Ken Black QA ch06

8/3/2019 Ken Black QA ch06

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Business Statistics, 5th ed.

by Ken Black

Chapter 6

ContinuousDistributions

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh,

Morehead State University


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Learning Objectives

Understand concepts of the uniformdistribution.

Appreciate the importance of the normaldistribution.

Recognize normal distribution problems, andknow how to solve them. Decide when to use the normal distribution to

approximate binomial distribution problems,and know how to work them.

Decide when to use the exponential distributionto solve problems in business, and know how towork them.


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Uniform Distribution

f xb a

for a x b

for

( )

1

0 all other values

Area = 1

f x( )

x

1

b a

a b


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Uniform Distribution of Lot Weights

f x

for x

for

( )

1

47 4141 47

0 all other values

Area = 1

f x( )

x

1

47 41

1

6

41 47


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Uniform Distribution Probability

P Xb a

x xx x( )

1 2

2 1

P X( )42 4545 42

47 41

1

2

42 45

f x( )

x41 47

45 42

47 41

1

2

Area

= 0.5


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Mean and Standard Deviation

Mean

=+

a b

2

Mean

=+

41 47

2

88

244

Standard Deviation

b a12

Standard Deviation

47 4112

63 464

1 732.

.


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Uniform Distribution of Assembly of

Plastic Modules

esother valuall0

392712

1

2739

1

)(

for

xfor

xf

Area = 1

f x( )

x27 39

12

1

2739

1


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Mean and Standard Deviation

Mean

=+

a b

233

2

72+93=

Mean

Standard Deviation

b a12

464.3464.312

122739

DeviationStandard


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Plastic Modules

4167.012

5

2739

3035)3530(

XP

42 45f x( )

x27 39

4167.02739

3035

30 35


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Plastic Modules

2500.012

3

2739

2730)30(

XP

f x( )

x27 3930

4167.02739

3035


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Normal Distribution

Probably the most widely known and used ofall distributions is the normal distribution.

It fits many human characteristics, such asheight, weigh, length, speed, IQ scores,scholastic achievements, and years of lifeexpectancy, among others.

Many things in nature such as trees, animals,insects, and others have many characteristics

that are normally distributed.


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Normal Distribution

Many variables in business and industry arealso normally distributed. For examplevariables such as the annual cost of householdinsurance, the cost per square foot of renting

warehouse space, and managers satisfactionwith support from ownership on a five-pointscale, amount of fill in soda cans, etc.

Because of the many applications, the normal

distribution is an extremely importantdistribution.


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Normal Distribution

Discovery of the normal curve of errors isgenerally credited to mathematician andastronomer Karl Gauss (17771855), whorecognized that the errors of repeatedmeasurement of objects are often normallydistributed.

Thus the normal distribution is sometimesreferred to as the Gaussian distribution or the

normal curve of errors.

In addition, some credit were also given toPierre-Simon de Laplace (17491827) andAbraham de Moivre (16671754) for thediscovery of the normal distribution.


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Properties of the Normal Distribution

The normal distribution exhibits the followingcharacteristics:

It is a continuous distribution.

It is symmetric about the mean. It is asymptotic to the horizontal axis.

It is unimodal.

It is a family of curves.

Area under the curve is 1. It is bell-shaped.


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Graphic Representation of the Normal

Distribution


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Probability Density of the Normal

Distribution


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Family of Normal Curves


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Standardized Normal Distribution

Since there is an infinite number ofcombinations for and , then we can generatean infinite family of curves.

Because of this, it would be impractical to dealwith all of these normal distributions.

Fortunately, a mechanism was developed bywhich all normal distributions can beconverted into a single distribution called the

z distribution. This process yields the standardized normal

distribution (or curve).


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The conversion formula for any x value of agiven normal distribution is given below. It iscalled the z-score.

A z-score gives the number of standarddeviations that a valuex, is above or below themean.

xz


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Ifx is normally distributed with a mean ofand a standard deviation of, then the z-scorewill also be normally distributed with a meanof 0 and a standard deviation of 1.

Since we can covert to this standard normal

distribution, tables have been generated for thisstandard normal distribution which will enableus to determine probabilities for normalvariables.

The tables in the text are set up to give the

probabilities between z = 0 and some other zvalue, z0 say, which is depicted on the nextslide.


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Z TableSecond Decimal Place in Z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.10 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.20 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.30 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.90 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.00 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.10 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.20 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

2.00 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

3.00 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.49903.40 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.49983.50 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998


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Applying the Z Formula

X is normally distributed with = 485, and = 105

P X P Z( ) ( . ) .485 600 0 1 10 3643

For X = 485,

Z =X -

485 485

1050

For X = 600,

Z =X -

600 485

1051 10.

Z 0.00 0.01 0.02

0.00 0.0000 0.0040 0.00800.10 0.0398 0.0438 0.0478

1.00 0.3413 0.3438 0.3461

1.10 0.3643 0.3665 0.3686

1.20 0.3849 0.3869 0.3888


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7123.)56.0()550(

100=and494,=withddistributenormallyisX

ZPXP

56.0100

494550-X=Z

550=XFor

0.5 + 0.2123 = 0.7123


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0197.)06.2()700(


ZPXP

06.2100

494700-X=Z

700=XFor

0.5 0.4803 = 0.0197


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94.1100

494300-X=Z

300=XFor

8292.)06.194.1()600300(


ZPXP

0.4738+ 0.3554 = 0.8292

06.1100

494600-X=Z

600=XFor


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These types of problems can be solved quiteeasily with the appropriate technology. Theoutput shows the MINITAB solution.

Demonstration Problem 6.9


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Normal Approximation

of the Binomial Distribution

The normal distribution can be used toapproximate binomial probabilities.

Procedure

Convert binomial parameters to normalparameters.

Does the interval 3 lie between 0 and n?If so, continue; otherwise, do not use thenormal approximation.

Correct for continuity.

Solve the normal distribution problem.


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Conversion equations

Conversion example:

Normal Approximation of Binomial:Parameter Conversion

n p

n p q

Given that X has a binomial distribution, find

andP X n p

n p

n p q

( | . ).

( )(. )

( )(. )(. ) .

25 60 30

60 30 18

60 30 70 3 55


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Normal Approximation of Binomial:

Interval Check

3 18 3 355 18 10 65

3 7 35

3 28 65

( . ) .

.

.

0 10 20 30 40 50 60n

70


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Graph of the Binomial Problem:

n = 60,p = 0.3

x

P(x)

3025201510

0.12

0.10

0.08

0.06

0.04

0.02

0.00


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Correcting for Continuity

ValuesBeing

DeterminedCorrection

XXXXX

X

+.50-.50-.50+.05

-.50 and +.50

+.50 and -.50

The binomial probability,

and

is approximated by the normal probabilit

P(X 24.5| and

P X n p( | . )

. ).

25 60 30

18 3 55


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Computations

252627282930313233

Total

0.01670.00960.00520.00260.00120.00050.00020.00010.00000.0361

X P(X)

The normal approximation,

P(X 24.5| and

18 355

24 5 18

355

183

5 0 183

5 4664

0336

. )

.

.

( . )

. .

. .

.

P Z

P Z

P Z


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Exponential Distribution

Continuous

Family of distributions

Skewed to the right

Xvaries from 0 to infinity Apex is always at X = 0

Steadily decreases asXgets larger

Probability function

f X XX

e( ) ,

for 0 0


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Different Exponential Distributions


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Exponential Distribution:

Probability Computation

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3 4 5

P X XX

P X

e

e

00

2 1212 2

0907

| .( . )( )

.


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Ken Black QA ch06

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