1. Chapter 5 Discrete Probability Distributions
Yandell – Econ 216

2. Chapter Goals After completing this chapter, you should be able to:
Distinguish between discrete and continuous probability distributions
Apply the Binomial probability distribution to find probabilities
Use PHStat to find Binomial probabilities
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3. Introduction to Probability Distributions
Random Variable
Represents a possible numerical value from a random event
Random
Variables
Discrete
Random Variable
Continuous
Random Variable
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4. Discrete Random Variables
Can only assume a countable number of values
Examples:
Roll a die twice
Let X be the number of times 4 comes up
(then X could be 0, 1, or 2 times)
Toss a coin 5 times.
Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)
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5. Discrete Probability Distribution
Experiment: Toss 2 Coins. Let X = # heads.
4 possible outcomes
Probability Distribution T T
X Value Probability
/4 = .25
/4 = .50
/4 = .25 T H
listen H T
.50
.25
Probability H H X
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6. Discrete Probability Distribution
A list of all possible [ Xi , P(Xi) ] pairs
Xi = Value of Random Variable (Outcome)
P(Xi) = Probability Associated with Value
Xi’s are mutually exclusive
(no overlap)
Xi’s are collectively exhaustive
(nothing left out)
0 £ P(Xi) £ 1 for each Xi
S P(Xi) = 1
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7. Discrete Random Variable Summary Measures
Expected Value of a discrete distribution
(Weighted Average)
E(X) = Xi P(Xi)
Example: Toss 2 coins,
X = # of heads,
compute expected value of X:
E(X) = (0 x .25) + (1 x .50) + (2 x .25)
= 1.0
X P(X)
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8. Discrete Random Variable Summary Measures
(continued)
Standard Deviation of a discrete distribution
where:
E(X) = Expected value of the random variable
X = Values of the random variable
P(X) = Probability of the random variable having the value of X
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9. Discrete Random Variable Summary Measures
(continued)
Example: Toss 2 coins, X = # heads,
compute standard deviation (recall E(X) = 1)
Possible number of heads = 0, 1, or 2
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10. Two Discrete Random Variables
Expected value of the sum of two discrete random variables:
E(X + Y) = E(X) + E(Y)
=  X P(X) +  Y P(Y)
(The expected value of the sum of two random variables is the sum of the two expected values)
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11. Covariance Covariance between two discrete random variables:
σXY =  [Xi – E(X)][Yj – E(Y)]P(XiYj)
where:
Xi = possible values of the X discrete random variable
Yj = possible values of the Y discrete random variable
P(Xi ,Yj) = joint probability of the values of Xi and Yj occurring
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12. Interpreting Covariance
Covariance between two discrete random variables:
XY > X and Y tend to move in the same direction
XY < X and Y tend to move in opposite directions
XY = X and Y do not move closely together
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13. Probability Distributions
Ch. 5
Discrete
Probability Distributions
Continuous
Probability Distributions
Ch. 6
Binomial
Normal
Poisson
Uniform
Hypergeometric
Exponential
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14. Continuous Probability Distributions
A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values)
thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value, depending only on the ability to measure accurately
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15. Discrete Probability Distributions
A discrete random variable is a variable that can assume only a countable number of values
Many possible outcomes:
number of complaints per day
number of TV’s in a household
number of rings before the phone is answered
Only two possible outcomes:
gender: male or female
defective: yes or no
spreads peanut butter first vs. spreads jelly first
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16. Application: Return on Investment
Let X measure the return on an investment measured in percent and let P(X) be the probability associated with each return.
The list of outcomes and their associated probabilities is given in the table on the next slide.
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17. A Discrete Probability Distribution
X (%)
P(X) 9
0.05 10
0.15 11
0.30 12
0.20 13 14
0.10 15
This row means that the probability of a 10% return is 15%.
Note that these probabilities sum to one, so this list is exhaustive.
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18. Graphically
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19. Problems The probability of at least a 14% return is:
P(X  14) = P(X=14) + P(X=15) =
= 0.15
X (%)
P(X) 9
0.05 10
0.15 11
0.30 12
0.20 13 14
0.10 15
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20. A Useful Trick
What is the probability of earning less than 14%? The hard way to find this is:
P(X < 14) = P(X=9) + P(X=10) + P(X=11)
+ P(X=12) + P(X=13)
The easy way is:
P(X < 14) = 1 - P(X  14) = = 0.85
X (%)
P(X) 9
0.05 10
0.15 11
0.30 12
0.20 13 14
0.10 15
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21. The Mean
The Mean of a discrete distribution is the weighted average of the outcomes:
Which is:
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22. Calculating the Mean listen X (%) P(X) X  P(X) 9 0.05 0.45 10 0.15
1.50 11
0.30
3.30 12
0.20
2.40 13
1.95 14
0.10
1.40 15
0.75
Mean = 11.75
listen
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23. The Variance
The Variance of a discrete distribution is the weighted average of the squared deviations from the mean:
Which becomes:
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24. Calculating the Variance
X - m
(X - m) 2
(X - m) 2  P(X)
7.5625
0.378
3.0625
0.459
0.5625
0.169
0.0625
0.013
1.5625
0.234
5.0625
0.506
0.528
Variance = 2.288
(PHStat will do the calculations for the mean and variance for us
when we use the Binomial distribution)
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25. The Standard Deviation
The standard deviation is the square root
of the variance:
Empirical Rule:
the mean  2 standard deviations
includes about 95% of the observations
Here,
So we know that about 95% of the values from this distribution fall between 8.75 and
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26. Mean  2 Standard Deviations
+/- 2σ m
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27. Practice
The probability distribution for damage claims paid per year for collision insurance is:
X ($)
P(X)
0.90
400
0.04
1000
0.03
2000
0.01
4000
6000
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28. Question 1
Find the expected collision payment to determine the annual collision insurance premium that will enable the company to break even.
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29. Question 2
The insurance company charges an annual rate of $260 for collision coverage. What is the expected value of the collision coverage to a policyholder? (expected payments from the company minus cost of the coverage)
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30. Another type of customer
Consider a different class of customer with the following claim distribution:
X ($)
P(X)
0.80
400
0.06
1000
0.05
2000
0.03
4000
6000
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31. Questions 3&4
How do your answers change for questions 1&2 using this different type of customer?
How do these customers compare?
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32. Probability Distributions
The Binomial Distribution
Probability Distributions
Discrete
Probability Distributions
Binomial
Poisson
Hypergeometric
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33. The Binomial Distribution
Characteristics of the Binomial Distribution:
A trial has only two possible outcomes – “success” or “failure”
There is a fixed number, n, of identical trials
The trials of the experiment are independent of each other
The probability of a success, π, remains constant from trial to trial
If π represents the probability of a success, then
(1- π) is the probability of a failure
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34. Binomial Distribution Settings
A manufacturing plant labels items as either defective or acceptable
A firm bidding for a contract will either get the contract or not
A marketing research firm receives survey responses of “yes I will buy” or “no I will not”
New job applicants either accept the offer or reject it
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35. Counting Rule for Combinations
A combination is an outcome of an experiment where x objects are selected from a group of n objects
where:
n! =n(n - 1)(n - 2) (2)(1)
x! = x(x - 1)(x - 2) (2)(1)
0! = 1 (by definition)
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36. Binomial Distribution Formula! x n - x
P(x) = π
(1- π)
x ! ( n - x ) !
listen
P(X = x) = probability of x successes in n trials,
with probability of success π on each trial
x = number of ‘successes’ in sample,
(x = 0, 1, 2, ..., n)
π = probability of “success” per trial
1 - π = probability of “failure”
n = number of trials (sample size)
Example: Flip a coin four times, let X = # heads:
n = 4
π = 0.5
1 - π = (1 - .5) = .5
x = 0, 1, 2, 3, 4
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37. Don’t Worry We will use PHStat to get Binomial Probabilities
More examples and sample PHStat output appear soon
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38. Binomial Distribution
The shape of the binomial distribution depends on the values of π and n
Mean
P(x)
n = 5, π = 0.1 .6 .4 .2
Here, n = 5 and π = .1 X 1 2 3 4 5
n = 5, π = 0.5
P(x)
listen .6 .4
Here, n = 5 and π = .5 .2 X 1 2 3 4 5
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39. Binomial Distribution Characteristics
Mean
Variance and Standard Deviation
Where n = sample size
π = probability of success
1 – π = probability of failure
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40. Binomial Characteristics
Examples
n = 5, π = 0.1
Mean
P(x) .6 .4 .2 X 1 2 3 4 5
n = 5, π = 0.5
P(x) .6 .4 .2 X 1 2 3 4 5
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41. Using Binomial Tables Examples:
n x
π=.15
π=.20
π=.25
π=.30
π=.35
π=.40
π=.45
π=.50 1 2 3 4 5 6 7 8 9 10
0.1969
0.3474
0.2759
0.1298
0.0401
0.0085
0.0012
0.0001
0.0000
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
π=.85
π=.80
π=.75
π=.70
π=.65
π=.60
π=.55
x n
Examples:
n = 10, π = .35, x = 3: P(X = 3|n =10, π = .35) = .2522
n = 10, π = .75, x = 2: P(X = 2|n =10, π = .75) = .0004
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42. Using PHStat In Excel 2007
Select PHStat / Probability & Prob. Distributions / Binomial…
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43. In Excel 2010 or 2013
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44. Using PHStat Enter desired values in dialog box Here: n = 10 π = .35
(continued)
Enter desired values in dialog box
Here: n = 10
π = .35
Output for X = 0
to X = 10 will be
generated by PHStat
Optional check boxes
for additional output
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45. PHStat Output P(X = 3 | n = 10, π = .35) = .2522
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46. Sample Problem: A Light Bulb Shipment
Consider a large shipment of light bulbs. We will select a sample of five bulbs and determine whether to accept or reject the whole shipment.
Binomial Assumptions are satisfied:
‘n’ Identical Trials
5 light bulbs randomly taken from a shipment
2 Mutually Exclusive Outcomes in Each Trial
defective or not defective light bulb
Constant Probability For Each Trial
each light bulb has the same probability, π,
of being not defective
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47. The Light Bulb Problem
(continued)
Suppose that we randomly select five (n = 5) light bulbs from a shipment. We will accept the shipment if at least four bulbs work. If each bulb has a π = .90 chance of working, what is the probability that we will accept the shipment?
That is, what is P(X  4) ?
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48. PhStat Demo
Click here to view the PHStat demo for the light bulb problem:
Click here to watch
the demonstration
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49. Probability Distributions
The Poisson Distribution
Probability Distributions
Discrete
Probability Distributions
Binomial
Poisson
Hypergeometric
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50. The Poisson Distribution
Characteristics of the Poisson Distribution:
The outcomes of interest are rare relative to the possible outcomes
The average number of outcomes of interest per unit is  (lambda)
The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest
The probability of that an outcome of interest occurs in a given segment is the same for all segments
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51. Poisson Distribution Formula
where:
x = number of successes per unit
 = expected number of successes
e = base of the natural logarithm system ( )
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52. Poisson Distribution Characteristics
Mean
Variance and Standard Deviation
where  = expected number of successes per unit
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53. Using Poisson Tables Example: Find P(X = 2) if  = .50 X  0.10 0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90 1 2 3 4 5 6 7
0.9048
0.0905
0.0045
0.0002
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.7408
0.2222
0.0333
0.0033
0.0003
0.6703
0.2681
0.0536
0.0072
0.0007
0.6065
0.3033
0.0758
0.0126
0.0016
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.4966
0.3476
0.1217
0.0284
0.0050
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
Example: Find P(X = 2) if  = .50
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54. Graph of Poisson Probabilities
Graphically:
 = .50 X
 =
0.50 1 2 3 4 5 6 7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
P(X = 2) = .0758
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55. Poisson Distribution Shape
The shape of the Poisson Distribution depends on the parameter  :
 = 0.50
 = 3.00
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56. Probability Distributions
The Hypergeometric Distribution
Probability Distributions
Discrete
Probability Distributions
Binomial
Poisson
Hypergeometric
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57. The Hypergeometric Distribution
“n” trials in a sample taken from a finite population of size N
Sample taken without replacement
Trials are dependent
Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population
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58. Hypergeometric Distribution Formula
(Two possible outcomes per trial)
Where
N = Population size
A = number of successes in the population
N – A = number of failures in the population
n = sample size
X = number of successes in the sample
n – X = number of failures in the sample
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59. Hypergeometric Distribution Formula
Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?
N = 10 n = 3
A = X = 2
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60. Hypergeometric Distribution in PHStat
Select:
Add Ins / PHStat / Probability & Prob. Distributions / Hypergeometric …
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61. Hypergeometric Distribution in PHStat
(continued)
Complete dialog box entries and get output …
N = 10 n = 3
A = 4 X = 2
P(X = 2) = 0.3
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62. Chapter Summary
Distinguished between discrete and continuous probability distributions
Examined discrete probability distributions and their summary measures (Binomial, Poisson, and Hypergeometric)
Yandell – Econ 216