Normal Distribution Presentation - Unitedworld School of Business

7/28/2019 Normal Distribution Presentation - Unitedworld School of Business

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Most popular continuous probability

distribution is normal distribution.It has mean & standard deviation

Deviation from mean x- Z = --------------------------- = --------------Standard deviation

Graphical representation of is called normalcurve


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Standard deviation( )

Standard deviation is a measure of spread

( variability ) of around

Because sum of deviations from mean is

always zero , we measure the spread by

means of standard deviation which is definedas square root of

(x- )2

(x- xbar)2

Variance (2) = -------------= ----------------N n-1

2 = variance


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Interpretation of sigma ( )

1.Sigma ( ) standard deviation is a measure ofvariation of population

2.Sigma ( ) is a statistical measure of the processs

capability to meet customers requirements

3.Six sigma ( 6 ) as a management philosophy4.View process measures from a customers point of

view

5.Continual improvement

6.Integration of quality and daily work

7.Completely satisfying customers needs profitably


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Use of standard deviation( )

Standard deviation enables us to determine , with agreat deal of accuracy , where the values of

frequency distribution are located in relation to mean.

1.About 68 % of the values in the population will fall

within +- 1 standard deviation from the mean

2.About 95 % of the values in the population will fall

within +- 2 standard deviation from the mean

3.About 99 % of the values in the population will fallwithin +- 3 standard deviation from the mean


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z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359

0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224

0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549

0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852

0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133

0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389


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1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621

1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830

1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177

1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441

1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545

1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633

1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706

1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767


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2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857

2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890

2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916

2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964

2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981

2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993

3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995

3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997

3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998


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SIGMA Mean CenteredProcess

Mean shifted( 1.5)

Defects/million

% Defects/million

%

1 317400 31.74 697000 69.0

2 45600 4.56 308537 30.8

3 2700 .26 66807 6.68

4 63 0.0063 6210 0.621

5 .57 0.00006

233 0.0233

6 .002 3.4 0.00034


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Thus , if t is any statistic , then by central limit

theoremvariable value Average

Z = --------------

Standard deviation or standard errorx- Z = --------------


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Properties

1. Perfectly symmetrical to y axis2. Bell shaped curve3. Two halves on left & right are same. Skewness

is zero4. Total area1. area on left & right is 0.55. Mean =mode = median , unimodal

6. Has asymptotic base i.e. two tails of the curveextend indefinitely & never touch x axis( horizontal )


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Importance of Normal Distribution

1. When number of trials increase , probabilitydistribution tends to normal distribution .hence, majority of problems and studies can be

analysed through normal distribution2. Used in statistical quality control for settingquality standards and to define control limits


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Hypothesis : a statement about the populationparameterStatistical hypothesis is some assumption or

statement which may or may not be true ,about a population or a probabilitydistribution characteristics about the givenpopulation , which we want to test on thebasis of the evidence from a random sample


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Testing of Hypothesis : is a procedure that

helps us to ascertain the likelihood ofhypothecated population parameter beingcorrect by making use of sample statistic

A statistic is computed from a sample drawnfrom the parent population and on the basis ofthis statistic , it is observed whether thesample so drawn has come from thepopulation with certain specifiedcharacteristic


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Procedure / steps for Testing a hypothesis

1. Setting up hypothesis2. Computation of test statistic3. Level of significance4. Critical region or rejection region5. Two tailed test or one tailed test6. Critical value7. Decision


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Hypothesis : two types

1. Null Hypothesis H02. Alternative Hypothesis H1

Null Hypothesis asserts that there is no difference

between sample statistic and population parameter& whatever difference is there it is attributable to

sampling errorsAlternative Hypothesis : set in such a way that rejection

of null hypothesis implies the acceptance ofalternative hypothesis


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Null Hypothesis

Say , if we want to find the population mean hasa specified value 0H0 : = 0

Alternative Hypothesis could bei. H1 : 0 ( i.e. > 0 or < 0 )ii. H1 : > 0

iii. H1 : < 0iv. R. A. Fisher Null Hypothesis is the

hypothesis which is to be tested for possible

rejection under the assumption that it is true


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4. level of significance :is the maximum probability ( ) ofmaking a Type I error i.e. : P [ Rejecting H

0when H

0is

true ]Probability of making correct decision is ( 1 - )Common level of significance 5 % ( .05 ) or 1 % ( .01 )

For 5 % level of significance ( = .05 ) , probability ofmaking a Type I error is 5 % or .05 i.e. : P [ Rejecting H0when H0 is true ] = .05Or we are ( 1 - or 1-0.05 = 95 % ) confidence that a

correct decision is madeWhen no level of significance is given we take = 0.05


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5.Critical region or rejection region :the value

of test statistic computed to test the nullhypothesis H0is known as critical value . Itseparates rejection region from the

acceptance region


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6.Two tailed test or one tailed test :

Rejection region may be represented by aportion of the area on each of the two sides

or by only one side of the normal curve ,

accordingly the test is known as two tailedtest ( or two sided test )or one tailed ( or one sided test )


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Two tailed test :where alternative

hypothesis is two sided or two tailede.g.Null Hypothesis

H0 : = 0Alternative HypothesisH1 : 0 ( i.e. > 0 or < 0 )


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Right tailed :

Null HypothesisH0 : = 0Alternative Hypothesis

H1 : > 0Left tailed :Null Hypothesis

H0 : = 0Alternative HypothesisH1 : < 0

Right tailed


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7.Critical value : value of sample statisticthat defines regions of acceptance andrejection

Critical value of z for a single tailed ( leftor right ) at a level of significance is thesame as critical value of z for two tailed test

at a level of significance 2 .


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Critical value

(Z )

Level of significance

1 % 5 % 10 %

Two tailed test [Z ] =2.58

[Z ] = 1.96 [Z ] = 1.645

Right tailedtest

Z = 2.33 Z = 1.645 Z= 1.28

Left tailed test Z = -

2.33

Z = -1.645 Z = - 1.28


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S.No. Confidence level(1- )

Value of confidencecoefficient Z ( two

tailed test)

1 90 % 1.64

2 95 % 1.96

3 98 % 2.334 99 % 2.58

5 Without anyreference to

confidence level

3.00

6

is level of significance which separates

acceptance & rejection level


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8. Decision :

1. if mod .Z < Zaccept Null Hypothesistest statistic falls in the region of

acceptance2. if mod .Z > Z

reject Null Hypothesis


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Q1.Given a normal distribution with mean 60 &standard deviation 10 , find the probability that xlies between 40 & 74

Given = 60 , =10P ( 40 < x < 74 ) = P ( -2 < z< 1.4 ) = P ( -2< z< 0 ) + P ( 0 < z < 1.4 )

= 0.4772+ 0.4192 = 0. 8964


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Q2.In a project estimated time ofcompletion is 35 weeks. Standarddeviation of 3 activities in critical pathsare 4 , 4 & 2 respectively . calculate theprobability of completing the project ina. 30 weeks , b. 40 weeks and c. 42weeks


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Test of significance

MeanNull there is no significance difference betweensample mean & population mean orThe sample has been drawn from the parentpopulation

Deviation from mean xbar- Z = --------------------------- = --------------

Standard Error Standard Error

xbar = sample mean

= population mean


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1. Standard Error of mean = / nWhen population standard deviation is known = standard deviation of the population

n = sample size2. Standard Error of mean = s / nWhen sample standard deviation is known

s = standard deviation of the samplen = sample size


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ProprtionNull there is no significance difference betweensample proportion & population proportion or

The sample has been drawn from a populationwith population proportion PNull hypothesis H0 : P = P0 where P0 is

particular value of PAlternate hypothesis H1 : P P0 ( i.e. P > P0 orP < P0 )


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P*(1-P)Standard error of proportion (S.E.(p)) = ------------

nDeviation from proprtion p-P

Test statistic Z = --------------------------- = --------

Standard Error (p) S.E.(p)


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Q3. a sample of size 400 was drawn andsample mean was 99. test whether thissample could have come from a normal

population with mean 100 & standarddeviation 8 at 5 % level of significance


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Ans. Given xbar = 99 , n = 400 = 100 , = 81.Null hypothesis sample has come from a normal

population with mean = 100 & s.d. = 8Null hypothesis H0 : = 100Alternate hypothesis H1 : 0 ( i.e. > 0 or < 0 )

Two tail test so out of 5% , 2.5 % on each side ( left hand &right hand)2.calculation of Test statistic

Standard error (s.e.) of xbar = / n = 8/ 400 = 8/20= 2/5

xbar- 99-100Test statistic Z = ----------- = -------- = -5/2 =- 2.5

S.E. 2/5


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Mod z = 2.5

3. level of significance 5 % i.e.value of = .05 ( hence ,level of confidence = 1- = 1-0.05 = 0.95 or 95%)4. Critical value (since it is Two tail test so out of 5% , wetake two tails on each side i.e.2.5 % on each side = 0.025 (

left hand & right hand)= read from z table value corresponding to area = 0.5-0.025 = 0.4750( 0.4750 on both sides i.e. 2* 0.4750 = 0.95 area which

means 95% confidence)= value of z corresponding to area is 1.96


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5. Decision since mod value of z is more thancritical valueNull Hypothesis is rejected & alternate hypothesis

is acceptedSample has not been drawn from a normalpopulation with mean 100 &s.d. 8


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Q4. the mean life time of a sample of 400fluorescent light tube produced by acompany is found to be 1570 hours with astandard deviation of 150 hrs. test the

hypothesis that the mean life time of thebulbs produced by the company is 1600 hrsagainst the alternative hypothesis that it is

greater than 1600 hrs at 1 &% level ofsignificance


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Ans. Given xbar = 1570 , n = 400 = 1600

, standard deviation of sample mean s= 150= 8Null hypothesis : mean life time of bulbs is1600 hrsi.e. Null hypothesis H0 : = 1600Alternate hypothesis H1 : > 1600i.e. it is a case of right tailed test

2.calculation of Test statisticStandard error (s.e.) of xbar = s / n150/ 400 = 150/20= 7.5


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xbar- m 1570-1600Z = ----------- = -------------- = -30/7.5= - 4

S.E. 7.5

Mod z = 43. level of significance 1 % i.e.value of = .01

( hence , level of confidence = 1- = 1-0.01 =0.99 or 99%)


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4. Critical value (since it is right tail test so out of 1% ,we take 1 % only one side= 0.01 on right hand)= read from z table value corresponding to area = 0.5 -0.01= 0.4900 ( 0.49 on one side together with +0.5 makes total

0.99 which means 99% confidence)= value of z ( corresponding to area 0 .49 is 2.335. Decision since mod value of z is more than critical value

Null Hypothesis is rejected & alternate hypothesis is

acceptedHence mean life time of bulbs is greater than 1600 hrs


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Q5.in a sample of 400 burners there were 12whose internal diameters were not within

tolerance . Is this sufficient to conclude thatmanufacturing process is turning out more than2 % defective burners. Take = .05


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Given P= 0.002 Q= 1-P =1-0.02 =0. 98

& p= 12/400 = 0.03Null hypothesis H0 : P = process is under controlP 0.02

Alternate hypothesis H1 : P > 0.02Left tail testCalculation of Standard error of proportion

P*(1-P) 0.02*0.98(S.E.(p)) = ------------ = --------------

n 400


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Deviation 0.03-0.02 0.001Z = ----------- = ------------ = -------- = 1.429

S.E.(p) (0.02*0.98) / 400 0.0073. level of significance 5 % i.e.value of = .05( hence , level of confidence = 1- = 1-0.01 = 0.95

or 95%)


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4. Critical value (since it is left tail test so out of5% , we take full 5 % only one side = 0.05 on lefthand) = read from z table value corresponding toarea = 0.5 -0.05 = 0.4500 ( 0.45 on one sidetogether with +0.5 makes total 0.95 which means

95% confidence) = value of z ( corresponding toarea 0 .45 is 1.6455. Decision since mod value of z is less than

critical valueNull Hypothesis is acceptedHence process is not out of control


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Q6. a manufacturer claimed that at least 95 %of the equipment which he supplied isconforming to specifications. A examination of

sample of 200 pieces of equipment revealedthat 18 were faulty. Test his claim at level ofsignificance i.) 0.05 ii.) 0.01


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Given P= 0.95 Q= 1-P =1-0.95 =0. 05n= 200p= 18/200 = - (200-18) / 200 = 182/200 =

0.91Null hypothesis H0 : P = process is undercontrol P = 0.95Alternate hypothesis H1 : P < 0.95Left tail test


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P*(1-P) 0.95*0.05S.E.(p) = ------------ = ---------

n2000.91-0.95 -0.04

Z = ------------ = --------- = -2.6 (0.02*0.98) / 200 0.0154


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3a. level of significance 5 % i.e.value of = .05 ( hence ,

level of confidence = 1- = 1-0.05 = 0.95 or 95%)4a. Critical value (since it is right tail test so out of 5% , full5 % on one side = 0.05 on right hand) = read from z tablevalue corresponding to area = 0.5 -0.05 = 0.4500 ( 0.45 on

one side together with +0.5 makes total 0.95 which means95% confidence) = value of z ( corresponding to area 0 .45is 1.645)

5a. Decision since mod value of z is more than critical

valueNull Hypothesis is rejectedManufacturers claim is rejected at 5 % level of significance


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3b. level of significance 1 % i.e.value of = .01

( hence , level of confidence = 1- = 1-0.01 =0.99 or 99%)4b. Critical value (since it is right tail test so out

of 1% , we take 1 % only one side = 0.01 on righthand) = read from z table value correspondingto area = 0.5 -0.01 = 0.4900 ( 0.49 on one side

together with +0.5 makes total 0.99 which means99% confidence) = value of z ( correspondingto area 0 .49 is 2.33


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5b. Decision since mod value of z ismore than critical valueNull Hypothesis is rejectedManufacturers claim is rejected at 1 %level of significance

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Normal Distribution Presentation - Unitedworld School of Business