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© 2003 Pearson Prentice Hall
Chapter 4
Discrete Random Variables
4-1
Learning Objectives
© 2003 Pearson Prentice Hall
1. Distinguish Between the Two Types of
Random Variables
2. Compute the Expected Value & Variance of
Discrete Random Variables
3. Describe the Binomial Distribution and
calculate probabilities for it
4. Understand derivation of formulas for mean
and variance of binomial distribution
4-2
Data Types
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Data
Numerical
Discrete
4-3
Continuous
Qualitative
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Discrete Random Variables
4-4
Random Variable
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Assign numeric values to outcomes of an
“experiment”
Random Variable takes on different
possible values corresponding to
experiment outcomes

Example: Number of Tails in 2 Coin
Tosses
4-5
© 2003 Pearson Prentice Hall

Possible values are discrete



Discrete Random
Variable
E.g., Whole Number (0, 1, 2, 3 etc.)
Obtained by Counting
Usually Finite Number of Values

4-6
But could be infinite (must be “countable”)
© 2003 Pearson Prentice Hall
Exercise 4.3: Discrete or
Continuous?
Number of misspelled words
Amount of water through Hoover dam in a day
How late for class
Number of bacteria in a water sample
Amount of CO produced from burning a gallon
of unleaded gas
Your weight
Number of checkout lanes at grocery store
Amount of time waiting in line at grocery store
4-7
© 2003 Pearson Prentice Hall
Discrete
Probability Distribution
1. List of All possible [x, p(x)] pairs


x = Value of Random Variable (Outcome)
p(x) = Probability Associated with Value
2. Mutually Exclusive (No Overlap)
3. Collectively Exhaustive (Nothing Left Out)
4. 0  p(x)  1
5.  p(x) = 1
4-8
© 2003 Pearson Prentice Hall
Discrete Probability
Distribution Example
Experiment: Toss 2 Coins. Count # Tails.
Probability Distribution
Values, x Probabilities, p(x)
© 1984-1994 T/Maker Co.
4-9
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Visualizing Discrete
Probability Distributions
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Listing
Table
# Tails
f(x)
Count
p(x)
0
1
2
1
2
1
.25
.50
.25
{ (0, .25), (1, .50), (2, .25) }
p(x)
.50
.25
.00
Graph
Equation
p ( x) 
x
0
4 - 10
1
2
n!
p x (1  p) n  x
x !(n  x)!
Summary Measures
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1. Expected Value


Mean of Probability Distribution
Weighted Average of All Possible Values


May not equal any of the possible values
 = E(X) = x p(x)
2. Variance


Weighted Average Squared Deviation about
Mean
2 = E[ (x (x  p(x)
4 - 11
Thinking Challenge
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You toss 2 coins. You’re
interested in the number
of tails. What are the
expected value &
standard deviation of
this random variable,
number of tails?
© 1984-1994 T/Maker Co.
4 - 12
© 2003 Pearson Prentice Hall
x
p(x)
0
.25
1
.50
2
.25
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Expected Value &
Variance Solution*
x p(x )
x-
(x -)
2
2
(x -) p( x )
© 2003 Pearson Prentice Hall
Expected Value &
Variance Solution*
x
p(x)
x p(x )
0
.25
0
1
.50
.50
2
.25
.50
 = 1.0
4 - 14
x-
(x -)
2
2
(x -) p( x )
© 2003 Pearson Prentice Hall
Expected Value &
Variance Solution*
2
x
p(x)
x p(x )
x-
(x -)
0
.25
0
-1.00
1.00
.25
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
 = 1.0
4 - 15
2
(x -) p( x )
 = .50
2
Exercise 4.25
© 2003 Pearson Prentice Hall
Values 0, 1, 2
Dist X has probabilities .3, .4, .3
Dist Y has probabilities .1, .8, .1
Without doing any calculations:


Which distribution has higher mean?
Which distribution has higher variance?
4 - 16
© 2003 Pearson Prentice Hall
Definition of
Independence
X and Y are independent if (and only if)



For any sets of outcome values A and B
P(X in A and Y in B) = P(X in A)P(Y in B)
That is, any pair of events defined by
outcome sets for X and Y are independent
events
4 - 17
Binomial Distribution
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1. Number of ‘Successes’ in a Sample of n
Observations (Trials)

# Reds in 15 Spins of Roulette Wheel

# Defective Items in a Batch of 5 Items

# Correct on a 33 Question Exam

# Customers Who Purchase Out of 100
Customers Who Enter Store

# of Bush-Cheney supporters in survey of 100
people
4 - 18
© 2003 Pearson Prentice Hall
Binomial Distribution
Properties
1. Sequence of n Identical Trials
2. Each Trial Has 2 Outcomes

‘Success’ (Desired/specified Outcome) or ‘Failure’
3. Constant Trial Probability
4. Trials Are Independent
5. # of successes in n trials is a binomial
random variable
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Exercise: Binomial?
© 2003 Pearson Prentice Hall
Pick 6 students from this class


Each flips a coin
Count # of heads
Pick 6 students from this class

X= # of 1st year students selected
Random digit dialing of 100 numbers

# of Bush-Cheney supporters
Random digit dialing of 100 numbers

Sum of ages of respondents
4 - 20
© 2003 Pearson Prentice Hall
Binomial Probability
Distribution Function
 n  x n x
n!
x
n x
p( x)    p q 
p (1  p)
x!(n  x)!
 x
p(x) = Probability of x ‘Successes’
n = Sample Size
p = Probability of ‘Success’
x = Number of ‘Successes’ in Sample
(x = 0, 1, 2, ..., n)
4 - 21
Binomial Probability
Distribution Example
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Experiment: Toss 1 Coin 5 Times in a Row.
Note # Tails. What’s the Probability of 3 Tails?
n!
x
n x
p( x ) 
p (1  p )
x !(n  x )!
5!
3
5 3
p(3) 
.5 (1 .5)
3 !(5  3)!
 0.3125
4 - 22
© 2003 Pearson Prentice Hall
Cumulative Distribution
Function
Fx a   Pr  X  a    p x 
xa
4 - 23
© 2003 Pearson Prentice Hall
Binomial Probability
Table (Portion)
n=5
p
k
.01
…
0.50
…
.99
0
.951
…
.031
…
.000
1
.999
…
.188
…
.000
2
1.000
…
.500
…
.000
3
1.000
…
.812
…
.001
4
1.000
…
.969
…
.049
Cumulative Probabilities FX(k)
4 - 24
© 2003 Pearson Prentice Hall
Binomial Distribution
Characteristics
Mean
  E ( x )  np
P(X)
.6
.4
.2
.0
  np (1  p)
X
0
Standard Deviation
P(X)
.6
.4
.2
.0
1
2
3
4
5
n = 5 p = 0.5
X
0
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n = 5 p = 0.1
1
2
3
4
5
Exercise 4.37
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N=4, x=2, q=.4– compute p(x)
4 - 26
Exercise 4.39
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N=25, p=.5

Compute E(X), Var(X), sigma
4 - 27
Exercise 4.51
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Pr(domestic abuse) = 1/3, or maybe 1/10
Sample of 15 women; 4 have been abused
If p=1/3, what is Pr(X>=4)?
If p=1/10, what is pr(X>=4)?
Given evidence from the sample, which abuse
rate seems more plausible?
Note: this is a preview of thinking about
sampling distributions
4 - 28
Exercise 4.109
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Response rate p=.4
If mail 20, what is Pr(X>12)?
If want 100 responses with Pr > .95, how
many should you mail?
4 - 29
4.109 Solution
© 2003 Pearson Prentice Hall
By empirical rule, 95% of observations are
within 2sigma




Find n such that mu-2sigma > 100
Mu = .4n
Sigma squared = npq=.24n
Solving for n  292
Note that for n=292 E(X) =292*.4=116.8

In order to be sure to get 100, need to have a
higher average number of returns
4 - 30
Useful Observation 1
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For any X and Y
E  X  Y      x  y  p( x  y)
x y
   xp( x  y )    yp( x  y )
x y
x y
  x p( x  y)   y  p( x  y)
x
y
y
  xp x    yp y 
x
4 - 31
y
 E  X   E Y 
x
One Binary Outcome
© 2003 Pearson Prentice Hall
Random variable X, one binary outcome
Code success as 1, failure as 0
P(success)=p, p(failure)=(1-p)=q
E(X) = p
Var  X   E  X  E  X 2

  E X 
E X
2
2
 p *1  q * 0   p 
2
2
 p  p  p1  p 
2
4 - 32
2

Mean of a Binomial
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Independent, identically distributed
X1, …, Xn; E(Xi)=p; Binomial X =  X i
E  X i 
  E X i 
 nE  X1 
 np
4 - 33
By useful
observation 1
Useful Observation 2
© 2003 Pearson Prentice Hall
For independent X and Y
E  XY      xy  p ( x  y )
x y
   xyp x  p y 
x y
  xp x  yp y    xp x E Y  
x
y
x
E Y  xp x   E Y E  X   E  X E Y 
x
4 - 34
Useful Observation 3
For independent X and Y


Var  X  Y   E  X  Y   E  X  Y 
2
2


2
2
 E X  E Y  2 E  XY 
 E X  Y  2 XY  E  X   E Y 
2
2
2
 E  X   E Y   2 E  X E Y 
2
2
2
2
2
2








 E X  E  X   E Y  E Y 
 2 E  XY   2 E  X E Y 
 
 
cancels by obs. 2
 E X 2  E  X 2  E Y 2  E Y 2  Var  X   Var Y 
Variance of Binomial
© 2003 Pearson Prentice Hall
Independent, identically distributed
X1, …, Xn; E(Xi)=p; Binomial X =  X i
Var  X i   Var  X i 
 nVar X i 
 np1  p 
4 - 36
© 2003 Pearson Prentice Hall
4 - 37
End of Chapter
Any blank slides that follow are
blank intentionally.
© 2003 Pearson Prentice Hall
Useful Observation 4
(We’ll use this later)
For any X


Var kX   E k 2 X 2  E kX 2
  kE X 
2
2
2
2
 k E X  k E  X 
2
k E X
2
 k Var  X 
2
4 - 39
2