chapter 5 student lecture notes 5-1 · probability distributions normal uniform exponential....

Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc.

Chapter 5 Student Lecture Notes 5-1

Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 5-1

Business Statistics: A Decision-Making Approach

6th Edition

Chapter 5Discrete and Continuous Probability Distributions


Chapter Goals

After completing this chapter, you should be able to:Apply the binomial distribution to applied problemsCompute probabilities for the Poisson and hypergeometric distributions

Find probabilities using a normal distribution table and apply the normal distribution to business problems

Recognize when to apply the uniform and exponential distributions


Probability Distributions

ContinuousProbability

Distributions

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

Normal

Uniform

Exponential




A discrete random variable is a variable that can assume only a countable number of valuesMany possible outcomes:

number of complaints per daynumber of TV’s in a householdnumber of rings before the phone is answered

Only two possible outcomes:gender: male or femaledefective: yes or nospreads peanut butter first vs. spreads jelly first

Discrete Probability Distributions


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 itemtime required to complete a tasktemperature of a solutionheight, in inches

These can potentially take on any value, depending only on the ability to measure accurately.


The Binomial Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions




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 trialsThe trials of the experiment are independent of each otherThe probability of a success, p, remains constant from trial to trialIf p represents the probability of a success, then (1-p) = q is the probability of a failure


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


Counting Rule for Combinations

A combination is an outcome of an experiment where x objects are selected from a group of n objects

)!xn(!x!nCn

x −=

where:n! =n(n - 1)(n - 2) . . . (2)(1)x! = x(x - 1)(x - 2) . . . (2)(1)

0! = 1 (by definition)




P(x) = probability of x successes in n trials,with probability of success p on each trial

x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n)

p = probability of “success” per trialq = probability of “failure” = (1 – p)n = number of trials (sample size)

P(x)n

x ! n xp qx n x!

( )!=

−−

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

Binomial Distribution Formula


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

Binomial Distribution

The shape of the binomial distribution depends on the values of p and n

Here, n = 5 and p = .1

Here, n = 5 and p = .5


Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

npE(x)µ ==

npqσ2 =

npqσ =

Where n = sample sizep = probability of successq = (1 – p) = probability of failure




n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

0.5(5)(.1)npµ ===

0.6708

.1)(5)(.1)(1npqσ

=

−==

2.5(5)(.5)npµ ===

1.118.5)(5)(.5)(1npqσ

=

−==

Binomial Characteristics

Examples


Using Binomial Tables

xp=.50p=.55p=.60p=.65p=.70p=.75p=.80p=.85

109876543210

0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010

0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003

0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001

0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000

0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000

0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000

0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000

0.19690.34740.27590.12980.04010.00850.00120.00010.00000.00000.0000

012345678910

p=.50p=.45p=.40p=.35p=.30p=.25p=.20p=.15x

n = 10

Examples: n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522

n = 10, p = .75, x = 2: P(x = 2|n =10, p = .75) = .0004


Using PHStat

Select PHStat / Probability & Prob. Distributions / Binomial…




Using PHStat

Enter desired values in dialog box

Here: n = 10p = .35

Output for x = 0 to x = 10 will be generated by PHStat

Optional check boxesfor additional output


P(x = 3 | n = 10, p = .35) = .2522

PHStat Output

P(x > 5 | n = 10, p = .35) = .0949


The Poisson Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions




The Poisson Distribution

Characteristics of the Poisson Distribution:The outcomes of interest are rare relative to the possible outcomesThe average number of outcomes of interest per time or space interval is λThe number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interestThe probability of that an outcome of interest occurs in a given segment is the same for all segments


Poisson Distribution Formula

where:t = size of the segment of interest x = number of successes in segment of interestλ = expected number of successes in a segment of unit sizee = base of the natural logarithm system (2.71828...)

!xe)t()x(P

tx λ−λ=


Poisson Distribution Characteristics

Mean

Variance and Standard Deviation

λtµ =

λtσ2 =

λtσ =where λ = number of successes in a segment of unit size

t = the size of the segment of interest




Using Poisson Tables

λt

0.90

0.40660.36590.16470.04940.01110.00200.00030.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.90480.09050.00450.00020.00000.00000.00000.0000

01234567

0.800.700.600.500.400.300.200.10X

Example: Find P(x = 2) if λ = .05 and t = 100

.07582!

e(0.50)!xe)t()2x(P

0.502tx

==λ

==−λ−


Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

01234567

X0.60650.30330.07580.01260.00160.00020.00000.0000

λt =0.50

P(x = 2) = .0758

Graphically:λ = .05 and t = 100


Poisson Distribution Shape

The shape of the Poisson Distribution depends on the parameters λ and t:

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

λt = 0.50 λt = 3.0




The Hypergeometric Distribution

Binomial

Poisson


DiscreteProbability

Distributions

Hypergeometric


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 “X” successes in the population


Hypergeometric Distribution Formula

Nn

Xx

XNxn

CCC)x(P

−−=

.

WhereN = Population sizeX = number of successes in the populationn = sample sizex = number of successes in the sample

n – x = number of failures in the sample

(Two possible outcomes per trial)




Hypergeometric Distribution Formula

0.3120

(6)(6)C

CCC

CC2)P(x 103

42

61

Nn

Xx

XNxn =====

−−

■ 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 = 3X = 4 x = 2


Hypergeometric Distribution in PHStat

Select:PHStat / Probability & Prob. Distributions / Hypergeometric …


Hypergeometric Distribution in PHStat

Complete dialog box entries and get output …

N = 10 n = 3X = 4 x = 2

P(x = 2) = 0.3

(continued)




The Normal Distribution


Distributions


Normal

Uniform

Exponential


The Normal Distribution

‘Bell Shaped’SymmetricalMean, Median and Mode

are EqualLocation is determined by the mean, µSpread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + ∞ to − ∞

Mean = Median = Mode

x

f(x)

µ

σ


By varying the parameters µ and σ, we obtain different normal distributions

Many Normal Distributions




The Normal Distribution Shape

x

f(x)

µ

σ

Changing µ shifts the distribution left or right.

Changing σ increases or decreases the spread.


Finding Normal Probabilities

Probability is the area under thecurve!

a b x

f(x) P a x b( )≤ ≤

Probability is measured by the area under the curve


f(x)

xµ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)xP( =∞<<−∞

0.5)xP(µ =∞<<0.5µ)xP( =<<−∞




Empirical Rules

µ ± 1σ encloses about 68% of x’s

f(x)

xµ µ+1σµ−1σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%


The Empirical Rule

µ ± 2σ covers about 95% of x’s

µ ± 3σ covers about 99.7% of x’s

xµ2σ 2σ

xµ3σ 3σ

95.44% 99.72%

(continued)


Importance of the Rule

If a value is about 2 or more standarddeviations away from the mean in a normaldistribution, then it is far from the mean

The chance that a value that far or farther away from the mean is highly unlikely, giventhat particular mean and standard deviation




The Standard Normal Distribution

Also known as the “z” distributionMean is defined to be 0Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values, values below the mean have negative z-values


The Standard Normal

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normaldistribution (z)

Need to transform x units into z units


Translation to the Standard Normal Distribution

Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation:

σµxz −

=




Example

If x is distributed normally with mean of 100and standard deviation of 50, the z value for x = 250 is

This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

3.050

100250σ

µxz =−

=−

=


Comparing x and z units

z100

3.00250 x

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)

µ = 100

σ = 50


The Standard Normal Table

The Standard Normal table in the textbook (Appendix D)

gives the probability from the mean (zero)up to a desired value for z

z0 2.00

.4772Example:P(0 < z < 2.00) = .4772




The Standard Normal Table

The value within the table gives the probability from z = 0up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = .4772

The row shows the value of z to the first decimal point

The column gives the value of z to the second decimal point

2.0

.

..

(continued)


General Procedure for Finding Probabilities

Draw the normal curve for the problem interms of x

Translate x-values to z-values

Use the Standard Normal Table

To find P(a < x < b) when x is distributed normally:


Z Table example

Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.120x8.68

05

88σ

µxz =−

=−

=

0.125

88.6σ

µxz =−

=−

=

Calculate z-values:




Z Table example

Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.120x8.68

P(8 < x < 8.6)

µ = 8σ = 5

µ = 0σ = 1

(continued)


Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

.0478.02

0.1 .0478

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.12)P(8 < x < 8.6)



Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(x < 8.6)

Z

8.68.0





Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(x < 8.6)

(continued)

Z

0.12

.0478

0.00

.5000P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= .5 + .0478 = .5478


Upper Tail Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(x > 8.6)

Z

8.68.0


Now Find P(x > 8.6)…(continued)

Z

0.120

Z

0.12

.0478

0

.5000 .50 - .0478 = .4522

P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= .5 - .0478 = .4522

Upper Tail Probabilities




Lower Tail Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(7.4 < x < 8)

Z

7.48.0


Lower Tail Probabilities

Now Find P(7.4 < x < 8)…

Z

7.48.0

The Normal distribution is symmetric, so we use the same table even if z-values are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= .0478

(continued)

.0478


Normal Probabilities in PHStat

We can use Excel and PHStat to quicklygenerate probabilities for any normaldistribution

We will find P(8 < x < 8.6) when x isnormally distributed with mean 8 andstandard deviation 5




PHStat Dialogue Box

Select desired options and enter values


PHStat Output


The Uniform Distribution


Distributions


Normal

Uniform

Exponential




The Uniform Distribution

The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable


The Continuous Uniform Distribution:

otherwise 0

bxaifab

1≤≤

−

wheref(x) = value of the density function at any x valuea = lower limit of the intervalb = upper limit of the interval

The Uniform Distribution(continued)

f(x) =


Uniform Distribution

Example: Uniform Probability DistributionOver the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

x

f(x)




The Exponential Distribution


Distributions


Normal

Uniform

Exponential



Used to measure the time that elapses between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at an unloading dockTime between transactions at an ATM MachineTime between phone calls to the main operator



aλe1a)xP(0 −−=≤≤

The probability that an arrival time is equal to or less than some specified time a is

where 1/λ is the mean time between events

Note that if the number of occurrences per time period is Poisson with mean λ, then the time between occurrences is exponential with mean time 1/ λ




Exponential Distribution

Shape of the exponential distribution(continued)

f(x)

x

λ = 1.0(mean = 1.0)

λ= 0.5 (mean = 2.0)

λ = 3.0(mean = .333)


Example

Example: Customers arrive at the claims counter at the rate of 15 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes?

Time between arrivals is exponentially distributed with mean time between arrivals of 4 minutes (15 per 60 minutes, on average)

1/λ = 4.0, so λ = .25

P(x < 5) = 1 - e-λa = 1 – e-(.25)(5) = .7135


Chapter Summary

Reviewed key discrete distributionsbinomial, poisson, hypergeometric

Reviewed key continuous distributionsnormal, uniform, exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems

chapter 5 student lecture notes 5-1 · probability distributions normal uniform exponential....

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