ch06busstat

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Business Statistics, 4e, by Ken Black. 2003 John Wiley & Sons.

6-1

Business Statistics, 4e

by Ken BlackChapter 6

ContinuousDistributions

In God we trust, others

must bring data(Robert Hayden, A USProfessor)

Discrete Distributions


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6-2

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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6-3

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

Uniform distribution, also called rectangular dist, in which

the function has same height over a range of values.


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6-4

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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6-5

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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6-6

Uniform Distribution

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: Applications

Variation of weights of a 400gms cereal box isbetween 380 to 420 gms. (or any bundle) couldhave uniform distribution. What is the probabilityof observing weights between 380 and 390 gms?

Variation in time that takes to complete 1000shirts. If it takes about 30 to 40 hours of work,how many would be completed between 30 to 31hours.

modeling time interval as a random variablewhere frequency could be no of customers.

As a crude measure of distribution to any probleminvolving continuous random variable.


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6-8

Characteristics of the Normal

Distribution

Continuous distribution Symmetrical distribution Asymptotic to the

horizontal axis

A family of curves Area under the curve

sums to 1. Area to right of mean is

1/2.

Area to left of mean is1/2.

1/2 1/2

X


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Normal Distribution: Example

Modern portfolio theory commonlyassumes that the returns of a diversifiedasset portfolio follow a normal distribution.

In operations management, processvariations often are normally distributed.

In human resource management, employeeperformance sometimes is considered to be

normally distributed.


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Normal Distribution: Examples

- daily changes in/ or closing prices of stocks- weight of cereal boxes or anything for that

matter,

- customer servicing time,

- measurement of length of a screw,

- errors, counts of WBC, blood pressuremeasure, job stress,

- petrol consumption of a car- Your grades,


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6-11

Probability Density Function

of the Normal Distribution

f xx

Where

e

e( )

:

1

2

1

2

2

mean of X

standard deviation of X

= 3.14159 . . .

2.71828 . . . X


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6-12

Normal Curves for Different

Means and Standard Deviations

20 30 40 50 60 70 80 90 100 110 120

5 5

10


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Normal Distribution


6-13

Any variable that can be described as having Normaldistribution has its probability distribution depended on only two

quantities: mean and variance; i.e. the knowledge of these two

quantities is enough to find out the probability of the r.v. from

corner to corner of its distribution. 68% within 1 sd from mean; 95% within 1.96 sd; 99.7% within

3 sd.


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6-14

A company produces lightbulbs whose lifetimes follow a

normal distribution with mean 1200 hours and standard

deviation 250 hours. If a lightbulb is chosen randomly

from the company s output, what is the probability that

its lifetime will be between 900 and 1300 hours?

How to find such probability?


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6-15

Standardized Normal Distribution

A normal distribution with a mean of zero, and

a standard deviation ofone

Z Formula standardizes any normaldistribution

Z Score computed by the Z

Formula the number of standard

deviations which a valueis away from the mean

Z X

1

0


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6-16

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 0.0359

0.10 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.20 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.30 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 0.3621

1.10 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830

1.20 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 0.4990

3.40 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998

3.50 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998


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6-17

-3 -2 -1 0 1 2 3

Table Lookup of a

Standard Normal Probability

P Z( ) .0 1 0 3413

Z 0.00 0.01 0.02

0.00 0.0000 0.0040 0.0080

0.10 0.0398 0.0438 0.0478

0.20 0.0793 0.0832 0.0871

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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6-18

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 0.0080

0.10 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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6-19

Normal Approximation

of the Binomial Distribution

The normal distribution can be used to approximatebinomial probabilities (with large n, binomial dist -> toN). (show simulation). For large n, binomial dist.cumbersome.

Translation Procedure

Convert binomial parameters to normal parameters

Does the interval lie between 0 and n? Ifso, continue; otherwise, do not use the normalapproximation.

Another rule of Thumb: use N if n.p>5 and n.q>5

Correct for continuity

Solve the normal distribution problem

3


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6-20

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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6-21

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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6-22


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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6-23

0

0.02

0.04

0.06

0.08

0.10

0.12

6 8 10 12 14 16 18 20 22 24 26 28 30


Graphs


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6-24


Computations

25

26

27

28

29

30

31

3233

Total

0.0167

0.0096

0.0052

0.0026

0.0012

0.0005

0.0002

0.00010.0000

0.0361

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

Business Statistics, 4e, by Ken Black. 2003 John Wiley & Sons. 6-25

If 2 accidents happen on average in 10 minutes, how much

time elapses on average before next accident happens?

It is closely related to Poisson. Poisson, a discrete

distribution, describes random occurrences over some interval.

Exponential, a continuous distribution, describes probability

distribution of time between random occurrences.

If arrivals follow a Poisson dist, time between arrivals will

follow an exponential dist.


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Business Statistics, 4e, by Ken Black. 2003 John Wiley & Sons. 6-26


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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6-27

Examples: time lapse between arrivals of customers, the

distance between major defects in a highway, time

between two successive breakdown of a machine, time

gap between two successive arrivals to a waiting line.

Also often used to model the failure time of manufactured

items; tells us that prob of survival of a product decays

exponentially, as f(x>x0)= e-x0

Mean of exponential distribution: E(x)= 1/

Standard deviation of exponential distribution= 1/


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Graphs of Selected Exponential

Distributions

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0 1 2 3 4 5 6 7 8

is the frequency with which the event occurs.


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Exponential distribution


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Say, arrivals at a bank are Poisson distributed with a of1.2 customers every minute. What is the mean interarrival

time? What is the probability that at least 2 minutes will

elapse between one arrival and the next arrival?

Interarrival times of random arrivals are exponentially

dist with mean and standard deviation both 1/=0.833

minutes. On average 0.833 minutes will elapse between

arrivals at the bank. The prob of an interval of 2 minutes

or more between arrivals can be calculated by


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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 12

12 2

0907

| .

( . )( )

.

About 9.07% of the time, 2minutes or more will elapse between

Arrivals.

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