1/71 Statistics Discrete Probability Distributions

1/71

Statistics

Discrete Probability Distributions

Contents

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Random Variables Discrete Probability Distributions Expected Value and Variance Binomial Distribution Poisson Distribution Hypergeometric Distribution

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.

A random variable is a numerical description of the outcome of an experiment.

Random Variables

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.

Example: Tossing a dicePossible outcomes: 1, 2, 3, 4, 5, and 6One can define random variables as 1 if outcome is greater than 3, and 0 if outcome is smaller or equal to 3.

Or1 if outcome is odd numbers0 if outcome is even numbers

Differences between outcomes and random variables

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.


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.


Examples of Discrete Random Variables

Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4)

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 a finite

number of values

Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .

Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .

Example: JSL Appliances Discrete random variable with an

infinite sequence of values

We can count the customers arriving, but thereis no finite upper limit on the number that might arrive.

Random Variables

Question Random Variable x Type

Familysize

x = Number of dependents reported on tax return

Discrete

Distance fromhome to store

x = Distance in miles from home to the store site

Continuous

Own dogor cat

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)

Discrete

Continuous Random VariablesExample 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

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.


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.

a tabular representation of the probability distribution for TV sales was developed.

Using past data on TV sales, …

Number Units Sold of Days

0 80 1 50 2 40 3 10 4 20

200

x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10 1.00

80/200


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0 1 2 3 4Values of Random Variable x (TV sales)

Pro

babili

ty


Graphical Representation of Probability Distribution

Discrete Uniform Probability Distribution

The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula.


the values of the random variable are equally likely

The discrete uniform probability function is

f(x) = 1/nwhere:

n = the number of values the random variable may assume


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.


x f (x)1 1/62 1/63 1/64 1/65 1/66 1/6

Example: Consider the random variable x with the

following discrete probability distribution.

This probability distribution can be defined by

the formula f (x) = x/ 10 for x = 1, 2, 3, or 4.


x f (x)1 1/102 2/103 3/104 4/10

Expected Value and Variance The expected value, or mean, of a random variable is a measure of its central location.

The variance summarizes the variability in the values of a random variable.

Var(x) = 2 = (x - )2f(x)Var(x) = 2 = (x - )2f(x)

E(x) = = x f(x)E(x) = = x f(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


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 118.125.1

Expected Value

expected number of TVs sold in a day

x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40

E(x) = 1.20


Variance and Standard Deviation

01234

-1.2-0.2 0.8 1.8 2.8

1.440.040.643.247.84

.40

.25

.20

.05

.10

.576

.010

.128

.162

.784

x - (x - )2 f(x) (x - )2f(x)

Variance of daily sales = s 2 = 1.660

x

TVssquared

Standard deviation of daily sales = 1.2884 TVs


Binomial Distribution Four Properties of a Binomial Experiment

3. The probability of a success, denoted by p, does not change from trial to trial.

4. The trials are independent.

2. Two outcomes, success and failure, are possible on each trial.

1. The experiment consists of a sequence of n identical trials.

stationarityassumption

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.

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.

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability Function


!!( )!

nx n x

( )(1 )x n xp p Probability of a particular sequence of trial outcomes with x successes in n trials

Number of experimental outcomes providing exactlyx successes in n trials

=

=

( )!( ) (1 )

!( )!x n xn

f x p px n x


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?


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.

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?

f xn

x n xp px n x( )

!!( )!

( )( )

1

1 23!(1) (0.1) (0.9) 3(.1)(.81) .243

1!(3 1)!f

Let: p = .10, n = 3, x = 1


1st Worker 1st Worker 2nd Worker2nd Worker 3rd Worker3rd Worker xx Prob.Prob.

Leaves (.1)Leaves (.1)

Stays (.9)Stays (.9)

33

22

00

22

22



S (.9)S (.9)



S (.9)S (.9)

S (.9)S (.9)

S (.9)S (.9)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1) .0010.0010

.0090.0090

.0090.0090

.7290.7290

.0090.0090

11

11

.0810.0810

.0810.0810

.0810.0810

11

Tree Diagram

Using Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .12501 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

p



(1 )np p

E(x) = = np

Var(x) = 2 = np(1 - p)

Expected Value

Variance

Standard Deviation


3(.1)(.9) .52 employees

E(x) = = 3(.1) = .3 employees out of 3

Var(x) = 2 = 3(.1)(.9) = .27

Expected Value

Variance

Standard Deviation

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

2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

1. The probability of an occurrence is the same for any two intervals of equal length.

Poisson Probability Function


f xex

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.

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. The random variable here is x = number of cars arriving in any 15-minute

period.


!

10)(

10

x

exf

x


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?


Using the Poisson Probability Function

4 33 (2.71828)(4) .1680

4!f

MERCY

= 6/hour = 3/half-hour, x = 4


Using Poisson Probability Tables

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

MERCY

Poisson Distribution of Arrivals

MERCY

Poisson Probabilities

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0.05

0.10

0.15

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0.25

0 1 2 3 4 5 6 7 8 9 10

Number of Arrivals in 30 Minutes

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y

actually, the sequencecontinues:11, 12, …


Poisson Distribution A property of the Poisson distribution is thatthe mean and variance are equal.

m = s 2


MERCY

Variance for Number of Arrivals

During 30-Minute Periods

m = s 2 = 3 m = s 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 Probability Function


n

N

xn

rN

x

r

xf )( for 0 < x < r

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.



( )

r N r

x n xf x

N

n

for 0 < x < r



number of ways x successes can be selected from a total of r successes in the population

number of waysa sample of size n can be selectedfrom a population of size N

x

r=

xn

rN=

number of ways n – x failures can be selected from a total of N – r failuresin the population

n

N=


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?


In this application, n = 3 and N = 12.

With r = 5 defective fuses in the box.

The probability of finding x = 1 defective fuse is

What is the probability of finding at least 1 defective fuse?


4773.220

)21)(5(

!9!3

!12!5!2

!7

!4!1

!5

3

12

2

7

1

5

)1(

f

The probability of x = 0 is

we conclude that the probability of finding at least 1 defective fuse must be 1 - .1591 = .8409.


1591.220

)35)(1(

!9!3

!12!4!3

!7

!5!0

!5

3

12

3

7

0

5

)0(

f

Hypergeometric Distribution Example: Neveready

Bob Neveready has removed twodead 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?

ZAP

ZA

P

ZAPZAP


Using the Hypergeometric Function2 2 2! 2!

2 0 2!0! 0!2! 1( ) .167

4 4! 62 2!2!

r N r

x n xf x

N

n

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


( )r

E x nN

2( ) 11

r r N nVar x n

N N N

Mean

Variance


22 1

4

rn

N

2 2 2 4 2 12 1 .333

4 4 4 1 3

Mean

Variance


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.


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


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).

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1/71 Statistics Discrete Probability Distributions