Slide 1

advertisement
ARE YOU READY FOR THE QUIZ TODAY
1.
2.
3.
Yes
14%
14%
No
With the little help from a friend
14%
14%
14%
14%
14%
Slide
1- 1
1
2
3
4
5
6
7
UPCOMING IN CLASS

Quiz 3 Thursday – February 21
 (HW4
and HW5)

Sunday Homework 6

Exam 1 – March 7th
CHAPTER 16
Random Variables
EVERYTHING
  E  X    x  P  x
  Var  X     x     P  x 
2
2
  SD  X   Var  X 
Slide
1- 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- 5
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- 6
FOR THE FOLLOWING PROBLEMS,
Let,
 Mean of X = 60
 SD of X = 10
 Mean of Y = 10
 SD of Y = 2

Slide
1- 7
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- 8
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- 9
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- 10
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- 11
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- 12
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- 13
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- 14
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- 15
*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- 16
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- 17
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- 18
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- 19
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- 20
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- 21
1
2
3
4
CEREAL BOWLS

Large Bowl
E(XL)=2.7 oz
 SD(XL)=0.2 oz


Small Bowl
E(XS)=1.7 oz
 SD(XS)=0.2 oz

Slide
1- 22
HOW MUCH MORE CEREAL DO YOU EXPECT TO
EAT BY USING THE LARGE BOWL? WHAT IS
THE SD?
1.
2.
3.
4.
Mean=2.7-1.7
SD=sqrt(0.22-0.22)
Mean=2.7-1.7
SD=sqrt(0.22+0.22)
Mean=2.7+1.7
SD=sqrt(0.22-0.22)
Mean=2.7+1.7
SD=sqrt(0.22+0.22)
25%
25%
25%
25%
Slide
1- 23
1
2
3
4
ASSUMING NORMALITY, WHAT IS THE PROB. THAT
THE SMALL BOWL CONTAIN MORE CEREAL?
1.
2.
3.
4.
-3.57
-3.25
0.0002
0.9998
25%
25%
25%
25%
Slide
1- 24
1.
2.
3.
4.
WHAT ARE THE MEAN AND SD OF THE
TOTAL AMOUNT IN THE TWO BOWLS?
1.
2.
3.
4.
Mean=2.7-1.7
SD=sqrt(0.22-0.22)
Mean=2.7-1.7
SD=sqrt(0.22+0.22)
Mean=2.7+1.7
SD=sqrt(0.22-0.22)
Mean=2.7+1.7
SD=sqrt(0.22+0.22)
25%
25%
25%
25%
Slide
1- 25
1
2
3
4
ASSUMING NORMALITY, WHAT’S THE PROB. YOU
POURED MORE THAN 4.7 OZ OF CEREAL
TOTAL?
1.
2.
3.
4.
1.0714
-1.0714
.8577
.1423
25%
25%
25%
25%
Slide
1- 26
1.
2.
3.
4.
UPCOMING IN CLASS

Quiz 3 Thursday – February 21
 (HW4
and HW5)

Sunday Homework 6

Exam 1 – March 7th
Download
Related flashcards

Statistical theory

24 cards

Colossal statues

38 cards

Plotting software

51 cards

Create Flashcards