StatChapter7._studentppt

8/8/2019 StatChapter7._studentppt

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CHAPTER 7CHAPTER 7

PROBABILITYPROBABILITY

DISTRIBUTIONDISTRIBUTION


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IntroductionIntroduction

AA discretediscrete randomrandom variablevariable assumesassumes eacheach ofof

itsits valuesvalues withwith aa certaincertain probabilityprobability.. ItIt isis

convenientconvenient toto representrepresent allall thethe probabilitiesprobabilities ofof aarandomrandom variable,variable, XX byby aa formulaformula.. TheThe formulaformula

asas aa functionfunction ofof thethe numericalnumerical values,values, xx shallshall

bebe denoteddenoted byby f(x),f(x), g(x)g(x) andand soso onon.. Thus,Thus, wewe

writewrite ff (x)(x) == P(P( X=X= xx )) ;; thatthat isis f(f(22)) == P(P( XX == 22))..TheThe setset of of orderedordered pairspairs (x,(x, f(x))f(x)) isis calledcalled thethe

probabilityprobability functionfunction oror probabilityprobability distributiondistribution ofof

thethe discretediscrete randomrandom variablevariable XX..


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Example 7Example 711

TwoTwo coinscoins areare tossedtossed togethertogether.. LetLet XX bebe aa randomrandomvariablevariable forfor havinghaving numbernumber ofof headsheads .. FindFind thethe

probabilityprobability distributiondistribution forfor randomrandom variablevariable X?X?

SolutionSolution

The space sample isThe space sample is

S =S = { , , , };{ , , , };

We can get eitherWe can get eitherno headno head ororoneone head orhead ortwotwoheadsheads from this trial. So, the discrete randomfrom this trial. So, the discrete random

variablevariable X,X, can take the value 0, 1 and 2. Therefore,can take the value 0, 1 and 2. Therefore,


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The discrete variable is associated with the

probabilities, is called discrete probability

distribution.Xdistribution could be shown ina histogram.


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From Figure 7.1 we can see that the areafor every bar in the histogram are the

probability value forX=x. The area of everybar is between 0 and 1 and the total areaof the bars are equal to 1. Thus, weunderstand the total probability of all the

random variables, X in the sample spaceshould equal to 1.

Probability for X is written within the range

, and the total = 1.


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Example 7.2Example 7.2

Two tetrahedrons, with faces labeled 1,Two tetrahedrons, with faces labeled 1,

2, 3 and 4, are thrown. Let X as the2, 3 and 4, are thrown. Let X as the

random variables which represent therandom variables which represent thetotal score on the tetrahedrons. Findtotal score on the tetrahedrons. Find

the random variablethe random variable XXand itsand its

probability distribution.probability distribution.


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SolutionSolution

The Sample Space,The Sample Space, SSof the total score on theof the total score on thetetrahedrons is as follow:tetrahedrons is as follow:


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Thus,Thus,

S =S = {2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8},{2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8},

The random variable X are { , , , , , , }The random variable X are { , , , , , , }


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To Calculate Mean andStandardTo Calculate Mean andStandard

Deviation From The ProbabilityDeviation From The Probability

DistributionDistributionMean and Standard Deviation can beMean and Standard Deviation can be

calculated from the Probability Distributioncalculated from the Probability Distribution

by using the following adjusted formula :by using the following adjusted formula :


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Find the mean and standardFind the mean and standard

deviation for the random variablesdeviation for the random variablesXX in the Example 7.2in the Example 7.2



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SolutionSolution


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ThisThis isis aa discretediscrete randomrandom variable,variable, wherewhere

thethe processprocess of of obtainingobtaining thethe BinomialBinomial

distributiondistribution isis calledcalled BernoulliBernoulli processprocess..AnAn experimentexperiment thatthat oftenoften consistsconsists of of

repeatedrepeated trials,trials, eacheach withwith twotwo possiblepossible

outcomes,outcomes, whichwhich couldcould bebe labeledlabeled asassuccesssuccess oror failurefailure.. ThisThis experimentexperiment isis

knownknown asas binomialbinomial experimentexperiment

Binomial DistributionBinomial Distribution


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A binomial experiment is one that possesses theA binomial experiment is one that possesses thefollowing properties:following properties:

1.1. The experiment consists ofThe experiment consists ofnn repeated trials.repeated trials.

2.2. Each trial has only 2 possible outcomes thatEach trial has only 2 possible outcomes that

can be classified ascan be classified as Success SuccessororFailure.Failure.3.3. The probability of a success and failure ,The probability of a success and failure ,denoted bydenoted bypp andand qq, remains constant from, remains constant fromtrial to trial.trial to trial.

4.4. The repeated, trials are independent.The repeated, trials are independent.


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Binomial FormulaBinomial Formula

The probability ofThe probability ofrrsuccess fromsuccess from nn trial is :trial is :

where;r - the number of successn the number of trialp the probability of success from one trialq the probability of failure from one trial.


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q = 1 - p

It is written as and read as X isBinomial distribution with parameters nsuccess and the probability of success, p.


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One coin is tossed 5 times. IfOne coin is tossed 5 times. IfXX is randomis randomvariable for number of heads. Find thevariable for number of heads. Find the

probability of gettingprobability of getting

(a)(a) no headsno heads(b)(b) one headone head

(c)(c) 3 heads3 heads

(d)(d) at least 3 headsat least 3 heads


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SolutionSolution

XX is the random variable represents theis the random variable represents the

number of heads.number of heads.

The possible outcome of one trial ( i.e oneThe possible outcome of one trial ( i.e one

tossed ) is atossed ) is a HeadHead or aor a TailTail..

ThusThus SS= { , },= { , },

pp is the probability of getting Head,is the probability of getting Head,pp ==

qq is the probability of not getting Head,is the probability of not getting Head,

qq ==nn is number of trial ( i.e number of tossed ),is number of trial ( i.e number of tossed ),

nn ==


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Thus, X~ B (5, )As t he number of trial, n = 5, we can obtain 0, 1, 2, 3, 4 or 5 heads from thetrials. Thus we can take the value of r as 1, 2, 3, 4 and 5.


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Example 7.5

In a process of producing screws, 10% of thescrew was rejected because the screws are too soft. What is t he probability for a sample

of 12 screws that contains

(a) 2 rejected screws(b) not more t han 2 rejected screws.


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Binomial Cumulative TableBinomial Cumulative Table


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Example 7.7

A new medicine is tested to t he patients andwas found to cure 35% on t he patients. If 10patients were chosen at random from t he

generalhospital and given t

he new medicine.Find the probability of

a) 2 patients will cure

b) at least 6 patients will curec) not more t han 5 patients will cure.d) more t han 5 patients will cure.


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Exercise 7.23. 25% of a local university w ho registered for the first year needs additional

class for mathematics. If 6 students are chosen at random, find the probability

a. 1 student needs t he additional classb. 2 students need t he additional classc. 3 students w ho need the additional class.

4. If X represents t he number of broken pencils from 5 pencils chosen at random

in a box of 100 pencils, wh

ere 10 are broken. Find th

e probability

a. P(X = 3) b. P(X 3) c. P(X 2)

5. In an examination of 10 objective questions, every one contains 5 answers butonly one t hat is correct. If one student t hat has never study, sit for examination could only answer by guessing the right answer. Find t heprobability

a. none of t he answer are correctb. only 3 answer are correct.


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Exercise 7.23. 25% of a local university w ho registered for the first year needs additional

class for mathematics. If 6 students are chosen at random, find the probability

a. 1 student needs t he additional class (0.3560)b. 2 students need t he additional class (0.2966)c. 3 students w ho need the additional class (0.1318)

4. If X represents t he number of broken pencils from 5 pencils chosen at randomin a box of 100 pencils, where 10 are broken. Find the probability

a. P(X = 3) 0.0081b. P(X 3) 0.0086c. P(X 2) 0.9914

5. In an examination of 10 objective questions, every one contains 5 answers butonly one t hat is correct. If one student t hat has never study, sit for examination could only answer by guessing the right answer. Find t heprobability

a. none of t he answer are correct 0.1074b. only 3 answer are correct. 0.2013


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EndofBinomialEndofBinomial

DistributionDistribution


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IfIf anan eventevent occursoccurs inin aa givengiven intervalinterval ofof timetime

oror scatteredscattered inin timetime (space),(space), andand XX variablevariable

representsrepresents thethe numbernumber ofof occurrencesoccurrences ininthethe givengiven interval,interval, hencehence XX isis thethe PoissonPoisson

randomrandom variablevariable..

Poisson DistributionPoisson Distribution


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If an event is randomly scattered in time (space) andIf an event is randomly scattered in time (space) and

represents the number of success, the probabilityrepresents the number of success, the probabilitydistribution for X Poisson random variable represents thedistribution for X Poisson random variable represents thenumber of success in a given time interval or space isnumber of success in a given time interval or space is

xx= 0, 1, 2, = 0, 1, 2,

WhereWhere is the mean for the number of success in theis the mean for the number of success in thegiven time interval or space. Poisson distribution for randomgiven time interval or space. Poisson distribution for randomvariable,variable, XX is written as foris written as forxx= 0,1 ,2,3.. . . .= 0,1 ,2,3.. . . .

Poisson DistributionPoisson Distribution

!)( r

e

rXP

rQ

Q

!!

Q

)(~ QoPX


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SolutionSolution


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SolutionSolution


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Example 7.10

The Country Council found that the average

number of fines to the night market hawkersis 8 fines a week. Find the probability that the

Country Council give 5 fines a week?


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Example 7.10

The Country Council found that the average number of

fines to the night market hawkers is 8 fines a week. Find

the probability that the Country Council give 5 fines a

week?


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Example 7.11


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Example 7.11: Solution


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Example 7.12


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The most important continuous probabilityThe most important continuous probability

distribution is the normal distribution.distribution is the normal distribution.

The probability density function (p.d.f) of theThe probability density function (p.d.f) of thestandard normal variable x is denoted bystandard normal variable x is denoted by

Normal DistributionNormal Distribution

2

2

1

2

1)(

!W

Q

TW

x

exf


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The probability of the certain range of random variables X isbetween 0 and 1. The area under the normal distribution curvegive the values of probability of the range of random variables X.For example if we want to find the probability of X between a andb, then the area is as the diagram below:

a b


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a b

The probability between a and b:

P(a < X < b) = P(a X b ) = P(a < X b) = P(a X < b)

!b

a

b

a

x

dxedxxf

2

21

2

1)( W

Q

TW

e an ar ormae an ar orma


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Example 7.12: SolutionExample 7.12: Solution


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a. Find P(-1.5 < Z < 0)

b. Find P (Z < 1.85)

c. Find P(Z < -2.23)


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a. Find P(-1.5 < Z < 0)


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b. Find P (Z < 1.85)


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c. Find P(Z < -2.23)

01287.0)23.2()23.2( !"! ZPZP


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Let say that the mark of 200 students in anexamination is normally distributed with mean 50and variance 16.

a. Find the probability of the students gotbetween 45 and 63 marks.

b. If t he passing mark is 40, find the number of

students who pass the examination.


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Exercise 7.4Exercise 7.4

1.1. FindFind

a.a. P(0 < Z < 1.85)P(0 < Z < 1.85)b.b. P(Z > 2.25)P(Z > 2.25)

c.c. P(Z P(Z > --1.04)1.04)

b.b. P(Z < 1.29)P(Z < 1.29)

c.c. P(Z >P(Z > --3.12)3.12)

d.d. P(Z >P(Z > --2.75)2.75)

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Exercise 7.4Exercise 7.4

1.1. FindFind

a.a. P(0 < Z < 1.85)P(0 < Z < 1.85) Ans: 0.4678Ans: 0.4678b.b. P(Z > 2.25)P(Z > 2.25) Ans: 0.0122Ans: 0.0122

c.c. P(Z P(Z > --1.04)1.04) Ans: 0.8508Ans: 0.8508

b.b. P(Z < 1.29)P(Z < 1.29) Ans: 0.9015Ans: 0.9015

c.c. P(Z >P(Z > --3.12)3.12) Ans: 0.9991Ans: 0.9991

d.d. P(Z >P(Z > --2.75)2.75) Ans: 0.997Ans: 0.997

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Exercise 7.4Exercise 7.44.4. Find the valuesFind the values

a.a. P(1.13 < Z < 2.83)P(1.13 < Z < 2.83)

b.b. P(P(--1.27 < Z P(Z > --0.83)0.83)

5.5. Given X~N(66, 4) find,Given X~N(66, 4) find,

a.a. P(X < 62)P(X < 62)b.b. P(66 < X < 70)P(66 < X < 70)

c.c. P(60 < X < 64)P(60 < X < 64)

d.d. P( X > 62)P( X > 62)

6.6. The marks of the statistics final examination with normal distribution haveThe marks of the statistics final examination with normal distribution havemean 70 and standard deviation 12. What is the probability of one studentmean 70 and standard deviation 12. What is the probability of one studentwho take the exam, scorewho take the exam, scorea.a. 90 or more90 or more

b.b. 60 or less60 or less

c.c. between 75 and 85between 75 and 85

d.d. if 10% from the students sit for the exam got A grade, what is theif 10% from the students sit for the exam got A grade, what is thelowest marks for Grade A students?lowest marks for Grade A students?

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Exercise 7.4Exercise 7.44.4. Find the valuesFind the values

a.a. P(1.13 < Z < 2.83)P(1.13 < Z < 2.83) Ans: 0.1269Ans: 0.1269

b.b. P(P(--1.27 < Z P(Z > --0.83)0.83) Ans: 0.7967Ans: 0.7967

5.5. Given X~N(66, 4) find,Given X~N(66, 4) find,

a.a. P(X < 62)P(X < 62) Ans: 0.02275Ans: 0.02275b.b. P(66 < X < 70)P(66 < X < 70) Ans: 0.47725Ans: 0.47725

c.c. P(60 < X < 64)P(60 < X < 64) Ans: 0.15735Ans: 0.15735

d.d. P( X > 62)P( X > 62) Ans: 0.97725Ans: 0.97725

6.6. The marks of the statistics final examination with normal distribution haveThe marks of the statistics final examination with normal distribution havemean 70 and standard deviation 12. What is the probability of one studentsmean 70 and standard deviation 12. What is the probability of one studentswho take the exam, scorewho take the exam, scorea.a. 90 or more90 or more Ans: 0.0475Ans: 0.0475

b.b. 60 or less60 or less Ans: 0.2033Ans: 0.2033

c.c. between 75 and 85between 75 and 85 Ans: 0.2316Ans: 0.2316

d.d. if 10% from the students sit for the exam got A grade, what is theif 10% from the students sit for the exam got A grade, what is thelowest marks for Grade A students?lowest marks for Grade A students? Ans: 85Ans: 85

R i i E i (Ch t 7)R i i E i (Ch t 7)


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Revision Exercise (Chapter7)Revision Exercise (Chapter7)

1. The probability that a component is not acceptable is 0.1. Ten1. The probability that a component is not acceptable is 0.1. Tencomponents are picked at random. What is the probability thatcomponents are picked at random. What is the probability that

i.i. at least three are not acceptableat least three are not acceptable

ii.ii. not more than five are not acceptable.not more than five are not acceptable.

2.2. A secretary in a company answers an average of 3 calls in 5A secretary in a company answers an average of 3 calls in 5minutes.minutes.

i.i. What is the probability that there is no call in 10 minutesWhat is the probability that there is no call in 10 minutes

ii.ii. What is the probability that the secretary answers more thanWhat is the probability that the secretary answers more than

15 calls in 20 minutes?15 calls in 20 minutes?

3.3. The average weight of a pack of wheat produced by a wheat factoryThe average weight of a pack of wheat produced by a wheat factory

is 5.6 kilogram. If the weight of a pack in normal distributions with theis 5.6 kilogram. If the weight of a pack in normal distributions with thestandard deviation of 0.5 kilogram. What is the probability of onestandard deviation of 0.5 kilogram. What is the probability of onepack wheat has the weightpack wheat has the weight

i.i. less than 5.5 kilogramless than 5.5 kilogram

ii.ii. between 5.3 and 5.7 kilogrambetween 5.3 and 5.7 kilogram

iii.iii. more than 6 kilogram or less than 5 kilogram?more than 6 kilogram or less than 5 kilogram?

R i i E i (Ch t 7)R i i E i (Ch t 7)


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Revision Exercise (Chapter7)Revision Exercise (Chapter7)

1. The probability that a component is not acceptable is 0.1. Ten1. The probability that a component is not acceptable is 0.1. Tencomponents are picked at random. What is the probability thatcomponents are picked at random. What is the probability that

i.i. at least three are not acceptableat least three are not acceptable (0.0702)(0.0702)

ii.ii. not more than five are not acceptable.not more than five are not acceptable. (0.9999)(0.9999)

2.2. A secretary in a company answers an average of 3 calls in 5 minutes.A secretary in a company answers an average of 3 calls in 5 minutes.

i.i. What is the probability that there is no call in 10 minutesWhat is the probability that there is no call in 10 minutes (0.003)(0.003)ii.ii. What is the probability that the secretary answers more thanWhat is the probability that the secretary answers more than

15 calls in 20 minutes?15 calls in 20 minutes? (0.1556)(0.1556)

3.3. The average weight of a pack of wheat produced by a wheat factory isThe average weight of a pack of wheat produced by a wheat factory is5.6 kilogram. If the weight of a pack in normal distributions with the5.6 kilogram. If the weight of a pack in normal distributions with the

standard deviation of 0.5 kilogram. What is the probability of one packstandard deviation of 0.5 kilogram. What is the probability of one packwheat has the weightwheat has the weight

i.i. less than 5.5 kilogramless than 5.5 kilogram (0.4207)(0.4207)

ii.ii. between 5.3 and 5.7 kilogrambetween 5.3 and 5.7 kilogram (0.305)(0.305)

iii.iii. more than 6 kilogram or less than 5 kilogram?more than 6 kilogram or less than 5 kilogram? (0.327)(0.327)

R i i E i (Ch 7)R i i E i (Ch 7)


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Revision Exercise (Chp 7)Revision Exercise (Chp 7)

4. Hupper Corporation produces many types of4. Hupper Corporation produces many types ofsoda drinks, including Orange Cola. The fillingsoda drinks, including Orange Cola. The fillingmachines are adjusted to pour 12 ounces ofmachines are adjusted to pour 12 ounces ofsoda in each 12soda in each 12--ounce can ofOrange Cola.ounce can ofOrange Cola.However, the actual amount of soda poured intoHowever, the actual amount of soda poured into

each can is not exactly 12 ounces; it varies fromeach can is not exactly 12 ounces; it varies fromcan to can. It has been observed that the netcan to can. It has been observed that the netamount of soda in such a can has a normalamount of soda in such a can has a normaldistribution with a mean of 12 ounces and adistribution with a mean of 12 ounces and astandard deviation of 0.015 ounces.standard deviation of 0.015 ounces.

What is the probability that a randomly selectedWhat is the probability that a randomly selectedcan ofOrange Cola contains 11.97 to 11.99can ofOrange Cola contains 11.97 to 11.99ounces of soda?ounces of soda?

R i i E i (Ch 7)R i i E i (Ch 7)


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4. Hupper Corporation produces many types of4. Hupper Corporation produces many types ofsoda drinks, including Orange Cola. The fillingsoda drinks, including Orange Cola. The fillingmachines are adjusted to pour 12 ounces ofmachines are adjusted to pour 12 ounces ofsoda in each 12soda in each 12--ounce can ofOrange Cola.ounce can ofOrange Cola.However, the actual amount of soda poured intoHowever, the actual amount of soda poured into

each can is not exactly 12 ounces; it varies fromeach can is not exactly 12 ounces; it varies fromcan to can. It has been observed that the netcan to can. It has been observed that the netamount of soda in such a can has a normalamount of soda in such a can has a normaldistribution with a mean of 12 ounces and adistribution with a mean of 12 ounces and astandard deviation of 0.015 ounces.standard deviation of 0.015 ounces.

What is the probability that a randomly selectedWhat is the probability that a randomly selectedcan ofOrange Cola contains 11.97 to 11.99can ofOrange Cola contains 11.97 to 11.99ounces of soda?ounces of soda? (Ans: 0.22865)(Ans: 0.22865)

Re ision E ercise (Chp 7)Re ision E ercise (Chp 7)


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5.5. The number of monthly breakdowns of the kind ofThe number of monthly breakdowns of the kind ofcomputer used by an office is a random variablecomputer used by an office is a random variablehaving the Poisson distribution with = 1.6.having the Poisson distribution with = 1.6.

Using the probability table, find the probabilities thatUsing the probability table, find the probabilities thatthis kind of computer will function for a monththis kind of computer will function for a month

i.i. without a breakdownwithout a breakdownii.ii. with one breakdownwith one breakdown

iii.iii. at most two will breakdownat most two will breakdown

Q



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5.5. The number of monthly breakdowns of the kind ofThe number of monthly breakdowns of the kind ofcomputer used by an office is a random variablecomputer used by an office is a random variablehaving the Poisson distribution with = 1.6.having the Poisson distribution with = 1.6.

Using the probability table, find the probabilities thatUsing the probability table, find the probabilities thatthis kind of computer will function for a monththis kind of computer will function for a month

i.i. without a breakdownwithout a breakdown (Ans: 0.2019)(Ans: 0.2019)ii.ii. with one breakdownwith one breakdown (Ans: 0.323)(Ans: 0.323)

iii.iii. at most two will breakdownat most two will breakdown (Ans: 0.7834)(Ans: 0.7834)

Q

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