QT1 Normal Distribution

8/8/2019 QT1 Normal Distribution

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Quantitative Techniques

Normal Distribution


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Learning Objectives

Appreciate the importance of the normaldistribution.

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

other distributions


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What we should know

Measure of dispersion

Data type and scales

Familiarity with SPSS and Excel Probability concepts

Various distribution

Parameters and statistic


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The Normal Distribution

The normal family of distributions occurs muchmore often in econometrics than any otherparametric family.

One reason for this is that the sum of a largenumber of independent random variables has anapproximately normal distribution.

Normal distributions are symmetrical about the

mean, and the normal probability curve is thefamiliar bell-shaped curve. The mean, median,and mode are equal for this family ofdistributions


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Shape of the Normal Distribution

0

0.1

0.2

0.3

y

-4 -2 2 4u


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Normal distribution bell-shaped

symmetrical about the mean total area under curve = 1 approximately 68% of distribution is

within one standard deviationone standard deviation of the mean

approximately 95% of distribution iswithin two standard deviationstwo standard deviations of themean

approximately 99.7% of distribution is

within 3 standard deviations3 standard deviations of the mean Mean = Median = Mode


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Characteristics of the Normal

Distribution

Continuous distribution Symmetrical distribution Asymptotic to the

horizontal axis

Unimodal A family of curves Area under the curve

sums to 1. Area to right of mean is

1/2. Area to left of mean is

1/2.

Q

1/2 1/2

X


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The Mathematical Model

21

2

2

1

2

: density o random variable

3.14159; 2.71828

: population mean

: population standard deviation

: value o random variable

X

f X e

f X X

e

X X

QW

TW

T

Q

W

!

! !

g g


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Many Normal Distributions

By varying the parameters W andQ, we obtain different

normal distributions

There are an infinite number ofnormal distributions


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Finding Probabilities

Probability is the

area under the

curve!

c dX

f(X)

?P c X de e !


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Which Table to Use?

An infinite number ofnormal distributions means an infinite

number oftables to look up!


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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 0.0359

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 0.3389

1.00 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 0.49903.40 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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-3 -2 -1 0 1 2 3

Table Lookup of a

Standard Normal Probability

P Z( ) .0 1 0 3413e e !

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 0.34

13 0.3438 0.346

1

1.10 0.3643 0.3665 0.3686

1.20 0.3849 0.3869 0.3888


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Applying the Z Formula

3643.)10.10()600485(

105=and485,=withddistributenormallyisX

!! ZPXP

WQ

For 485,

Z-QW

!

!485 485

1050

For 600,

Z-QW

!

!600 485

105110.

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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The Cumulative StandardizedNormal Distribution

Z .00 .01

0.0 .5000.5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.5478.02

0.1 .5478

Cumulative Standardized Normal

Distribution Table (Portion)

Probabilities

ShadedArea

Exaggerated

Only One Table is Needed

0 1Z Z

Q W! !

Z = 0.12

0


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Standardizing Example

6.2 50.12

10

XZ

Q

W

! ! !

Normal Distribution StandardizedNormal Distribution

ShadedArea Exaggerated

10W !1

ZW !

5Q !6.2 X Z

0Z

Q !0.12


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Example:



10W !1

ZW !

5Q !7.1 X Z

0Z

Q !0.21

2.9 5 7.1 5.21 .21

10 10

X XZ Z

Q QW W

! ! ! ! ! !

2.9 0.21

.0832

2.9 7.1 .1664P Xe e !

.0832


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Z .00 .01

0.0 .5000.5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.5832.02

0.1 .5478



ShadedArea

Exaggerated

0 1Z Z

Q W! !

Z = 0.21

Example:

2.9 7.1 .1664P Xe e !(continued)

0


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Z .00 .01

-03 .3821 .3783 .3745

.4207 .4168

-0.1 .4602 .4562 .4522

0.0 .5000 .4960 .4920

.4168.02

-02 .4129



ShadedArea

Exaggerated

0 1Z Z

Q W! !

Z = -0.21

Example:

2.9 7.1 .1664P Xe e !(continued)

0


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Example:

8 .3821P X u !



10W !1

ZW !

5Q !8 X Z

0Z

Q !0.30

8 5.30

10

XZ

QW

! ! !

.3821


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Example:

8 .3821P X u !(continued)

Z .00 .01

0.0 .5000.5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.6179.02

0.1 .5478



ShadedArea

Exaggerated

0 1Z Z

Q W! !

Z = 0.30

0


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.6217

Finding Z Values for KnownProbabilities

Z .00 0.2

0.0 .5000 .5040 .5080

0.1 .5398 .5438 .5478

0.2 .5793 .5832 .5871

.6179 .6255

.01

0.3


Distribution Table (Portion)What is ZGiven

Probability = 0.6217 ?

ShadedArea

Exaggerated

.6217

0 1Z ZQ W! !

.31Z !

0


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Recovering X Values for KnownProbabilities

5 .30 10 8X ZQ W! ! !

Normal Distribution Standardized

Normal Distribution10W !

1Z

W !

5Q ! ? X Z0ZQ ! 0.30

.3821.6179


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Assessing Normality

Not all continuous random variables arenormally distributed

It is important to evaluate how well the dataset seems to be adequately approximated bya normal distribution


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Assessing Normality Construct charts

For small- or moderate-sized data sets, do stem-and-leaf display and box-and-whisker plot looksymmetric?

For large data sets, does the histogram or polygonappear bell-shaped?

Compute descriptive summary measures Do the mean, median and mode have similar

values? Is the range approximately 6 W?

(continued)


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Assessing Normality

Observe the distribution of the data set

Do approximately 2/3 of the observations liebetween mean 1 standard deviation?

Do approximately 4/5 of the observations liebetween mean 1.28 standard deviations?

Do approximately 19/20 of the observations liebetween mean 2 standard deviations?

Evaluate normal probability plot Do the points lie on or close to a straight line?

(continued)

s

s

s


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Assessing Normality

Normal probability plot

Arrange data into ordered array

Find corresponding standardized normal quantilevalues

Plot the pairs of points with observed data valueson the vertical axis and the standardized normal

quantile values on the horizontal axis Evaluate the plot for evidence of linearity

(continued)


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Lowest Stock decision at post office

The manager of a small postal substation is tryingto quantify the variation in the weekly demand formailing envelops. She has decided to assume that

this demand is normally distributed. She knowsthat on an average 100 envelops are purchasedweekly and that 90 percent of the time, weeklydemand is below 115. The manager wants to stockenough mailing envelops each week so that thepercentage of running out of envelops is no higher

than 5 percent. Can you suggest her the lowestsuch stock level?


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Prediction of number of spectators in a match

Mr. Sanga Kumar, the McDonald stand manager for the Micromax tri-series ODI at DambullaStadium, SriLanka, just had two cancellation on hiscrew. This means that if more than 72,000 peoplecome to watch todays cricket match, the line forhot-dogs will constitute a disgrace to Mr. Kumarand will harm business at the future games. Mr.

Kumar knows from his experience that number ofspectators who come to the game is normallydistributed with mean 67,000 and standard deviation

4,000 people. Mr. Kumar has an option to hire twotemporary employees to ensure the business wontbe harmed in the future at an additional cost of$200. If he believes the future harm to business ofhaving more than 72,000 fans at the match would be$ 5000, what would you suggest him to go for?


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Inspection Shop

On the basis of past experience, automobileinspectors in Maruti Udyog Limited in Gurgaon,have noticed that 5 percent of the cars coming in for

their annual inspection fail to pass. Find the probability that between 7 and 18 of the next 200cars to enter the Inspection shop will fail in theinspection.


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Normal Approximation

of the Binomial Distribution The normal distribution can be used to

approximate binomial probabilities

Procedure

Convert binomial parameters to normal parameters Does the n*p is greater than 5? If so, continue;

otherwise, do not use the normal approximation.

Correct for continuity

Solve the normal distribution problem


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Conversion equations

Conversion example:

Normal Approximation of Binomial:ParameterConversion

Q

W

!

!

n p

n p q

Given that has a binomial distribution , ind

andP X n p

n p

n p q

( | . ).

( )(. )

( )(. )(. ) .

u ! !

! ! !

! ! !

25 60 30

60 30 18

60 30 70 3 55

Q

W


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Normal Approximation of Binomial:

I

ntervalC

heck

Q W

Q W

Q W

s ! s ! s

!

!

3 18 3 355 18 1065

3 7 35

3 2865

( . ) .

.

.

0 10 20 30 40 50 60n

70


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C

orrecting forC

ontinuity

ValuesBeing

Determined

Correction

X"

Xu

X

Xe

eXe

X

+.50

-.50

-.50

+.05

-.50 and +.50

+.50 and -.50

).55.3and18|24.5(X

yprobabilitnormalby theedapproximatis

)30.and60|25(

y,probabilitbinomialThe

!!u

!!u

WQ

pnXP


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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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Computations

25

2627

28

29

30

31

32

33

Total

0.0167

0.00960.0052

0.0026

0.0012

0.0005

0.0002

0.0001

0.0000

0.0361

X P(X)

The normal approximation,

(X 24.5| andu ! !

! u

! u

! e e

! !

Q W18 355

24 5 18

355

183

5 0 183

5 4664

0336

. )

.

.

( . )

. .

. .

.

P Z

P Z

P Z


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Normal Distribution with Excel

NORMDIST(x, mean, sd, cumulative)

NORMSDIST(z)

NORMINV(probability, mean, sd)

NORMSINV(probability)

STANDARDIZE(x, mean, sd)


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Thanks


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QT1 Normal Distribution

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