE MEAN AND SD IF THE
COMPANY WRITE 3 POLICIES?
1.
2.
3.
4.
Mean=3*130
SD=sqrt(3*4,000)
Mean=130+130+130
SD=sqrt(3*4,0002)
Mean=3*130
SD=3*4,000
Mean=130+130+130
SD=3*4,0002
25%
25%
25%
25%
Slide
1- 25
1
2
3
4
WHAT IS THE MEAN AND SD IF THE
COMPANY WRITE 10,000 POLICIES?
1.
2.
3.
4.
Mean=10,000*130
SD=sqrt(10,000*4,0002)
Mean=130+130+130
SD=sqrt(10,000*4,0002)
Mean=10,000*130
SD=10,000*4,000
Mean=130+130+130
SD=10,000*4,000
25%
25%
25%
25%
Slide
1- 26
1
2
3
4
WHAT ASSUMPTION DID YOU MAKE?
1.
2.
3.
4.
The annual profit on a policy is a continuous
random variable
Losses are dependent
Losses are independent of each other
The annual profit on a policy is a discrete
random variable.
25%
25%
25%
25%
Slide
1- 27
1
2
3
4
WILL THE COMPANY BE PROFITABLE?
1.
2.
3.
4.
No. The variance is larger than the mean.
Yes, $0 is 3.25 standard deviations below the
mean for 10,000 policies
Yes, the expected value is greater than zero.
No. Catastrophes are far too unpredictable to
expect a profit.
25%
25%
25%
25%
Slide
1- 28
1
2
3
4
UPCOMING IN CLASS

Sunday Homework 5

Quiz 3 – Sept. 25th

Exam 1 – October 9th