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Statistics
Discrete Probability Distributions
1/71
Contents






Random Variables
Discrete Probability Distributions
Expected Value and Variance
Binomial Distribution
Poisson Distribution
Hypergeometric Distribution
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STATISTICS in PRACTICE


Citibank makes available a wide range of
financial services.
Citibanking’s automatic teller
machines (ATMs) located in
Citicard Banking Centers
(CBCs), let customers do all
their banking in one place with
the touch of a finger.
STATISTICS in PRACTICE


Periodic CBC capacity studies are
used to analyze customer waiting times and
to determine whether additional ATMs are
needed.
Data collected by Citibank showed that the
random customer arrivals followed a
probability distribution known as the
Poisson distribution.
Random Variables



A random variable is a numerical
description of the outcome of an experiment.
A discrete random variable may assume
either a finite number of values or an infinite
sequence of values.
A continuous random variable may assume
any numerical value in an interval or
collection of intervals.
Differences between outcomes
and random variables
Example: Tossing a dice
Possible outcomes: 1, 2, 3, 4, 5, and 6
One can define random variables as
1 if outcome is greater than 3, and
0 if outcome is smaller or equal to 3.
 Or
1 if outcome is odd numbers
0 if outcome is even numbers

Discrete Random Variables

Example 1.
The certified public accountant (CPA)
examination has four parts.
Define a random variable as x = the number of
parts of the CPA examination passed and It is
a discrete random variable because it may
assume the finite number of values 0, 1, 2, 3,
or 4.
Discrete Random Variables
Example 2.
An experiment of cars arriving at a
tollbooth. The random variable is x = the
number of cars arriving during a one-day
period. The possible values for x come from
the sequence of integers 0, 1, 2, and so on.
x is a discrete random variable assuming
one of the values in this infinite sequence.
Discrete Random Variables

Examples of Discrete Random Variables
Example: JSL Appliances

Discrete random variable with a finite
number of values
Let x = number of TVs sold at the store in
one day, where x can take on 5 values
(0, 1, 2, 3, 4)
Example: JSL Appliances

Discrete random variable with an
infinite sequence of values
Let x = number of customers arriving in one day,
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there
is no finite upper limit on the number that might
arrive.
Random Variables
Question
Family
size
Random Variable x
x = Number of dependents
reported on tax return
Distance from x = Distance in miles from
home to store
home to the store site
Own dog
or cat
Type
Discrete
Continuous
Discrete
x = 1 if own no pet;
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
Continuous Random Variables
Example 1.
Experimental outcomes based on
measurement scales such as time, weight,
distance, and temperature can be described
by continuous random variables.
Continuous Random Variables
Example 2.
An experiment of monitoring incoming
telephone calls to the claims office of a
major insurance company. Suppose the
random variable of interest is x = the time
between consecutive incoming calls in
minutes. This random variable may
assume any value in the interval x ≥ 0.
Continuous Random Variables

Example of Continuous Random Variables
Discrete Probability Distributions

The probability distribution for a random
variable describes how probabilities are
distributed over the values of the random
variable.

We can describe a discrete probability
distribution with a table, graph, or equation.
Discrete Probability Distributions

The probability distribution is defined by a
probability function, denoted by f(x), which
provides the probability for each value of
the random variable.
The required conditions for a discrete
probability function are:
f(x) > 0
f(x) = 1
Discrete Probability Distributions

Example: Probability Distribution for the
Number of Automobiles Sold During a Day
at Dicarlo Motors.
Discrete Probability
Distributions
Using past data on TV sales, …
 a tabular representation of the probability
distribution for TV sales was developed.
Number
x
f(x) 80/200

Units Sold
0
1
2
3
4
of Days
80
50
40
10
20
200
0
1
2
3
4
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Discrete Probability
Distributions
Graphical Representation of Probability
Distribution
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Probability

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Values of Random Variable x (TV sales)
Discrete Uniform Probability
Distribution
The discrete uniform probability distribution
is the simplest example of a discrete
probability distribution given by a formula.
Discrete Uniform Probability
Distribution
The discrete uniform probability function is
f(x) = 1/n
the values of the
random variable
are equally likely
where:
n = the number of values the random
variable may assume
Discrete Uniform Probability
Distribution

Example:
An experiment of rolling a die we define
the random variable x to be the number of
dots on the upward face. There are n = 6
possible values for the random variable;
x = 1, 2, 3, 4, 5, 6. The probability function
for this discrete uniform random variable is
f (x) = 1/6 x = 1, 2, 3, 4, 5, 6.
Discrete Uniform Probability
Distribution
x
1
2
3
4
5
6
f (x)
1/6
1/6
1/6
1/6
1/6
1/6
Discrete Uniform Probability
Distribution

Example:
Consider the random variable x with the
following discrete probability distribution.
x
1
2
3
4
f (x)
1/10
2/10
3/10
4/10
This probability distribution can be defined by
the formula f (x) = x/ 10 for x = 1, 2, 3, or 4.
Expected Value and Variance

The expected value, or mean, of a random
variable is a measure of its central location.
E(x) =  = x f(x)

The variance summarizes the variability in the
values of a random variable.
Var(x) =  2 = (x - )2f(x)
Expected Value and Variance


The standard deviation,  , is defined
as the positive square root of the variance.
Here, the expected value and variance
are computed from random variables
instead of outcomes
Expected Value and Variance

Example: Calculation of the Expected
Value for the Number of Automobiles Sold
During A Day at Dicarlo Motors.
Expected Value and Variance

Example: Calculation of the Variance for
the Number of Automobiles Sold During A
Day at Dicarlo Motors.
The standard deviation is   1.25  1.118
Expected Value and Variance

Expected Value
x
0
1
2
3
4
expected number of
TVs sold in a day
f(x)
xf(x)
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.10
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E(x) = 1.20
Expected Value and Variance

Variance and Standard Deviation
x
x-
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
(x - )2
1.44
0.04
0.64
3.24
7.84
f(x)
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(x - )2f(x)
.576
.010
.128
.162
.784
Variance of daily sales =  2 = 1.660
TVs
squared
Standard deviation of daily sales = 1.2884 TVs
Binomial Distribution

Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
2. Two outcomes, success and failure, are possible
on each trial.
3. The probability of a success, denoted by p, does
not change from trial to trial.
stationarity
assumption
4. The trials are independent.
Binomial Distribution

Our interest is in the number of successes
occurring in the n trials.

We let x denote the number of successes
occurring in the n trials.
Binomial Distribution

Binomial Probability Function
where:
f(x) = the probability of x successes in
n trials,
n = the number of trials,
p = the probability of success on any
one trial.
Binomial Distribution
Number of experimental
n!
x !(n  x )! = outcomes providing exactly
x successes in n trials
(nx )
p (1  p)
x
= Probability of a particular
sequence of trial outcomes
with x successes in n trials
Binomial Distribution

Binomial Probability Function
Binomial Distribution


Example:
The experiment of tossing a coin five times
and on each toss observing whether the
coin lands with a head or a tail on its
upward face. we want to count the number
of heads appearing over the five tosses.
Does this experiment show the properties of
a binomial experiment?
Binomial Distribution

Note that:
1. The experiment consists of five identical
trials; each trial involves the tossing of one
coin.
2. Two outcomes are possible for each trial: a
head or a tail. We can designate head a
success and tail a failure.
Binomial Distribution
Note that:
3. The probability of a head and the probability of
a tail are the same for each trial, with p = .5
and 1- p = .5.
4. The trials or tosses are independent because the
outcome on any one trial is not affected by
what happens on other trials or tosses.

Binomial Distribution

Example: Evans Electronics
Evans is concerned about a low
retention rate for employees. In recent
years, management has seen a turnover of
10% of the hourly employees annually.
Thus, for any hourly employee chosen at
random, management estimates a
probability of 0.1 that the person will not be
with the company next year.
Binomial Distribution

Using the Binomial Probability Function
Choosing 3 hourly employees at random,
what is the probability that 1 of them will
leave the company this year?
Let: p = .10, n = 3, x = 1
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
3!
f (1) 
(0.1)1 (0.9)2  3(.1)(.81)  .243
1!(3  1)!
Binomial Distribution

T
r
e
e
D
i
a
g
r
a
m
1st Worker
2nd Worker
Leaves (.1)
Leaves
(.1)
3rd Worker
L (.1)
x
3
Prob.
.0010
S (.9)
2
.0090
L (.1)
2
.0090
S (.9)
1
.0810
L (.1)
2
.0090
S (.9)
1
.0810
1
.0810
0
.7290
Stays (.9)
Leaves (.1)
Stays
(.9)
L (.1)
Stays (.9)
S (.9)
Binomial Distribution

Using Tables of Binomial Probabilities
p
n
x
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.50
3
0
1
2
3
.8574
.1354
.0071
.0001
.7290
.2430
.0270
.0010
.6141
.3251
.0574
.0034
.5120
.3840
.0960
.0080
.4219
.4219
.1406
.0156
.3430
.4410
.1890
.0270
.2746
.4436
.2389
.0429
.2160
.4320
.2880
.0640
.1664
.4084
.3341
.0911
.1250
.3750
.3750
.1250
Binomial Distribution

Expected Value
E(x) =  = np

Variance
Var(x) =  2 = np(1  p)

Standard Deviation
  np(1  p )
Binomial Distribution

Expected Value
E(x) =  = 3(.1) = .3 employees out of 3

Variance
Var(x) =  2 = 3(.1)(.9) = .27

Standard Deviation
  3(.1)(.9)  .52 employees
Poisson Distribution

A Poisson distributed random variable is often
useful in estimating the number of occurrences
over a specified interval of time or space

It is a discrete random variable that may
assume an infinite sequence of values
(x = 0, 1, 2, . . . ).
Poisson Distribution

Examples of a Poisson distributed random variable:

the number of knotholes in 14 linear feet of
pine board

the number of vehicles arriving at a
toll booth in one hour
Poisson Distribution

Two Properties of a Poisson Experiment
1. The probability of an occurrence is the
same for any two intervals of equal length.
2. The occurrence or nonoccurrence in any
interval is independent of the occurrence
or nonoccurrence in any other interval.
Poisson Distribution

Poisson Probability Function
f ( x) 
 x e
x!
where:
f(x) = probability of x occurrences in
an interval,
 = mean number of occurrences in
an interval,
e = 2.71828.
Poisson Distribution

Example: We are interested in the number of
arrivals at the drive-up teller window of a
bank during a 15-minute period on weekday
mornings. Assume that the probability of a
car arriving is the same for any two time
periods of equal length and that the arrival or
nonarrival of a car in any time period is
independent of the arrival or nonarrival in any
other time period.
Poisson Distribution
The Poisson probability function is applicable.
Suppose that the average number of cars arriving
in a 15-minute period of time is 10; in this case,
the following probability function applies.

x
10
10 e
f ( x) 
x!
The random variable here is
x = number of cars arriving in any 15-minute
period.
Poisson Distribution

Example: Mercy Hospital
Patients arrive at the emergency
room of Mercy Hospital at the average rate of
6 per hour on weekend evenings.
What is the probability of 4 arrivals in 30
minutes on a weekend evening?
MERCY
Poisson Distribution

Using the Poisson Probability Function
 = 6/hour = 3/half-hour, x = 4
3 4 (2.71828)3
f (4) 

4!
.1680
MERCY
Poisson Distribution

Using Poisson Probability Tables

x
0
1
2
3
4
5
6
7
8
2.1
.1225
.2572
.2700
.1890
.0992
.0417
.0146
.0044
.0011
2.2
.1108
.2438
.2681
.1966
.1082
.0476
.0174
.0055
.0015
2.3
.1003
.2306
.2652
.2033
.1169
.0538
.0206
.0068
.0019
2.4
.0907
.2177
.2613
.2090
.1254
.0602
.0241
.0083
.0025
2.5
.0821
.2052
.2565
.2138
.1336
..0668
.0278
.0099
.0031
2.6
.0743
.1931
.2510
.2176
.1414
.0735
.0319
.0118
.0038
2.7
.0672
.1815
.2450
.2205
.1488
.0804
.0362
.0139
.0047
2.8
.0608
.1703
.2384
.2225
.1557
.0872
.0407
.0163
.0057
2.9
.0550
.1596
.2314
.2237
.1622
.0940
.0455
.0188
.0068
3.0
.0498
.1494
.2240
.2240
.1680
.1008
.0504
.0216
.0081
MERCY
Poisson Distribution
Poisson Distribution of Arrivals
Poisson Probabilities
0.25
Probability

0.20
actually,
the sequence
continues:
11, 12, …
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
Number of Arrivals in 30 Minutes
10
Poisson Distribution
A property of the Poisson distribution is that
the mean and variance are equal.

=2
MERCY
Poisson Distribution

Variance for Number of Arrivals
During 30-Minute Periods
=2=3
Hypergeometric Distribution
The hypergeometric distribution is closely related
to the binomial distribution.
However, for the hypergeometric distribution:
the trials are not independent, and
the probability of success changes from trial
to trial.
Hypergeometric Distribution

Hypergeometric Probability Function
 r  N  r 
 

x  n  x  for 0 < x < r

f ( x) 
N
 
n
where: f(x) = probability of x successes in n trials,
n = number of trials,
N = number of elements in the population,
r = number of elements in the population
labeled success.
Hypergeometric Distribution

Hypergeometric Probability Function
for 0 < x < r
Hypergeometric Distribution

Hypergeometric Probability Function
r 
number of ways x successes can be
  = selected from a total of r successes
 x
in the population
 N  r  number of ways n – x failures can be

 = selected from a total of N – r failures
 n  x  in the population
number of ways
N
  = a sample of size n can be selected
n 
from a population of size N
Hypergeometric Distribution


Example: A Quality Control Application.
Electric fuses produced by Ontario Electric are
packaged in boxes of 12 units each. Suppose an
inspector randomly selects 3 of the 12 fuses in a
box for testing.
If the box contains exactly 5 defective fuses,
what is the probability that the inspector will find
exactly 1 of the 3 fuses defective?
Hypergeometric Distribution

In this application, n = 3 and N = 12.
With r = 5 defective fuses in the box.
Hypergeometric Distribution

The probability of finding x = 1 defective
fuse is

What is the probability of finding at least 1
defective fuse?
Hypergeometric Distribution

The probability of x = 0 is

we conclude that the probability of finding
at least 1 defective fuse must be 1 - .1591 =
.8409.
Hypergeometric Distribution

Example: Neveready
Bob Neveready has removed two
dead batteries from a flashlight
and inadvertently mingled
them with the two good
batteries he intended
as replacements. The four batteries look
identical. Bob now randomly selects two of
the four batteries. What is the probability
he selects the two good batteries?
Hypergeometric Distribution

Using the Hypergeometric Function
where:
x = 2 = number of good batteries selected
n = 2 = number of batteries selected
N = 4 = number of batteries in total
r = 2 = number of good batteries in total
Hypergeometric Distribution

Mean

Variance
Hypergeometric Distribution

Mean

Variance
Hypergeometric Distribution


Consider a hypergeometric distribution with
n trials and let p = (r/n) denote the probability
of a success on the first trial.
If the population size is large, the term
(N – n)/(N – 1) approaches 1.
Hypergeometric Distribution

The expected value and variance can be written
E(x) = np and Var(x) = np(1 – p).

Note that these are the expressions for the
expected value and variance of a binomial
distribution.
continued
Hypergeometric Distribution

When the population size is large, a
hypergeometric distribution can be approximated
by a binomial distribution with n trials and a
probability of success p = (r/N).