23 4 14 prob distribution

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Statistics for Business and

Economics

ChapterRandom Variables &

Probability Distributions


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Content

1. Two Types of Random Variables

2. Probability Distributions for DiscreteRandom Variables

3. The Binomial Distribution

4. Poisson and yper!eometric Distributions

". Probability Distributions for #ontinuousRandom Variables

$. The %ormal Distribution


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Content (continued)

&. Descripti'e (ethods for )ssessin! %ormality

*. )ppro+imatin! a Binomial Distribution witha %ormal Distribution

,. -niform and +ponential Distributions

1/. 0amplin! Distributions11. The 0amplin! Distribution of a 0ample

(ean and the #entral imit Theorem


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2/11 Pearson ducation nc

Learnin !b"ecti#es

1. De'elop the notion of a random 'ariable

2. earn that numerical data are obser'ed 'alues ofeither discrete or continuous random 'ariables

3. 0tudy two important types of random 'ariablesand their probability models5 the binomial andnormal model

4. Define a samplin! distribution as the probability ofa sample statistic

". earn that the samplin! distribution of follows anormal model

x


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$hin%in Challene

• You’re taking a 33 question multiplechoice test. Each question has 4choices. Clueless on 1 question, you

decide to guess. What’s the chanceyou’ll get it right?

• If you guessed on all 33 questions, hatould !e your grade? Would you pass?


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Two Types of

Random Variables


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Random Variable

) random variable is a 'ariable that assumes

numerical 'alues associated with the random

outcomes of an e+periment where one 6and onlyone7 numerical 'alue is assi!ned to each sample

point.


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Discrete

Random Variable

Random 'ariables that can assume a countable

number 6finite or infinite7 of 'alues are calleddiscrete.


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Discrete Random Variable

Examples

+perimentRandom

Variable

Possible

Values

#ount #ars at TollBetween 115// 8 15//

9 #ars)rri'in!

/ 1 2 ... :

(a;e 1// 0ales #alls 9 0ales / 1 2 ... 1//

nspect &/ Radios 9 Defecti'e / 1 2 ... &/

)nswer 33


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Continuous

Random Variable

"andom #aria!les that can assume

#alues corresponding to any of thepoints contained in one or moreinter#als $i.e., #alues that are in%niteand uncounta!le& are calledcontinuous.


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Continuous RandomVariable Examples

(easure TimeBetween )rri'als

nter=)rri'alTime

/ 1.3 2.&* ...

ExperimentRandom

Variable

Possible

Values

>ei!h 1// People >ei!ht 4".1 &* ...

(easure Part ife ours ,// *&"., ...

)mount spent on food ? amount "4.12 42 ...


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Probability Distributionsfor Discrete Random

Variables


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Discrete

Probability Distribution

'he probability distribution of a

discrete random ariable is agraph, ta!le, or formula that speci%esthe pro!a!ility associated ith each

possi!le #alue the random #aria!lecan assume.


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Reuirements for the

Probability Distribution of aDiscrete Random Variable x

1. p6 x7 @ / for all 'alues of x

2. Σ p6 x7 A 1

where the summation of p6 x7 is o'er all possible

'alues of x.


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Discrete Probability

Distribution E'ample

Probability DistributionValues, x Probabilities, p( x )

0 1/4 !"#

1 "/4 !#0

" 1/4 !"#

+periment5 Toss 2 coins. #ount number of

tails.


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Visualiin Discrete

Probability Distributions$istin% &able

'ormula

&ails( x )

*ountp( x )

0 1 !"#

1 " !#0

" 1 !"#

p x n

x!(n – x)!( )

+ p x ( 1 – p)n – x

rap-

!00!"#

!#0

0 1 "

x

p( x )

. (0, !"#), (1, !#0), (", !"#)


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Summary easures

1. +pected Value 6(ean of probability distribution7 >ei!hted a'era!e of all possible 'alues

µ A E 6 x7 A Σ x p6 x7

2. Variance

>ei!hted a'era!e of sCuared de'iation about

mean

σ 2 A E 6 x − µ )2] = Σ 6 x − µ )2 p6 x73. 0tandard De'iation

2σ σ =E


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Summary easures

Calculation $able

x p(x) x p(x) x – µ

&otal Σ6 x − µ )2 p6 x7

(x – µ) 2 (x – µ) 2 p(x)

Σ x p6 x7


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Thin!in" Challen"e

You toss ( coins. You’re interested in

the num!er of tails. What are the

expected alue, ariance, and

standard deiation of this random

#aria!le, num!er of tails?


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E)pected *alue + *ariance

olution

0 !"# 1!00 1!00

1 !#0 0 0

" !"# 1!00 1!00

0

!#0

!#0

1!0

x p(x) x p(x) x – µ (x – µ) 2 (x – µ) 2 p(x)

!"#

0

!"#

σ

2

= 50

σ

= 71


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Probability Rules for Discrete

Random Variableset x be a discrete random 'ariable with probability

distribution p6 x7 mean µ and standard de'iation σ .Then dependin! on the shape of p6 x7 the followin!

probability statements can be made5

#hebyshe'Fs Rule mpirical Rule

P x −σ


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The #inomialDistribution


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Binomial Distribution

%umber of GsuccessesF in a sample of n

obser'ations 6trials7

%umber of reds in 1" spins of roulette wheel

%umber of defecti'e items in a batch of " items

%umber correct on a 33 Cuestion e+am

%umber of customers who purchase out of 1//customers who enter store 6each customer is

eCually li;ely to pyrchase7


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Binomial Probability

*-aracteristics o a inomial Experiment1! The e+periment consists of n identical trials.

"! There are only two possible outcomes on each trial. >e

will denote one outcome by S 6for success7 and the other

by F 6for failure7.

2! The probability of S remains the same from trial to trial.

This probability is denoted by p and the probability of

F is denoted by q. %ote that q A 1 H p.4! The trials are independent.

#! The binomial random 'ariable x is the number of S Fs in

n trials.


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Binomial Probability

DistributionI6 7 61 7I 6 7I

x n x x n xn n

p x p q p p x x n x

− − = = − ÷ −

p( x ) Probability o x 3uccesses5 p Probability o a 3uccess5 on a sin%le trial

q 1 p

n 6umber o trials

x 6umber o 3uccesses5 in n trials

( x 0, 1, ", !!!, n)

n – x 6umber o ailures in n trials


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#inomial Probability

Distribution Example

3 " 3

I6 7 61 7I6 7I

"I637 ." 61 ."73I6" 37I

.312"

x n xn p x p p x n x

p

−

−

= −−

= −−

=

+periment5 Toss 1 coin " times in a row. %ote

number of tails. >hatFs the probability of 3 tailsJ


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Binomial Probability Table (Portion)

n = 5 p

k ./1 K /."/ K .,,

/ .,"1 K ./31 K .///

1 .,,, K .1** K .///

2 1./// K ."// K .///

3 1./// K .*12 K .//1

4 1./// K .,$, K ./4,

#umulati'e Probabilities

p( x 7 2) p( x 7 ") !81" !#00 !21"


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Binomial Distribution

Characteristicsn # p 0!1

n # p 0!#

µ = E 6 x7 = np

9ean

tandard Deviation

σ = npq


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#inomial DistributionThin!in" Challen"e

LouFre a telemar;eter sellin! ser'ice contracts for

(acyFs. LouF'e sold 2/ in your last 1// calls 6 p

!"07. f you

call 1" people toni!ht whatFs the probability of ). %o salesJ

B. +actly 2 salesJ

#. )t most 2 salesJD. )t least 2 salesJ


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Binomial Distribution Solution*

n 1", p !"0

:. p6/7 A !0;8


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4!4

=t-er Discrete Distributions>

Poisson and ?yper%eometric


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Poisson Distribution

1. %umber of e'ents that occur in an inter'al e'ents per unit

O Time en!th )rea 0pace

2. +amples %umber of customers arri'in! in 2/ minutes

%umber of stri;es per year in the -.0.

%umber of defects per lot 6!roup7 of DVDFs


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Characteristics of a Poisson

Random Variable

1. #onsists of countin! number of times an e'entoccurs durin! a !i'en unit of time or in a !i'enarea or 'olume 6any unit of measurement7.

2. The probability that an e'ent occurs in a !i'en unitof time area or 'olume is the same for all units.

3. The number of e'ents that occur in one unit oftime area or 'olume is independent of the numberthat occur in any other mutually e+clusi'e unit.

4. The mean number of e'ents in each unit is denoted by λ.


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Poisson Probability

Distribution +unction

µ = λ σ 2 = λ

p6 x7 A Probability of x !i'en λ

λ A (ean 6e+pected7 number of e'ents in unite A 2.&1*2* . . . 6base of natural lo!arithm7

x A %umber of e'ents per unit

p x x

6 7I

= xλ λe H

6 x A / 1 2 3 . . .7


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Poisson Probability

Distribution +unctionλ

0!#

λ ;

9ean

tandard Deviation

σ λ =

µ = E ( x) = λ


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Poisson Distribution E'ample

#ustomers arri'e at a rate of &2 per hour. >hat

is the probability of 4 customers arri'in! in 3

minutesJ


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Poisson Distribution Solution


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Poisson Probability Table (Portion)

#umulati'e Probabilities

p( x 7 4) p( x 7 2) !


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$hin%in Challene

You ork in -uality ssurance for an

in#estment %rm. clerk enters $%

ords per minute ith & errors per

hour. What is the pro!a!ility of '

errors in a (%%)word !ond

transaction?


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Poisson Distribution Solution,

+indin λ* &" wordsmin A 6&" wordsmin76$/ minhr7

A 4"// wordshr

$ errorshr A $ errors4"// wordsA !00122 errorsword

n a "##=word transaction 6inter'al75

λ A (!00122 errorsword 76"## words7A .34 errors2""=word transaction


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Poisson Distribution Solution:

Finding p()!

( )

=

/ =.34

6 7

I.34

6/7 .&11*/I

xe

p x

xe

p

λ λ =

= =


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Characteristics of a*yper"eometricRandom Variable

1. The e+periment consists of randomly drawin! n

elements without replacement from a set of N elements r of which are S Fs 6for success7 and 6 N Hr 7 of which are F Fs 6for failure7.

2. The hyper!eometric random 'ariable x is the

number of S Fs in the draw of n elements.


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"ypergeometric Probability

Distribution Function

where . . .

x A (a+imum / n H 6 N H r 7 K

(inimum 6r n7N

p x( )=

r

x

N − r n − x

N

n

µ = nr

N σ 2

= r N − r ( )n N − n( )

N 2 N −1( )


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/ypergeometric 0ro!a!ility

1istri!ution 2unction

N 3 'otal num!er of elements

r 3 4um!er of S’s in the N elements

n 3 4um!er of elements dran

x 3 4um!er of S’s dran in the n

elements


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Probability

Distributions forContinuous Random

Variables


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Continuous Probability

Density +unctionThe !raphical form of the probability distribution for acontinuous random 'ariable x is a smooth cur'e


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Continuous Probability Density

+unction

This cur'e a function of x is denoted by the symbol f 6 x7

and is 'ariously called a probability density unction

(pd), a reBuency unction or a probability

distribution.

The areas under a probability

distribution correspond to

probabilities for x. The area A beneath the cur'e between two

points a and b is the probability

that x assumes a 'alue between a and b.


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The ,ormalDistribution


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@mportance o

6ormal Distribution

1. Describes many random processes or

continuous phenomena

2. #an be used to appro+imate discrete probability distributions

+ample5 binomial

3. Basis for classical statistical inference


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%ormal Distribution

1. GBell=shapedF 8

symmetrical

2. (ean median mode

are eCual x

f ( x )

9ean

9edian

9ode


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Probability Density 'unction

where µ A (ean of the normal random 'ariable x

σ A 0tandard de'iation

π A 3.141" . . .

e A 2.&1*2* . . . P 6 x Q a7 is obtained from a table of normal

probabilities

f ( x) =1

σ 2π e

−1

2

x− µ σ

2


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Eect o Varyin%

Parameters ( C σ )


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6ormal Distribution

Probability

c d x

f ( x )

Probability is

area under

curve+P(c ≤ x ≤ d ) = f ( x)

c

d

∫ dx?


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0tandard %ormal Distribution

'he standard normal distribution is a normaldistri!ution ith µ 3 5 and σ 3 6. random#aria!le ith a standard normal distri!ution,

denoted !y the sym!ol z , is called a standardnormal random #aria!le.


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z 0

σ

1

1!A;

./4 ./"

1.* .4$&1 .4$&* .4$*$

.4&3* .4&44

2./ .4&,3 .4&,* .4*/3

2.1 .4*3* .4*42 .4*4$

The 0tandard %ormal Table5

P 6/ Q z Q 1.,$7

!0;

1!A .4&"/

0tandardiSed %ormal Probability

Table 6Portion7

Probabilities

!4


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&-e tandard 6ormal &able>

P (1!"; ≤ z ≤ 1!";)

z 0

σ 1

1!";

tandardied 6ormal Distribution

-aded area exa%%erated

!2A;"

1!";

!2A;" P (1!"; 7 7 1!";)

!2A;" !2A;"

!


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P ( z F 1!";)

z 0

σ 1


1!";

P ( z F 1!";)

!#000 !2A;"

!1028

!2A;"

!#000

&- t d d 6 l & bl


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P ("!


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P ( z F "!12)

z 0

σ 1

"!12


-aded area exa%%erated

P ( z F "!12)

!4824 !#000

!A824

!#000!4824

6 d d 6 l


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x

( x )

6onGstandard 6ormal

Distribution

6ormal distributions dier by

mean C standard deviation!

Eac- distribution Hould

reBuire its oHn table!

&-at5s an infinite

number o tables+


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Property o 6ormal Distribution

f x is a normal random 'ariable with mean μ and

standard de'iation σ then the random 'ariable z defined by the formula

has a standard normal distribution. The 'alue z describesthe number of standard de'iations between x and µ.

z = x − µσ

t d di t-


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tandardie t-e

6ormal Distribution

6ormal

Distribution

x

σ

One table!

0

σ 1

z

tandardied 6ormal

Distribution

z = x − µ σ

'i di P b bilit * di t


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'indin% a Probability *orrespondin% to a

6ormal Random Variable

1. 0;etch normal distribution indicate mean and shadethe area correspondin! to the probability you want.

2. #on'ert the boundaries of the shaded area from x

'alues to standard normal random 'ariable z

z = x − µσ

0how the z 'alues under correspondin! x 'alues.

3. -se Table V in )ppendi+ ) to find the areas

correspondin! to the z 'alues. -se symmetry when

necessary.

6on standard 6ormal μ # σ 10>


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z 0

σ 1

!1"

tandardied 6ormal

Distribution

0haded area e+a!!erated

!04

P (# I x I ;!")

6ormal

Distribution

x #

σ 10

;!"

z = x − µ

σ = $.2 − "

1/= .12

6 t d d 6 l # 10


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z 0

σ 1

G!1"

tandardied 6ormal

Distribution

6onGstandard 6ormal J #, K 10>

P (2!8 ≤ x ≤ #)

6ormal

Distribution

x #

σ 10

2!8

!04


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0

σ 1

G!"1 z !"1

tandardied 6ormal

Distribution

6onGstandard 6ormal μ #, σ 10>

P ("!A ≤ x ≤

#

σ 10

"!A

6ormal

Distribution

!1;;4

!082"!082"

0haded area e+a!!erated

z = x − µ

σ = 2., − "

1/= −.21

z = x − µ

σ = &.1− "

1/= .21

6on standard 6ormal # 10> P(


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6onGstandard 6ormal μ #, σ 10> P ( x

≥ 8)

x #

σ 10

8

6ormal

Distribution

z 0 !20

tandardied 6ormal

Distribution

σ 1

!28"1

!#000

!11


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0

σ 1

!20 z !"1

tandardied 6ormal

Distribution

6on standard 6ormal J #, K 10>

P(

#

σ 10

8

6ormal

Distribution

!11


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,ormal DistributionThin!in" Challen"e

Lou wor; in hatFs the probability that a bulb will

last

). between "000 and "400

hoursJ

B. less than 14


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tandardied 6ormal

Distribution

z 0

σ 1

"!0

olutionL P ("000 ≤ x ≤ "400)

6ormal

Distribution

x "000

σ "00

"400

!4


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z 0

σ 1

"!;#

tandardied 6ormal

Distribution

olutionL P ( x ≤ 14


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'indin% z GValues

or MnoHn Probabilities

N-at is Z , %iven

P ( z ) !1"1


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'indin% x Values

or MnoHn Probabilities

6ormal Distribution

x #

σ 10

O

!1"1<


0haded areas e+a!!erated

z 0

σ 1

!21

!1"1<

8!1

x = µ + z ⋅σ = 5 + .31( ) 10( )=


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4!<

Descriptive 9et-ods or:ssessin% 6ormality

Determinin% N-et-er t-e Data


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:re rom an :pproximately

6ormal Distribution

1. #onstruct either a histo!ram or stem=and=leaf

display for the data and note the shape of the !raph.f the data are appro+imately normal the shape of

the histo!ram or stem=and=leaf display will be

similar to the normal cur'e.

Determinin" -hether the


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"Data .re from an

.pproximately ,ormalDistribution

2. #ompute the inter'als and

determine the percenta!e of measurements fallin!in each. f the data are appro+imately normal the

percenta!es will be appro+imately eCual to $*U

,"U and 1//U respecti'ely from the mpirical

Rule 6$*U ,"U ,,.&U7.

x ± s, x ± 2s, and x ± 3s,


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Determinin% N-et-er t-e Data :re rom

an :pproximately 6ormal Distribution

3. Wind the interCuartile ran!e


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2/11 Pearson ducation nc


:re rom an :pproximately

6ormal Distribution

4. +amine a normal

probability plot for thedata. f the data are

appro+imately normal

the points will fall

6appro+imately7 on a

strai!ht line.=bserved value

E x p

e c t e d

z 0 s c o r e


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6ormal Probability Plot

normal probability plot for a dataset is a scatterplot ith the ranked data#alues on one a)is and their

corresponding e)pected z 7scores from astandard normal distri!ution on the othera)is. 84ote9 Computation of the e)pectedstandard normal z 7scores are !eyond thescope of this te)t. 'herefore, e ill relyon a#aila!le statistical softare packagesto generate a normal pro!a!ility plot.:


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.pproximatin" a#inomial Distribution

with a ,ormalDistribution

6ormal :pproximation o


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6ormal :pproximation o

inomial Distribution

1. -seful because not all

binomial tables e+ist

2. ReCuires lar!e sample

siSe

3. i'es appro+imate

probability only

4. %eed correction for

continuity

n 10 p 0!#0

!0

!1!"!2

0 " 4 ; 8 10 x

p( x )


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!0

!1

!"

!2

0 " 4 ; 8 10 x

p( x )

N-y Probability, is :pproximate

Binomial Probability5

Bar ei!ht

%ormal Probability5 )rea -nder

#ur'e from 3." to 4."

Probability )dded

by %ormal #ur'e

Probability ost by

%ormal #ur'e

# ti f # ti it


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#orrection for #ontinuity

1. ) 12 unit adYustment to

discrete 'ariable

2. -sed when appro+imatin!a discrete distribution

with a continuous

distribution

3. mpro'es accuracy

4!#

(4 !#)

2!#

(4 !#)4


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-sin! a %ormal Distribution to

)ppro+imate Binomial Probabilities

1. Determine n and p for the binomial distribution

then calculate the inter'al5

f inter'al lies in the ran!e / to n the normal

distribution will pro'ide a reasonableappro+imation to the probabilities of most

binomial e'ents.

µ ± 3σ = np ± 3 np 1− p( )


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sin% a 6ormal Distribution to

:pproximate inomial Probabilities

2. +press the binomial probability to be

appro+imated by the form

Wor e+ample

P x ≤ a( ) or P x ≤ b( )− P x ≤ a( )

P x < 3( )= P x ≤ 2( )P x ≥ 5( )= 1− P x ≤ 4( )

P 7 ≤ x ≤ 10( )= P x ≤ 10( )− P x ≤ 6( )



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3. Wor each 'alue of interest a the correction for

continuity is 6a M ."7 and the correspondin!

standard normal z ='alue is

z = a + .5( )− µ

σ



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4. 0;etch the appro+imatin! normal distribution and

shade the area correspondin! to the e'ent of

interest. -sin! Table V and the z ='alue 6step 37

find the shaded area.

This is the appro+imate

probability of the

binomial e'ent.

6 l : i ti E l


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!0

!1

!"

!2

0 " 4 ; 8 10

x

P( x )

6ormal :pproximation Example

2!# 4!#

>hat is the normal appro+imation of p6 x A 47!i'en n A 1/ and p A /."J

6 l : i ti l ti


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6ormal :pproximation olution

1. #alculate the inter'al5

2. +press binomial probability in form5

P x = 4( )= P x ≤ 4( )− P x ≤ 3( )

np ± 3 np 1− p( ) = 10 0.5( )± 3 10 0.5( ) 1− 0.5( )

=5

±4.74

=0.26, 9.74

( )nter'al lies in ran!e / to 1/ so normalappro+imation can be used

6 l : i ti l ti


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3. #ompute standard normal z 'alues5

z ≈ a + .5( )− n ⋅ p

n ⋅ p 1− p( ) =3.5 −10 .5( )10 .5( ) 1− .5( ) = −0.95

z ≈ a

+.5

( )− n

⋅ p

n ⋅ p 1− p( ) =4.5

−10 .5

( )10 .5( ) 1− .5( ) = −0.32


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6 l : i ti l ti


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!0

!1

!"

!2

0 " 4 ; 8 10

x

p( x )

". The e+act probability from the binomial formula is

/.2/"1 6'ersus .2/347


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/ther ContinuousDistributions0

niform andExponential

-niform Distribution


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-niform Distribution

#ontinuous random 'ariables that appear to ha'eeCually li;ely outcomes o'er their ran!e of possible

'alues possess a uniorm probability distribution.

0uppose the random

'ariable x can assume

'alues only in an

inter'al ! Z x Z " .

Then the uniorm

reBuency unction

has a rectan!ular

shape.

Probability Distribution for a


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Probability Distribution for a

-niform Random Variable x

(ean5

f 6 x7 =

1

" − !! ≤ x ≤ " Probability density function5

σ = d − c

12 µ =

c + d 2

0tandard De'iation5

P a


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niform DistributionExample

You’re production manager of a soft drink !ottling

company. You !elie#e that hen a machine is set

to dispense 6( o;., it really dispenses !eteen

112% and 1(2% o;. inclusi#e. uppose the

amount dispensed has a uniform distri!ution.

What is the pro!a!ility that less than 112 o;. is

dispensed?

-niform Distribution 0olution


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-niform Distribution 0olution

P (11!#≤

x ≤

11!8) (ase)/(?ei%-t)

(11!8 11!#)/(1) !20

11!# 1"!#

f ( x )

x 11!8

1 1

12." 11."

1

1./1

" !=

− −

= =

1!0

E'ponential Distribution


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E'ponential Distribution

'he length of time !eteen emergencyarri#als at a hospital, the length of time!eteen !reakdons of manufacturingequipment, and the length of time !eteen

catastrophic e#ents $e.g., a stockmarketcrash&, are all continuous randomphenomena that e might ant to descri!epro!a!ilistically.

'he length of time or the distance !eteen

occurrences of random e#ents like thesecan often !e descri!ed !y the exponentialprobability distribution. 2or this reason,the e)ponential distri!ution is sometimes

called the waitin")time distribution.

Probability Distribution


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Probability Distributionfor an Exponential Random

Variable x

(ean5

f 6 x7 = 1

θ e

− xθ x > /( )Probability density function5

σ = θ

µ = θ

0tandard De'iation5


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Exponential Distribution


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Exponential DistributionExample

(ean5 f 6 x7 =

1

θ e

− x

θ

x > /( )

0uppose the len!th of time 6in hours7 between

emer!ency arri'als at a certain hospital is modeled as

an e+ponential distribution with θ A 2. >hat is the probability that more than " hours pass without an

emer!ency arri'alJ

σ = θ = 2

µ = θ = 2

0tandard De'iation5

Exponential Distribution


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Exponential Distributionolution

Wrom Table V5

A = e−a θ

= e− " 2( )

= e−2."

Probability is the area A to

the ri!ht of a A ".

Probability that more than "

A = e−2."

= ./*2/*"

hours pass between emer!ency arri'als is about ./*.


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4!10

amplin% Distributions

Parameter C tatistic


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Parameter C tatistic

) parameter is a numerical descripti'e measureof a population. Because it is based on all the

obser'ations in the population its 'alue is

almost always un;nown.

) sample statistic is a numerical descripti'e

measure of a sample. t is calculated from theobser'ations in the sample.

*ommon tatistics C Parameters


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*ommon tatistics C Parameters

0ample 0tatistic Population Parameter

Variance s# σ 2

0tandardDe'iation s σ

(ean

x µ

Binomial

Proportion p pQ

amplin% Distribution


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The samplin% distribution of a sample statistic

calculated from a sample of n measurements is

the probability distribution of the statistic.


Deelopin"


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p "amplin" Distributions

Population siSe N A 4

Random 'ariable x

Values of x5 1 2 3 4

-niform distribution

uppose &-ere5s a Population !!!

Population *-aracteristics


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Population *-aracteristics

µ = x

$$=1

N

∑ N

= 2."

Population Distributionummary 9easure

!0!1!"

!2

1 " 2 4

P6 x7

x

:ll Possible amples


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p

o ie n "

ample Hit- replacement

1./ 1." 2./ 2."

1." 2./ 2." 3./

2./ 2." 3./ 3."2." 3./ 3." 4./

1; amples

1st

Obs

11 12 13 14

21 22 23 24

31 32 33 3441 42 43 44

"nd =bservation

1 " 2 4

1

"

24

"nd =bservation

1 " 2 4

1

"

24

1st

Obs

1; ample 9eans



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p %

o :ll ample 9eans

1./ 1." 2./ 2."

1." 2./ 2." 3./

2./ 2." 3./ 3."

2." 3./ 3." 4./

"nd =bservation

1 " 2 41

"

2

4

1st

Obs

1; ample 9eans amplin% Distribution

o t-e ample 9ean

!0

!1

!"

!2

1!0 1!# "!0 "!# 2!0 2!# 4!0

P( x )

x

ummary 9easure o


112/132

y

:ll ample 9eans

µ %

= x

$$=1

N

∑

N

=1./ + 1." + ...+ 4./

1$

= 2."

*omparison


113/132

*omparison

Population amplin% Distribution

2." x µ =

!0!1

!"!2

1 " 2 4

2." µ =

!0

!1!"

!2

1!0 1!# "!0 "!# 2!0 2!# 4!0

P( x )

x

P6 x7

x


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'he ampling 1istri!utionof a ample


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p p %

Distribution o x

1. (ean of the samplin! distribution eCuals mean of

sampled population[ that is

µ x = E x( )= µ .2. 0tandard de'iation of the samplin! distribution eCuals

Standard deviation of sampled population

Square root of sample size

σ x =σ

n.That is

tandard Error o t-e 9ean


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tandard Error o t-e 9ean

The standard de'iation is often referred to

as the standard error o t-e mean.

σ x

&-eorem 4!1


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&-eorem 4!1

f a random sample of n obser'ations is selected from

a population with a normal distribution the samplin!

distribution of will be a normal distribution. x

amplin% rom


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n 1;σ

x "!#

6ormal Populations *entral &endency

Dispersion

H 0amplin! withreplacement

µ

#0

σ 10

x

n 4σ

x #

µ

x #0G x


Population Distribution

x µ µ =

x

n

σ σ =

tandardiin% t-e amplin%


119/132

Distribution o x

tandardied 6ormal

Distribution

0

σ 1

z

z = x − µ xσ

x

= x − µ

σ

namplin%Distribution

x x

σ

x

&-inin% *-allen%e


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&-inin% *-allen%e

You’re an operations analyst for'+'. =ong7distance telephone callsare normally distri!uted ith 5

min. and 5 ( min. If you selectrandom samples of (% calls, hatpercentage of the sample means

ould !e !eteen $2 + 2( minutes?

amplin% Distribution olutionL


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p %

amplin%

Distribution

8

σ

x !4

= −."/

z = x − µ

σ

n

=

*.2 − *

2

2"

= ."/

amplin% rom


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6onG6ormal Populations

*entral &endency

Dispersion

H 0amplin! withreplacement

Population Distribution


n 20σ

x 1!8n 4

σ

x #

#0

σ 10

x

x #0G x

x µ µ =

x

n

σ σ =

*entral $imit &-eorem


123/132

#onsider a random sample of n obser'ations selectedfrom a population 6an& probability distribution7 with

mean μ and standard de'iation σ . Then when n issufficiently lar!e the samplin! distribution of will be

appro+imately a normal distribution with mean

and standard de'iation The lar!er the

sample siSe the better will be the normal appro+imation

to the samplin! distribution of .

µ x = µ x

σ x = σ n .

x



124/132


x

:s sample

sie %ets

lar%e

enou%-(n ≥ 20) !!!

samplin%

distributionbecomes almost

normal!

x µ µ =

xn

σ σ =

*entral $imit &-eorem Example


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*entral $imit &-eorem Example

'he amount of soda in cans of aparticular !rand has a mean of 1( o;and a standard de#iation of 2( o;. If

you select random samples of %' cans, hat percentage of the samplemeans ould !e less than 1126% o;?

*entral $imit &-eorem olutionL


126/132

amplin%

Distribution

1"

σ

x !02

11!A#

x 0

σ 1

1!


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y

Properties o Probability Distributions

Discrete Distributions

1. p6 x7 @ /

".

*ontinuous Distributions

1. P 6 x A a7 A /

". P 6a Q x Q b7 A area under cur'e between a and b

p x( )= 1all x∑

.ey /deas


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y

6ormal :pproximation to inomial

x is binomial 6n p7

P x ≤ a( )≈ P z < a + .5( )− µ{


129/132

.ey /deas


130/132

y

9et-ods or :ssessin% 6ormality

"! Sem+an"+leaf "$spla&

1 &

2 33*,

3 24"$&&

4 1,

" 2

.ey /deas


131/132

y

9et-ods or :ssessin% 6ormality

2. 6 IQ7 ,S X 1.3

4. Nrmal prbab$l$& pl

.ey /deas


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y

eneratin% t-e amplin% Distribution o x

23 4 14 prob distribution

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