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HOW MUCH SLEEP DID YOU GET?
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UPCOMING IN CLASS

Sunday Homework 5

Exam 1 – October 9th
CHAPTER 14.1-14.2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson
Addison-Wesley
Random Variables
THEORY OF A RANDOM VARIABLE
  E  X    x  P  x
  Var  X     x     P  x 
2
2
  SD  X   Var  X 
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RANDOM VARIABLES

A random variable is a variable whose number
value is based on the outcome of a random event.
We use a capital letter, like X, to denote a random
variable.
 A particular value of a random variable will be
denoted with a lower case letter, in this case x.

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TYPE OF RANDOM VARIABLES

There are two types of random variables:

Discrete random variables can take one of a finite
number of distinct outcomes.


Example: Number of credit hours
Continuous random variables can take any numeric
value within a range of values.

Example: Cost of books this term
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EXPECTED VALUE: CENTER

A probability model for a random variable
consists of:
The collection of all possible values of a random
variable, and
 the probabilities that the values occur.


Of particular interest is the value we expect a
random variable to take on, notated μ (for
population mean) or E(X) for expected value.
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EXPECTED VALUE: CENTER (CONT.)

The expected value of a (discrete) random
variable can be found by summing the products
of each possible value and the probability that
it occurs:
  E  X    x  P  x

Note: Be sure that every possible outcome is
included in the sum and verify that you have a
valid probability model to start with.
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PLAYING CARDS
You draw a card from a deck.
 If you get a red card, you win nothing.
 If you get a spade, you win $7
 For any club, you get $10 plus an extra $20 for
the Ace of clubs.

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CREATE A PROBABILITY MODEL FOR THE
AMOUNT YOU WIN AT THIS GAME
1.
2.
3.
Red 26/52,
Spade 13/52,
Ace Club 1/52,
Other Club 12/52
Red 26/52,
Spade 13/52,
Ace Club 1/52,
Other Club 13/52
Red 26/52,
Spade 13/52,
Ace Club 1/52,
Other Club 26/52
33%
33%
33%
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1
2
3
FIND THE EXPECTED AMOUNT THAT YOU
WIN.
1.
2.
3.
½ * 0 + ¼ * 7 + 1/52 * 30 + ¼ * 10
½ * 0 + ¼ * 7 + 1/52 * 30 + 3/13 * 10
½ * 0 + ¼ * 7 + 1/52 * 10 + 3/13 * 10
0%
1
0%
2
0%
3
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HOW MUCH WOULD YOU PAY TO PLAY IF
YOU WERE CONTINUALLY PLAYING?
1.
2.
3.
4.
5.
6.
$0
$1
$2
$3
$4
$5
17%
17%
17%
17%
17%
17%
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1
2
3
4
5
6
HOW MUCH SHOULD YOU PAY TO PLAY IF
YOU WERE PLAYING A FEW TIMES?
1.
2.
3.
4.
5.
6.
$0
$1
$2
$3
$4
$5
17%
17%
17%
17%
17%
17%
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2
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HOW MUCH SHOULD YOU PAY TO PLAY?
1.
2.
3.
4.
At least the expected value, and much more
than that if planning to play many times
No more than the expected value, and much
less than that if planning to play only a few
times
No more than the expected value, and much
less than that if planning to play many times
At least the expected value, and much more
than that if planning to play only a few times
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NOT JUST FOR PAYOFFS

The expected value of a (discrete) random
variable can be found by summing the products
of each possible value and the probability that
it occurs:
  E  X    x  P  x

Note: Be sure that every possible outcome is
included in the sum and verify that you have a
valid probability model to start with.
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THE PROBLEM WITH CHILDREN


A couple plans to have children until they get a
girl, but they agree they will not have more than
three children, even if all are boys.
Assume that the probability of having a girl is
0.49
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CREATE A PROBABILITY MODEL FOR THE
NUMBER OF CHILDREN THEY’LL HAVE
1.
2.
3.
4.
Zero .49; One .49; Two .49; Three .49;
One .49, Two .49, Three .49
One .49, Two .51*.49, Three .51*.51*.49
One .49, Two .51*.49, Three (.51*.51*.49+.51*.51*.51)
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0%
1
0%
2
0%
3
0%
4
FIND THE EXPECTED NUMBER OF
CHILDREN
1.
2.
3.
1*.49 + 2 *.49 +3 *.49
1*.49 + 2*.51*.49 + 3 *.51*.51*.49
1*.49 + 2*.51*.49 + 3 *.51*.51*.49 +3*.51*.51*.51
0%
1
0%
2
0%
3
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FIND THE EXPECTED NUMBER OF BOYS
1.
2.
3.
1*.49 + 2 *.51*.49 + 3*.51*.51*.49 + 3*.51*.51*.51
0*.49 + 1 *.51*.49 + 2*.51*.51*.49 + 3*.51*.51
0*.49 + 1 *.51*.49 + 2*.51*.51*.49
0%
1
0%
2
0%
3
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FIRST CENTER, NOW SPREAD…
For data, we calculated the standard deviation by first
computing the deviation from the mean and squaring
it. We do that with discrete random variables as well.
 The variance for a random variable is:

  Var  X     x     P  x 
2

2
The standard deviation for a random variable is:
  SD  X   Var  X 
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FIND THE STANDARD DEVIATION OF THE
RANDOM VARIABLE X
If x= 43, then P(X=x)=0.4
 If x= 30, then P(X=x)=0.3
 If x= 34, then P(X=x)=0.2
 If x= 46, then P(X=x)=0.1

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PLAYING CARDS
You draw a card from a deck.
 If you get a red card, you win nothing.
 If you get a spade, you win $6.
 For any club, you win $12 plus an extra $5 for
the ace of clubs.

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FIND THE STANDARD DEVIATION OF THE
AMOUNT YOU MIGHT WIN DRAWING A CARD.
1.
2.
3.
4.
(E(x) -0)*1/2 + (E(x) -6)*1/4 + (E(x) -12)*3/13 +(E(x) 17)*1/52
(E(x) -0)2 *1/2 + (E(x) -6) 2*1/4 + (E(x) -12) 2*3/13
+(E(x) -17) 2*1/52
Sqrt ((E(x) -0)*1/2 + (E(x) -6) *1/4 +
(E(x) - 12)*3/13 +(E(x) -17)*1/52)
Sqrt ((E(x) -0)2 *1/2 + (E(x) -6) 2*1/4
+ (E(x) -12) 2*3/13 +(E(x) -17) 2*1/52)
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BASEBALL PROBLEM
You play two games against the same opponent.
 The probability you win the first game is 0.7.
 If you win the first game, the probability you win
the second is 0.6.
 If you lose the first games, the probability you
win the second is 0.5.

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BASEBALL PROBLEM

Find the expected number of games won.

Find the standard deviation.
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ADDITION AND SUBTRACTION OF A
CONSTANT:
MEANS AND VARIANCES

Adding or subtracting a constant from data shifts
the mean but doesn’t change the variance or
standard deviation:
E(X ± c) = E(X) ± c Var(X ± c) = Var(X)

Example: Consider everyone in a company receiving
a $5000 increase in salary.
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MULTIPLICATION AND DIVISION OF A
CONSTANT:
MEANS AND VARIANCES

In general, 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)

Example: Consider everyone in a company receiving
a 10% increase in salary.
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FOR THE FOLLOWING PROBLEMS,
Let,
 Mean of X = 60
 SD of X = 10
 Mean of Y = 10
 SD of Y = 2

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FIND THE MEAN AND STANDARD
DEVIATION FOR THE RANDOM VARIABLE
1.
2.
3.
4.
3X
3*60 and 3*10
3*60 and 32*10
3*60 and 10
60 and 10
0%
1
0%
2
0%
3
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0%
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4
FIND THE MEAN AND SD FOR THE
RANDOM VARIABLE 3Y+4
1.
2.
3.
4.
3*10 +4 and 3*2 +4
3*10 and 3*2
3*10+4 and 3*2
3*10 and 3*2+4
0%
0%
0%
0%Slide
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1
2
3
4
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
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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
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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
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1
2
3
4
EVERYTHING
  E  X    x  P  x
  Var  X     x     P  x 
2
2
  SD  X   Var  X 
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RULES FOR E(X) AND VAR(X)
Adding or subtracting a constant from data shifts
the mean but doesn’t change the variance or
standard deviation:
E(X ± c) = E(X) ± c Var(X ± c) = Var(X)
 In general, 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)

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UPCOMING IN CLASS

Sunday Homework 5

Exam 1 – October 9th