Module 5: Discrete Distributions
This work by Linda S. Williams is licensed under a Creative Commons Attribution 4.0 International License.
Discrete vs. Continuous
• Discrete Random Variables:
• Result of “counting” something
• They must have “space” between them
• Continuous Random Variables:
• Results from a “measurement”
• Assumes one of an infinite number of values, within a relevant range
• Random Variable:
• The particular outcome of an experiment, determined by chance
• Discrete Probability Distribution:
•
•
•
•
Displays all possible outcomes and the probability of each outcome
Sum of the probabilities of all possible outcomes must equal 1.
The individual P(X) must be between 0 and 1
The outcomes must not overlap (discrete outcomes)
Mean or Expectation
• Mean (Expectation) of a Discrete Distribution
• Measures the center of the distribution
• Represents the theoretical long-run average of the random
variable
• Weighted average where each outcome is weighted by the
probability of the outcome or X * P(X)
• Also known as the “Expected Value” or “Expectation.”
• Calculated as follows:
𝜇 = Σ [𝑥 ∗ 𝑃 𝑥 ]
X
$ 200
$ 25
$ (85)
P(X)
0.30
0.50
0.20
X * P(X)
$ 60.00
$ 12.50
$(17.00)
μ = $ 55.50
Mean of Discrete Distribution
Variance of Discrete Distribution
• The variance of the distribution describes the “spread”
• The standard deviation of the distribution can be used to
compare two distributions with the same mean to determine
which is the most variable
𝜎2 =
X
$ 200
$ 25
$ (85)
P(X)
X * P(X)
0.30 $ 60.00
0.50 $ 12.50
0.20 $(17.00)
μ = $ 55.50
[ 𝑥 − 𝜇
X-μ
$ 144.50
$ (30.50)
$(140.50)
2
𝑥 𝑃(𝑋)]
(X - μ)2
$20,880.25
$
930.25
$19,740.25
(X - μ)2 x P(X)
$
$
$
$
6,264.08
465.13
3,948.05
10,677.25
$
103.33
Variance
Standard
Deviation
The Binomial Distribution
The binomial experiment is a probability
experiment that satisfies these
requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain
the same for each trial.
The Binomial Distribution
In a binomial experiment, the probability of
exactly X successes in n trials is
n!
X
n X
P X  
 p q
 n - X ! X !
•
•
•
•
•
n = the number of trials
X = number of successes desired
p = probability of success
q = probability of failure
Where p + q = 1
The Binomial Distribution
A local textbook broker determines that 30% of all students purchase
used textbooks for their classes. If 5 students come into the store,
what is the probability that at least 3 will purchase used books?
n  5, p  0.30,"at least 3"  X  3, 4,5
5!
3
2
P  3 
  0.30    0.70   0.132
2!3!
5!
4
1
P  4 
  0.30    0.70   0.028
1!4!
5!
5
0
P  5 
  0.30    0.70   0.002
0!5!
The Binomial Distribution
To determine the probability of “at least” three,
determine the cumulative probabilities of:
[P(X) = 3] + [P(X) = 4] + [P(X) = 5]
P  X  3  0.132
0.028
0.002
 0.162
Answer: The probability that at least 3 out of 5 students will purchase used
textbooks is 16.2%.
The Binomial Distribution
To determine the probability of “at least” three,
determine the cumulative probabilities of:
[P(X) = 3] + [P(X) = 4] + [P(X) = 5]
P  X  3  0.132
0.028
0.002
 0.162
Answer: The probability that at least 3 out of 5 students will purchase used
textbooks is 16.2%.
The Binomial Distribution
1. Open MINITAB
2. Select “Calc” from Menu
3. Select “Probability
Distributions”
4. Select “Binomial”
The Binomial Distribution
1. Select “Cumulative
Distribution”
2. Number of Trials = “n”
3. Event probability = “p”
4. Input Constant – The value
and all below that value to
be EXCLUDED from the
calculation of the Binomial
Probability.
MINTAB RESULTS: P(X = 0, 1 and 2) = 0.83692
Solving for P(X ≥ 3) as 1.00 – 0.83692 = 0.16308 ≈ 16.31% probability that
at least 3 students out of the 5 students rented their textbooks.
The Binomial Distribution
1.
2.
3.
4.
Select “Probability”
Number of Trials = “n”
Event probability = “p”
Input Constant – The single
value for which you are
finding the probability such
as P(X = 3), or the
probability of exactly 3.
MINTAB RESULTS: P(X = 3) = 0.1323 ≈ There is a 13.23% probability that
exactly 3 students out of the 5 students rented their textbooks.
The Binomial Distribution
Find the probability for the following:
1. n = 10
p = .87 P(X = 6)
2. n = 15
p = .20 P(4 < X < 7)
3. n = 15
p = .20 P(X = 0)
4. n = 18
p = .30 P(X > 8)
5. n = 20
p = .90 P(X < 16)
6. n = 20
p = .20 P(X > 8)
7. n = 20
p= .70
P(X < 12)
The Binomial Distribution
The mean, variance, and standard deviation
of a variable that follows a binomial
distribution :
Mean:   np
Variance:   npq
2
Standard Deviation:   npq
The Binomial Distribution
The Automobile Dealers Association reports that in
2013 of the first 8,000 hybrid vehicles produced 2%
of them had a fault in the recharger circuit. What is
the mean, variance and standard deviation for this
data?
  np  8000  0.02   160
 2  npq  8000  0.02  0.98  156.8  157
  npq  8000  0.02  0.98  12.5  13
The Poisson Distribution
The Poisson distribution describes events
that will occur over an interval: time, distance,
area or volume. In a Poisson Distribution:
1.
The events are independent
2.
The probability is proportional to the interval
or “the longer the interval the higher the
probability.”
It is also considered the “probability of
improbable events” where n is large and p is
small
The Poisson Distribution
In a Poisson Distribution, the probability of exactly X
events over the interval of interest is:
e   X
P X ; 
X!
• X = number of occurrences of events or
“successes” desired
• λ = mean number of occurrences over
the interval
• e = Base of Napierian Logarithm, a
constant equal to 2.71828
The Poisson Distribution
The mean (λ) of a Poisson Distribution is found as:
𝜇 =𝑛∗𝑝
n = number of trials and p = probability of success
• The variance of a Poisson Distribution is equal to the
mean
• The random variable X in a Poisson Distribution can
assume an infinite number of values, thereby having
no upper limit
• The Poisson Distribution is positively skewed
The Poisson Distribution
The mean (λ) of a Poisson Distribution is proportional to the
interval of occurrence, which requires an adjustment of the value
of λ for intervals OTHER than the “original” interval.
Example:
λ = 10 defects per 5,000 units produced
to find P(Defects) per 2,500 units produced:
to find P(Defects) per 1,000 units produced:
to find P(Defects) per 10,000 units produced:
to find P(Defects) per 15,000 units produced:
λ=5
λ=2
λ = 20
λ = 30
The Poisson Distribution
If there are 200 typographical errors randomly distributed in a 500page manuscript, find the probability that a given page contains
exactly 3 errors.
First, find the mean number of errors. With 200 errors distributed
over 500 pages, each page has an average of how many
errors per page.

e 
P X ; 
X!
X

e
0.4
 0.4 
3!
3
 0.0072
Thus, there is less than 1% chance that any given page
will contain exactly 3 errors.
The Poisson Distribution
1. Open MINITAB
2. Select “Calc” from Menu
3. Select “Probability
Distributions”
4. Select “Poisson”
The Poisson Distribution
1. Select “Cumulative
Distribution”
2. Mean = “λ”
3. Input Constant – The value
and all below that value to
be EXCLUDED from the
calculation of the Poisson
Probability.
MINTAB RESULTS: P(X = 0, 1 and 2) = 0.99274
Solving for P(X ≥ 3) as 1.00 – 0.99274 = 0.00726 ≈ 0.726% probability that
there will be at least 3 errors on a given page of a 500 page manuscript.
The Poisson Distribution
1. Select “Probability”
2. Mean = λ
3. Input Constant – The single
value for which you are
finding the probability such
as P(X = 3), or the
probability of exactly 3.
MINTAB RESULTS: P(X = 3) = 0. 0071501 ≈ There is a .715 % probability
that there will be exactly 3 errors on any given page of a 500 page
manuscript.
The Poisson Distribution
• The Securities and Exchange Commission has determined that the number
of companies listed in NYSE declaring bankruptcy is approximately a
Poisson distribution with a mean of 2.6 per month. Find the probability
that exactly 4 bankruptcies occur next month.
• Assume the number of trucks passing an intersection has a Poisson
distribution with mean of 5 trucks per minute. What is the probability of
0 or 1 trucks in one minute?
•
During off hours, cars arrive at a tollbooth on the East-West toll road
at an average rate of 0.5 cars per minute. The arrivals are distributed
according to the Poisson distribution. What is the probability that
during the next five minutes three cars will arrive?