1. Discrete Random Variables
Discrete Probability Distributions
Binomial Distribution
Poisson Distribution
Hypergeometric Distribution
Econ10/Mgt Stuffler

2. Discrete Probability Distributions
Discrete Random Variables – Sample Space includes all mutually exclusive outcomes
Probabilities – from subjective, frequency or subjective methods
Two conditions apply:
Econ10/Mgt Stuffler

3. Discrete Probability Distributions
Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household (millions) from US survey data:
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4. Discrete Probability Distributions
Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data:
1,218 ÷ 101,501 = 0.012
Econ10/Mgt Stuffler

5. Discrete Probability Distributions
Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data:
EX: P(X=4) = P(4) = = 7.6%
Econ10/Mgt Stuffler

6. Discrete Probability Distributions
What is the probability there is at least one television
but no more than three in any given household?
Econ10/Mgt Stuffler

7. Discrete Probability Distributions
What is the probability there is at least one television
but no more than three in any given household?
Econ10/Mgt Stuffler

8. Discrete Probability Distributions
What is the probability there is at least one television
but no more than three in any given household?
“at least one television but no more than three”
P(1 ≤ X ≤ 3) = P(1) + P(2) + P(3) = = .884
Econ10/Mgt Stuffler

9. Discrete Probability Distributions
Assume a mutual fund
salesman knows
that there is 20% chance
of closing a sale on each
call he makes.
Econ10/Mgt Stuffler

10. Discrete Probability Distributions
What is the probability distribution of the number of sales if he plans to call three customers?
Let S denote success, making a sale:
P(S) = .20,
then Sʹ, not making a sale:
P(Sʹ) = .80
Econ10/Mgt Stuffler

11. Discrete Probability Distributions
Developing a Probability Distribution Tree
Sales Call 1
P(S)=.2
P(S’)=.8
Econ10/Mgt Stuffler

12. Discrete Probability Distributions
Developing a Probability Distribution Tree
Sales Call 1
Sales Call 2
P(S)=.2
P(S’)=.8
P(S)=.2
P(S’)=.8
Econ10/Mgt Stuffler

13. Discrete Probability Distributions
Developing a Probability Distribution Tree
Sales Call 1
Sales Call 2
Sales Call 3
S S S
S S S’
S S’ S
S S’ S’
S’ S S
S’ S S’
S’ S’ S
S’ S’ S’
P(S)=.2
P(S’)=.8
P(S)=.2
P(S’)=.8
P(S)=.2
P(S’)=.8
Econ10/Mgt Stuffler

14. Discrete Probability Distributions
Developing a Probability Distribution Tree
Sales Call 1
Sales Call 2
Sales Call 3
S S S
S S S’
S S’ S
S S’ S’
S’ S S
S’ S S’
S’ S’ S
S’ S’ S’
P(S)=.2
P(S’)=.8
P(S)=.2
P(S’)=.8
P(S)=.2
P(S’)=.8
Econ10/Mgt Stuffler

15. Discrete Probability Distributions
Developing a Probability Distribution
Sales Call 1
Sales Call 2
Sales Call 3
S S S
S S S’
S S’ S
S S’ S’
S’ S S
S’ S S’
S’ S’ S
S’ S’ S’
P(S)=.2
P(S’)=.8
P(S)=.2
‘)=.8
P(S’)=.8
X P(x)
.23 = .008
3(.032)=.096
3(.128)=.384
0 .83 = .512
P(S)=.2
P(S’)=.8
Econ10/Mgt Stuffler

16. Discrete Probability Distributions
Explain how to derive .032 and .128
Sales Call 1
Sales Call 2
Sales Call 3
S S S
S S SC
S SC S
S SC SC
SC S S
SC S SC
SC SC S
SC SC SC
P(S)=.2
P(SC)=.8
P(S)=.2
P(SC)=.8
X P(x)
.23 = .008
3(.032)=.096
3(.128)=.384
0 .83 = .512
P(S)=.2
P(SC)=.8
Econ10/Mgt Stuffler

17. Discrete Probability Distributions
The P(X=2) is:
Sales Call 1
Sales Call 2
Sales Call 3
(.2)(.2)(.8)= .032
S S S
S S SC
S SC S
S SC SC
SC S S
SC S SC
SC SC S
SC SC SC
P(S)=.2
P(SC)=.8
P(S)=.2
P(SC)=.8
X P(x)
.23 = .008
3(.032)=.096
3(.128)=.384
0 .83 = .512
P(S)=.2
P(SC)=.8
Econ10/Mgt Stuffler

18. Population Mean and Population Variance
Discrete Probability Distributions
A discrete probability distribution represents a population
Example: Population of number of TVs per household
Example: Population of sales call outcomes
Since we have populations, we can describe
them by computing various parameters:
Population Mean and Population Variance
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19. Discrete Probability Distributions
Population Mean (Expected Value)
- Weighted average of all values with the weights
being the probabilities
- Expected value of X, E(X)
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20. Discrete Probability Distributions
Population variance - weighted average of the
squared deviations from the mean.
“Short cut” formula for the variance
Standard deviation formula
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21. Discrete Probability Distributions
Find the mean, variance, and standard deviation for the population of the number of color televisions per household.
= 0(.012) + 1(.319) + 2(.374) + 3(.191) + 4(.076) + 5(.028)
= 2.084
Econ10/Mgt Stuffler

22. Discrete Probability Distributions
Find the mean, variance, and standard deviation for the population of the number of color televisions per household.
= (0 – 2.084)2(.012) + (1 – 2.084)2(.319)+…+(5 – 2.084)2(.028)
= 1.107
Econ10/Mgt Stuffler

23. Discrete Probability Distributions
Find the mean, variance, and standard deviation for the population of the number of color televisions per household.
= 1.052
Econ10/Mgt Stuffler

24. Discrete Probability Distributions
Special application of Expected Value
Suppose the probability that an insurance agent makes a sale is .20 and after costs earns a commission of $525.
If he/she does not make a sale, they must pay $75 in costs.
What is their expected value from a sales call?
Does the benefit exceed the cost or is the opposite true?
Econ10/Mgt Stuffler

25. Discrete Probability Distributions
Special application of Expected Value
Let X be the discrete random variable of
making a sale call
x P(x) x•P(x)
Sale $ $ 105
No Sale - $ $ 60
E(x) = μ = ∑ xP(x) = $ 45
Econ10/Mgt Stuffler

26. Binomial Distribution
Binomial distribution is the probability distribution that results from doing a “binomial experiment” which have the properties:
Fixed number of identical trials, represented as n.
Each trial has two possible outcomes:
a “success” or “failure”
For all trials, the probability of success, P(success)=p, and the probability of failure, P(failure)=1–p=q, are constant.
The trials are independent
Econ10/Mgt Stuffler

27. Several Binomial Distributions
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28. Binomial Distribution
Success and failure: labels for binomial experiment outcomes, no value judgment is implied.
EX: Coin flip results in either heads or tails.
If we define “heads” as success, then
“tails” is considered a failure.
Other binomial examples:
An election candidate wins or loses
An employee is male or female
A worker is employed or unemployed
Econ10/Mgt Stuffler

29. Binomial Distribution
Binomial Formula:
where x = number of successes in n trials,
n – x = number of failures in n trials,
px = the probability of success raised to the number of successes, and
qn-x = probability of failure raised to the number of failures
Econ10/Mgt Stuffler

30. Binomial Distribution
The random variable of a binomial experiment is defined as the number of successes in the n trials, and is called the binomial random variable.
EX: Flip a fair coin 10 times
1) Fixed number of trials  n=10
2) Each trial has two possible outcomes  {heads (success),
tails (failure)}
3) P(success)= 0.50; P(failure)=1–0.50 = 0.50 
4) The trials are independent  (i.e. the outcome of heads on the
first flip will have no impact on subsequent coin flips).
Flipping a coin ten times is a binomial
experiment since all conditions are met.
Econ10/Mgt Stuffler

31. Binomial Distribution
Another sales call example:
Assume a mutual fund salesman knows that there is
20% chance of closing a sale on each call he makes.
We want to determine the probability of making two sales in three calls:
P(sale) = .2
P(no sale) = .8
P(X=2) = ! = .096
2!(3-2)!
Econ10/Mgt Stuffler

32. Binomial Distribution
Another sales call example:
Assume a mutual fund salesman knows that there is
20% chance of closing a sale on each call he makes
We want probability of making two sales in three calls
P(sale) = .2
P(no sale) = .8
GOOD NEWS!
Megastat→Probability→Discrete Distributions→Binomial
Econ10/Mgt Stuffler

33. Binomial Distribution
Econ10/Mgt Stuffler

34. Pat Statsly Pat Statsly is a student (not a good student) taking a
statistics course. Pat’s exam strategy is to rely on luck
for the first test. The test consists of 10 multiple-choice
questions. Each question has five possible answers, only
one of which is correct. Pat plans to guess the answer
to each question.
What is the probability that Pat gets no answers correct?
What is the probability that Pat gets two answers correct?
Econ10/Mgt Stuffler

35. Pat Statsly n=10 P(correct) = p = 1/5 = .20 P(wrong) = q = .80
Is this a binomial experiment?
Check the conditions
Econ10/Mgt Stuffler

36. Pat Statsly n=10 P(correct) = p = 1/5 = .20 P(wrong) = q = .80
Is this a binomial experiment? Check the conditions:
 There is a fixed finite number of trials (n=10).
 An answer can be either correct or incorrect.
 The probability of a correct answer (P(success)=.20) does not change from question to question.
 Each answer is independent of the others.
Econ10/Mgt Stuffler

37. What’s the interpretation of this result?
Pat Statsly
n=10, and P(success) = .20
What is the probability that Pat gets no answers correct?
EX: P(x=0)
What’s the interpretation of this result?
Econ10/Mgt Stuffler

38. Pat Statsly n=10, and P(success) = .20 EX: P(x=0)
What is the probability that Pat gets no answers correct?
EX: P(x=0)
Pat has about an 11% chance of getting no answers correct
using the guessing strategy.
Econ10/Mgt Stuffler

39. Pat Statsly
Econ10/Mgt Stuffler

40. Pat Statsly n=10, and P(success) = .20
What is the probability that Pat gets two answers correct?
EX: P(x=2)
Econ10/Mgt Stuffler

41. Pat Statsly
We have been using the binomial probability distribution to find probabilities for individual values of x.
To answer the question:
“Find the probability that Pat fails the quiz”
requires a cumulative probability, that is, P(X ≤ x)
If a grade on the quiz is less than 50% (i.e. 5 questions
out of 10), that’s considered a failed quiz.
Econ10/Mgt Stuffler

42. Pat Statsly
We have been using the binomial probability distribution to find probabilities for individual values of x.
To answer the question:
“Find the probability that Pat fails the test”
requires a cumulative probability, that is, P(X ≤ x)
If a grade on the test is less than 50% (i.e., 5 questions
out of 10), that’s considered a failed test.
We want to know what is: P(X ≤ 4) to answer
Econ10/Mgt Stuffler

43. Pat Statsly P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4) = .9672
What is the interpretation of this result?
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44. Pat Statsly fail the test using the luck strategy
Its about 97% probable that Pat will
fail the test using the luck strategy
and guessing at answers.
Econ10/Mgt Stuffler

45. Binomial Table
Calculating binomial probabilities by hand is tedious and
error prone. There is an easier way. Refer to Table E-6 in
the Appendices. For the Pat Statsly example, n=10, so go to
the n=10 table:
Look in the Column, p=.20 and substitute into:
P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)
P(X ≤ 4) = = .9672
The probability of Pat failing the test is 96.72%
Econ10/Mgt Stuffler

46. Poisson Distribution Named for Simeon Poisson, Poisson distribution -
Discrete probability distribution
There there are no trials
Number of independent events (successes) occurring in a fixed time period or region of space that occur with a known average rate such as, arrivals, departures, or accidents, number of baskets in a quarter, etc.
Econ10/Mgt Stuffler

47. Poisson Distribution For example:
The number of cars arriving at a service station in 1 hour. (The interval of time is 1 hour.)
The number of flaws in a bolt of cloth. (The specific region is a bolt of cloth.)
The number of accidents in 1 day on a particular stretch
of highway. (The interval is defined by both time, 1 day, and space, the particular stretch of highway.)
Econ10/Mgt Stuffler

48. Poisson Distribution
Poisson random variable - number of successes that occur in a period of time or an interval of space
EX: On average, 96 trucks arrive at a border crossing
every hour.
EX: Number of typographical errors in a new textbook
edition averages 1.5 per 100 pages.
Econ10/Mgt Stuffler

49. Poisson Distribution…
The Poisson random variable is the number of successes that occur in a period of time or an interval of space in a Poisson experiment.
EX: On average, 96 trucks arrive at a border crossing
every hour.
E.g. The number of typographic errors in a new textbook edition averages 1.5 per 100 pages.
successes
time period
successes (?!)
interval
Econ10/Mgt Stuffler

50. Poisson Distribution
Poisson experiment has four defining characteristics or properties:
The number of successes that occur in any interval is independent of the number of successes that occur in any other interval
The probability of a success in an interval is the same for all equal-size intervals
The probability of a success is proportional to the size of the interval
Only one value is required to determine the probability of a designated number of events occurring during an interval
Econ10/Mgt Stuffler

51. µ = n(p) = number of successes times the probability of success
Poisson Distribution
The probability that a Poisson random variable assumes a value of x is given by:
µ = n(p) = number of successes times the probability of success
The unit of time or the interval should be short enough so that the mean rate is not large (µ < 20)
YOUR TEXTBOOK USES λ (lambda) REPRESENT THE MEAN NUMBER OF SUCCESSES IN THE INTERVAL
Econ10/Mgt Stuffler

52. Poisson Distribution EXAMPLE
The number of typographical errors in new editions of textbooks varies considerably from book to book.
After some analysis she concludes that the number of errors is Poisson distributed with a mean of 1.5 per 100 pages.
The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos?
MEGASTAT→PROBABILITY→DISCRETE DISTRIBUTIONS→POISSON
AND ENTER THE MEAN VALUE: 1.5, CLICK OK
Econ10/Mgt Stuffler

53. Poisson Distribution
Econ10/Mgt Stuffler

54. Mean and Variance of a Poisson Random Variable
If x is a Poisson random variable with parameter , then
Mean number of events per unit of time or space = µ
Variance = V(X) = σ2 = μ
Econ10/Mgt Stuffler

55. Poisson Distribution
As mentioned on the first Poisson Distribution slide:
The probability of a success is proportional
to the size of the interval
Thus, knowing an error rate of 1.5 typos per 100 pages, we can determine a mean value for a 400 page book as:
μ = n(p) = 1.5(4) = 6 typos / 400 pages.
Econ10/Mgt Stuffler

56. Poisson Distribution Interpretation?
For a 400 page book, what is the probability that there are
no typos?
P(X=0) =
Interpretation?
Econ10/Mgt Stuffler

57. For a mean of 6,the Excel result is 0.00248
GOOD NEWS!
Go to Excel, Megastat, Probability, Discrete, Poisson.
Type in the mean, μ, 6, then OK.
For a mean of 6,the Excel result is
Econ10/Mgt Stuffler

58. Poisson Distribution P(X≤5) = P(0) + P(1) + … + P(5)
For a 400 page book, what is the probability that there are five or less typos?
P(X≤5) = P(0) + P(1) + … + P(5)
Another alternative is to refer to Table E-7 in the Appendices
For x = 5, μ =6, and P(X ≤ 5) = P(0) + … + P(5) = .4456
There is about a 45% chance that there will be 5 or less typos
Econ10/Mgt Stuffler

59. Hypergeometric Distribution
If a sample is taken from a finite
population without replacement, the
probability of X number of successes,
follows the hypergeometric distribution.
NOTE: When you see “without replacement,” this indicates that hypergeometric trials are dependent.
Econ10/Mgt Stuffler

60. Hypergeometric Distribution
For both the Binomial Distribution
and the Hypergeometric
Distribution, there are two possible outcomes for the random discrete variable:
Success or Failure
Econ10/Mgt Stuffler

61. Hypergeometric Distribution
The difference is that for
Hypergeometric each trial is
“without replacement,” so the probability changes from one draw to the next
For Binomial, the probability of success and failure remain constant since each trial or draw is independent
Econ10/Mgt Stuffler

62. Hypergeometric Distribution
1. Trials are NOT independent
2. Finite population
3. Sample without replacement
Probability of success and failure
changes from trial to trial
Econ10/Mgt Stuffler

63. Hypergeometric Distribution
r N – r μ = n r
P(x) = x n – x N N
n σ2 = r (N-r) n (N-n)
N2 (N-1)
where N = Total number in the population
n = number in the sample,
x = number selected or drawn,
r = number of successes in N
Econ10/Mgt Stuffler

64. Hypergeometric Distribution
EXAMPLE: Assume that 3 stocks are chosen randomly from a list of 10 stocks and that of the 10 stocks 4 stocks pay dividends. The number (x) of the three selected stocks that pay a dividend is a hypergeometric random variable.
Econ10/Mgt Stuffler

65. Hypergeometric Distribution
N=10, n=3, r = 4 and
x = number of the 3 stocks selected
that pay a dividend
Calculate the mean, the variance and the probability that none of the 3 stocks pay dividends?
Econ10/Mgt Stuffler

66. Hypergeometric Distribution
– ! !
P(0) = = 0!(4-0)! 3!(6-3)! = 1 !
!(10-3)!
μ = (3)(4) = 1.2 10
σ2 = 4(10-4)3(10-3) = .56, σ = .75
(10)2 (10-1)
Econ10/Mgt Stuffler

67. Hypergeometric Distribution
GOOD NEWS!
Go to Excel→Megastat→Probability →Discrete Probability Distributions
→Hypergeometric, enter the required data
Econ10/Mgt Stuffler

68. Hypergeometric Distribution
Econ10/Mgt Stuffler

69. Discrete Probability WHICH DISTRIBUTION?
Sony issued a recall for their plasma screen TVs. They choose 25 TVs and 7 do not operate. If the inspector chooses 3 at random from the 25, what is the probability that 2 of the 3 do not operate?
Econ10/Mgt Stuffler

70. Discrete Probability
WHICH DISTRIBUTION?
When Wet Water drills a well, their success rate is .55 and their profit is $2,550. If they do not find water, they lose $1,500.
Econ10/Mgt Stuffler

71. Discrete Probability
WHICH DISTRIBUTION?
The US unemployment rate is calculated from survey data of 60,000 households. What is the probability that no more than 5,000 respondents are unemployed?
Econ10/Mgt Stuffler

72. Discrete Probability
WHICH DISTRIBUTION?
US Transportation and Safety Board reports the annual average number of plane crashes is For a given month, what is the probability that at 2 crashes will occur? That 3 crashes will occur?
Econ10/Mgt Stuffler

73. Discrete Probability
WHICH DISTRIBUTION?
The US Statistical Abstract of the US tracks many data items such as, the average number of automobiles per household. The average for 2009 is 2.5 autos. What is the probability that the 2010 average is at least 3 autos?
Econ10/Mgt Stuffler

74. Discrete Random Variables
Summary:
Two Types of Random Variables
Discrete Probability Distributions
The Binomial Distribution
The Poisson Distribution
The Hypergeometric Distribution
Econ10/Mgt Stuffler