ProbabilityDistributions...... Find the number of expected Jazz records when 4 records are selected...

Chapter 5Probability Distributions

Recall that one of the objectives of statistics is to make generalizations re-garding a specified population. And these generalizations are always subjectto uncertainties due to the limited information that can be obtained fromsample observations. One of the ways to deal with this problem is throughthe study of probability theories. Probability theory provides a way to con-struct a model that theoretically describes the behavior of a population thatis associated with the statistical experiment involved.

5.1 Random Variables

We recall that a statistical experiment is yields random outcomes. And thereare instances that we are just interested some of the details of the outcomes.For instance an experiment of tossing a fair die twice, there would be 36possible outcomes. If we are just interested in the number of heads in theoutcome of the toss, then we are only considering once characteristic of theoutcome of the experiment. And since the outcomes can vary from sampleto sample we may consider this characteristic our variable. Thus we definewhat we mean by a random variable.

59

CHAPTER 5. PROBABILITY DISTRIBUTIONS 60

Definition

A random variable is defined to be a function whose value is a real numberdetermined by each element in the sample space is called a random variable.A random variable is usually denoted by a capital letter and specific values ofthe random variable are represented by a small letter.

Example

For instance, in an experiment of tossing a coin thrice and we are only con-cerned with the outcome of the number of heads occurring in the experimentwe may associate the number 0, 1, 2, and 3 to the number of head that mayoccur in a particular outcome. To represent these values we may want to usea variable, a random variable. Random since we are not definite about thevalues of our variable. We just know the possible values it may take.If we let X be the random variable that represents the number of tails in theoutcome then we have the following: Given sample space

S = HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Sample points TTT HTT, THT, TTH HHT, HTH, THH HHHx 0 1 2 3

We recall that the sample space of a given experiment is the set of allpossible outcomes. And so if we define our random variable based on thatsample space we can categorize a random variable in the following manner:Discrete and Continuous.Definition

Discrete Sample Space - If a sample space contains a finite number ofpossibilities or an unending sequence with as many elements as there wholenumbers, it is called a discrete sample space.Continuous Sample Space - If a sample space contains an infinite numberof possibilities equal to a number of points on a line segment, it is called acontinuous sample space.

Remark

A Discrete Random Variable is a random variable which is defined on adiscrete sample space while a Continuous Random Variable is a randomvariable defined on a continuous sample space.


Example

Classify the following random variables as discrete or continuous.

1. the number of automobile accidents each year in Virginia Ans. discrete

2. the length of time to play 18 hole of golf Ans. continuous

3. the amount of milk produced by a certain cow per month Ans. continuous

4. the number of eggs laid each month by a specific hen Ans. discrete

5. the weight of grain in pound produced per acre Ans. continuous

Definition

Discrete Probability Distribution - A table or a formula listing all possiblevalues that a discrete random variable can assume, along with the associatedprobabilities, is called a discrete probabilities distribution.

Example

In an experiment of tossing a coin three times the following sample space isobtained:

S = HHH,HHT,HTH, THH,HTT, THT, TTH, TTT .

We define the random variable X, as the number of head in an outcome. Wesummarize the result of the experiment and identify the values of our randomvariable as well as the associated probability with each value of the randomvariable.

Sample points TTT HTT, THT, TTH HHT, HTH, THH HHH

x 0 1 2 3

P (X) = f(x)1

8

3

8

3

8

1

8


Example

Find the probability distribution given a random variable X defined as thesum of the numbers when a pair of dice is tossed. The following diagramillustrates all possible outcomes when a pair of dice is tossed and theassociated value of the random variable X.

x 2 3 4 5 6 7 8 9 10 11 12f(x) 1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

Usually it is convenient to represent all the probabilities associated witheach value of a random variable by a formula. And such formulas are calledprobability functions or probability distributions.Definition

Probability Density Function - The function with values f(x) is called aprobability density function for a continuous random variable x if the totalarea under its curve and above the x−axis is equal to 1 and if the area underthe curve between any two ordinates a and b gives the probability that therandom variable xbetween aand b.


5.2 Mean and Variance of Discrete Random

Variables

Definition

Given a discrete random variable X with a probability distribution f(x),themean or expected value of the random variable X is given by

µ = E[X] =n

∑

i=1

xi · f(xi)

And the variance of X is given by

σ2 = V ar[X] = E[

(X − µ)2]

=n

∑

i=1

(xi − µ)2 f(xi)

Example

Find the mean and variance of H, where H is a random variable which repre-sents the number of automobiles that are used for official business purpose onany given workday by a certain company. The probability distribution for His as follows:

H 1 2 3f(H) 0.3 0.4 0.3

µ = E[X] =n

∑

i=1

xi · f(xi) = (1)(0.3) + (2)(0.4) + (3)(0.3) = 2

V ar[X] =n

∑

i=1

(xi − µ)2 f(xi) = (1−2)2(0.3)+(2−2)2(0.4)+(3−2)2(0.3) = 0.6


Example

A shipment of 7 television sets contains 2 defectives were delivered by TanElectronic Company at a certain mall in Manila. If Dusit Hotel makesa random purchase of 3 of the sets. If G is the number of defective setspurchased by the hotel, find the mean and variance of G.Solution: First we have to determine the probabilities associated with eachvalue of the random variable G and obtain their corresponding probabilities.The values of G can only be 0, 1, and 2. We now have to find the probabilitieswhere the g = 0, 1, and 2. f(0) is the probability that the hotel purchases3 television sets where none of the set is defective. f(1) is the probabilitythat the hotel purchases 3 television sets where one of the set is defective.And so on, using our notions of finding probabilities we have the followingcomputation:

f(g) =n

N, where n is the number of outcomes in the event where the hotel

purchases g defective television sets and N is the total number of waysof selecting 3 television sets out of the 7 sets that were shipped; so wehave,N =7 C3 = 35.Suppose g = 0.Since in this event we are selecting none of the defective and 3 out of the5 television sets that are not defective we have the following computation;n =5 C3 ·2 C0 = 10. Thus we have f(0) = 10

35.

Suppose g = 1.Since in this event we are selecting one of the defective and 2 out of the5 television sets that are not defective we have the following computation;n =5 C2 ·2 C1 = 20. Thus we have f(1) = 20

35.

Suppose g = 2.Since in this event we are selecting two of the defective and 1 out of the5 television sets that are not defective we have the following computation;n =5 C1 ·2 C2 = 5. Thus we have f(2) = 5

35.

g 0 1 2f(g) 10

35

20

35

5

35

Now, computation of the mean and variance of G gives us µ = 6

7and σ2 =

0.4081653265 respectively.


5.3 Properties of the Mean and Variance of

a Random Variable

Theorem

If a and b are consants, then

µaX+b = E [aX + b] = aE [X] + b = aµX + b

and

σ2

aX+b = a2σ2

X .

The above theorem can ge generalized so that we have a function ofa random variable say, g(X) instead of a linear expression of the randomvariable X.

Theorem

Suppose we have a function g(x) of a discrete random variable X, we have

E [g(X)] =i=0∑

n

xif(xi).

We note that for a countable sample space, the limits of the summation nota-

tion should be up to infinity.

Exercises

(1) In an experiment of selecting 3 persons to form a committee from a set of4 boys and 3 girls. Let H represent the number of boys on the committee.

(2) Find the number of expected Jazz records when 4 records are selected atrandom from a collection consisting of 5 jazz records, 2 classical records,and 3 polka records.

(3) A coin is tossed three times. Let Y be the random variable that representsthe number of tails. Find the probability distribution of Y. Find the meanand variance of the probability distribution of the random variable Y.


(4) In an experiment of tossing a dice first and then tossing a coin, wherethe coin is tossed once if the dice resulted in an even number and twiceif the dice resulted to an odd number. Find the probability distributionof the random variable Y, where Y represents the number or heads inthe outcome.

(5) A stockroom clerk returns three safety helmets at random to three steelmill employees who had previously checked them. If Smith , Jones, andBrown, in that order, receive one of the three hats, list the sample spacepoints for the possible orders of returning the helmets, and find thevalue m of the random variable M that represents the number of correctmatches.

(6) A shipment of 8 similar microcomputers to a retail outlet contains 3that are defective. If a school makes a random purchase of 2 of thesecomputers, find the probability distribution for the number of defectives.

(7) An attendant at a car wash is paid according to the number of carsthat pass through. Suppose the probabilities are 1

12, 1

12, 14, 14, 16and 1

6,

respectively, and that the attendant receives $7, $9, $11, $13, $15, or $17between 4:00PM and 5:00PM on any sunny Friday. Find the attendant’sexpected earnings for this particular period.

(8) By investing in a particular stock, a person can make a profit in one yearof $4000 with probability 0.3 or take a loss of $1000 with probability 0.7.What is this person’s expected gain?

Chapter 6Discrete Probability Distributions

Probability distributions describe the behavior of our random variable andthis is presented either in a tabular form like the probability histogram orin a tabular form. And often we just need to generalize and summarize howto describe the distribution of the random variable. This is obtained by rep-resenting the probability distribution by means of a mathematical functionor formula. And in practice, we only need a handful of important discreteprobability distributions that would describe most random variable that canbe encountered in real world applications.Some of the discrete probability distributions are the following: Discrete

Uniform, Binomial , Negative Binomial, Geometric, Hypergeo-

metric , and Poisson distribution.

6.1 Discrete Uniform Probability Distributions

The simplest of all disrete probability distribution is one where the values ofthe random variable have equal probabilities.Definition

If the random variable X assumes the values x1, x2, . . . , xk with equal proba-bilities, then the discrete uniform distribution is given by

f(x; k) =1

k, x = x1, x2, . . . , xk.

67

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 68

Example

Suppose that a box contains 4 light bulbs with the following watts: 40-watt, 60-watt, 75-watt and a 100-watt. Consider the experiment of selectin a light bulbfrom this box. We now have a uniform distribution with x ∈ {40, 60, 75, 100}each with equal probabilities and hence,

f(x; 4) =1

4x = 40, 60, 75, 100.

Example

When a die is tossed, each element of the sample space S = {1, 2, 3, 4, 5, 6}occurs with probability 1

6. Therefore, we have a uniform distribution, with

f(x; 6) =1

6, x = 1, 2, 3, 4, 5, 6.

6.2 Binomial Probability Distributions

Definition

A binomial experiment or sometimes referred to as the Bernoulli Process hasthe following properties:

1. The experiment consists of n repeated trials

2. Each trial results in an outcome that may be classified as a success or afailure

3. The probability of a success, denoted by p, remains constant from trialto trial.

4. The repeated trials are independent.

Usually if the first 3 conditions are already met, the last condition is pre-sumably a forgone conclusion.


Definition

The number x of success of a random variable X in n trials of a binomialsexperiment is called a binomial random variable.If a binomial experiment can result in a success with probability p and thefailure with the probability q = 1− p, then the probability distribution of thebinomial random variable X, as the number of success in n independent trialsis

b(x;n; p) =

(

n

x

)

pxqn−x for x = 0, 1, 2 . . . , n.

The mean and variance of the binomial random variable are given by

µ = np and σ2 = npq.


Example

Find the probability of obtaining exactly three 2’s if an ordinary dice is tossed5 times.Solution: Suppose X is the random variable representing the number of 2’soccurring in tossing a dice 5 times. Check if the conditions of the binomialexperiment are satisfied.

1. The experiment consists of n repeated trials: There are 5 repeatedtrials of tossing a dice

2. Each trial results in an outcome that may be classified as asuccess or a failure: The outcome can be classified as a success whenthe result of the dice is 2 and a failure if the outcome is not 2.

3. The probability of a success, denoted by p, remains constantfrom trial to trial: The probability of a success on each of the 5 trials

is1

6and the probability of failure is

5

6.

4. The repeated trials are independent: We conclude that the trialsare independent from one another since the result of the first toss doesnot affect of the result of the next toss.

From this we see that the experiment satisfies the conditions of a binomialexperiment. Thus the random variable X is a binomial random variable with

n = 5; p =1

6; q =

5

6; and x = 3.

Thus the probability of obtaining exactly three 2’s if an ordinary dice is tossed5 times is given by

b(x;n; p) =

(

n

x

)

pxqn−x =

(

53

)(

1

6

)3 (

5

6

)2

= 0.032


6.3 Hypergeometric Probability Distributions

Definition

A Hypergeometric experiment has the following properties:

1. A random sample of size n is selected from a population of N items.

2. k of the N items may be classified as success and N − k as failures.

Definition

A random variable defined as the number of successes in a Hypergeometricexperiment is called a Hypergeometric Random Variable.If the population of size contains k items labeled as ”success” and N−k itemslabeled as ”failures” then the probability distribution of the Hypergeometricrandom variable X, the number of successes in a random sample of size n, is

h (x;N, n, k) =

(

k

x

)(

N − k

n− x

)

(

N

n

) for x = 0, 1, 2, . . . , n

The mean and variance of the hypergeometric random variable are given

µ =nk

Nand σ2 =

(

N − n

N − 1

)(

nk

N

)(

N − k

N

)


6.4 Negative Binomial Probability Distribu-

tions

Consider an experiment in which the properties are the same as those listedfor a binomial experiment, with the exception that the trials will be repeateduntil a fixed number of successes occur. From this, instead of finding theprobability of x number of successess in n trials where n is fixed, we are nowinterested in the probability that the kth succcess to occurr on the xth trial.This kind of experiments are called negative binomial experiments.Definition

If repeated independent trials can result in a success with probability p anda failure with probability q = 1 − p, then the probability distribution of therandom variable X, the number of the trial on which the kth success occurs isgiven by

b∗(x; k, p) =

(

x− 1k − 1

)

pkqx−k x = k, k + 1, k + 2, . . .

Example

Find the probability that a person tossing three coins will get either all headsor all tails for the second time on the fifth toss.

b∗(5; 2,1

4) =

37

256.

6.5 Geometric Probability Distributions

If we consider the special case of the negative binomial distribution wherek = 1, we have a probability distribution for the number of trials requiredfor a single/first success. This is called the geometric distribution.

Definition

If repeated independent trials can result in a success with probability p anda failure with probability q = 1 − p, then the probability distribution of therandom variable X, the number of the trial on which the first success occursis given by

g(x; p) = pqx−1 x = 1, 2, 3, . . .


Example

Find the probability that a person flipping a balanced coins requires 4 tossesto get a head.

g(4;1

2) =

1

16.

Example

In a certain manufacturing process it is known that, on the average 1 in every100 items is defective. What is the probability that the fifth item inspected isthe first defective item found?

We have x = 5, p = 0.01, thus we have g(4;1

2) =

1

16.

6.6 Poisson Probability Distributions

Definition

A poisson experiment has the following properties:

1. The number of outcomes occurring in one time interval or a specifiedregion is independent of the number of outcomes that occur in any otherdisjoint time interval or region space.

2. The probability that a single outcome will occur during a very short timeinterval or in a small region is proportional to the length of time intervalor the size of the region and does not depend on the outcomes occurringoutside this time interval or region.

3. The probability that more than one outcome will occur in a very shorttime interval or a small region is very small and can be assumed to benegligible.

Definition


The number X of success in a poisson experiment is called a poisson ran-

dom variable. The probability distribution of a poisson random variable X

representing the number of outcomes occurring in the given time interval orspecified region is

p(x;µ) =e−µµx

x!for x = 0, 1, 2, . . .

Where µ is the average number of outcomes occurring in the given time intervalor specified region and e = 2.71828 . . . is the natural number.

Example

The average number of days school is closed due to floods during the rainyseason in a city in Pampanga is 4. What is the probability that the schools inthis particular city in Pampanga will close for 6 days during a winter?

Solution: p(x = 6;µ = 4) =e−446

6!≈ 0.1042


Exercises

(1) On the average, the intersection of Taft Avenue and Buendia results in3 traffic accidents per month. What is the probability that in any givenmonth at this intersection

(a) exactly 5 accidents will occur?

(b) less than 3 accidents will occur?

(c) at least 2 accidents will occur?

(2) A basketball player’s shooting average is 0.25, what is the probabilitythat he gets exactly 1 shoot in his next 5 times attempt to shoot theball?

(3) A multiple-choice quiz has 10 questions, each with 4 possible answers ofwhich only one is correct. What is the probability that sheers guess workyields from 3 to 6 correct answers?

(4) If probability that a patient recovers from a leukemia is 0.4. And if 15people are known to have contracted this disease, what is the probabilitythat

(a) at least 13 survive?

(b) from 3 to 5 person survive?

(c) exactly 5 survive?

(5) In a Metro Manila, MMDA says that the need for money to by drugs isgiven as the reason for 55% of all thefts. What is the probability thatexactly 2 of the next 4 theft cases-reported to MMDA resulted from theneed for money to buy drugs?

(6) A homeowner plants 5 bulbs selected at random from a box containing5 rose bulbs and 4 sampaguita bulbs. What is the probability that heplanted 2 sampaguita bulbs and 3 rose bulbs?

(7) A professor in biology gave a multiple choice quiz with 10 items, eachwith 5 possible answers and only one of which is correct.

(a) What is the probability that a student took the test my merely guess-ing and got a score of 5?


(b) What is the probability that merely guessing the answers from thetest would yield a score of 4 to 8?

(c) What is the probability that merely guessing the answers from thetest would yield a score of at least 5?

(8) What is the probability that a waiter will refuse to serve alcoholic drinksto only 2 minors if he randomly checks the Identification cards of 5students from among the 10 students where 4 of which are not of legalage?

(9) The average number of patients arriving at the emergency room of Philip-pine General Hospital (PGH) on Monday nights between 9:00 pm up to12:00 midnight is 5. If we assume that the patients arrive at random andindependently, what is the probability that less than 5 patients arrive atthe emergency room of PGH on a Monday night from 9:00 pm to 12:00midnight?

(10) A box contains10 red marbles and 15 blue marbles and 5 marbles areselected at random from the box.

(a) What is the probability of obtaining at least 3 red marbles?

(b) What is the probability of obtaining at most 2 blue marbles?

(c) What is the probability of obtaining exactly 1 red marble?

(11) Suppose that the average number of earthquakes experienced in Min-danao is 10 per year. What is the probability that on a given year,Mindanao will experience at least 5 earthquakes?

(12) In certain computer shop, the typist commits on the average two typo-graphical error per page. What is the probability that the typist makes

(a) 3 or more errors?

(b) at least 1 error?

(c) no errors?

(13) In Davao, the probability that a household has a Pomelo tree in theirbackyard is 0.35. Find the probability that 4 out of the 10 randomlyselected houses has a Pomelo tree in their backyard.


(14) Batanes is hit by 8 storms per year on the average. What is the proba-bility that on a certain year, Batanes will be hit by at least 5 storms?

(15) Warranty records show that the probability that a new car needs repairin the first 90 days is 0.10. If a sample of ten new cars is selected,

(a) what is the probability that none needs a warranty repair?

(b) what is the probability that at least 3 needs a warranty repair?

(c) what is the probability that from 5 to 8 (inclusive) needs a warrantyrepair?

(d) what is the probability that at most 6 needs a warranty repair?

(16) The quality control manager of Mandy’s Cookies is inspecting a batchof chocolate chip cookies that has just been baked. If the productionprocess is in control, the average number of chip parts per cookie is 6.0.What is the probability that in any particular cookie being inspected,

(a) exactly 5 chip parts will be found?

(b) more than 3 chip parts will be found?

(c) less than 7 chip parts will be found?

Chapter 7Continuous Probability Distributions

A continuous random variable has a probability of zero of assuming exactlyany of its values. And due to the nature of the random variable, we can-not enumerate all of its possible values. Thus when we consider continuousrandom variables and their probabilities, we only look at probabilities ofthe random variable have a value in a specified interval. However, we willonly consider one type of continuous random variable, the Normal randomvariable and its associated probability distribution.

7.1 Normal Probability Distributions

The normal distribution is one of the most important continuous distributionin the entire field of statistics. And the graph of this distribution is calledthe normal curve. This distribution is sometimes called the Gaussian

distribution in honor of Karl Friedrich Gauss, who derived its equation.

78

CHAPTER 7. CONTINUOUS PROBABILITY DISTRIBUTIONS 79

Remark

Properties of the normal curve:

1. It has a bell-shaped curve.

2. The mode, which is the point on the horizontal axis where the curve isa maximum, occurs at x = µ .

3. The curve is symmetric about a vertical axis through the mean, µ .

4. The normal curve approaches the horizontal axis asymptotically as weproceed in either direction away from the mean. (The graph approachesthe x-axis but the graph will never intersect the x-axis).

5. The total area under the curve and above the horizontal axis is equalto 1.

Definition

A continuous random variable X having the bell-shaped distribution is calleda normal random variable. The mathematical equation for the probabilitydistribution of the normal random variable depends on two parameters µ andσ ; its mean and standard deviation. Thus we denote the probability densityof X by N(x;µ; σ).If X is a normal random variable with mean µ and variance σ2 , then theequation of the normal curve is

N(x;µ; σ) =1√2πσ

e1

2(x−µ

σ)2

for −∞ < x <∞.

Remark

It is difficult to compute for the probabilities of a normal random variableusing the above formula. However, another way of calculating such prob-abilities is through the transformation of a normal random variable to itscorresponding standard normal random variable. By transforming a normalrandom variable to a standard normal random variable we can now deter-mine probabilities of the said random variable. Thus we define the standardnormal random variable and its distribution.


Definition

The distribution of a normal random variable with mean µ = 0 and standarddeviation σ = 1 is called a standard normal distribution. In order totransform a normal random variable to a standard normal one, we use thefollowing formula:

Z =X − µ

σ.

By using the table for the standard normal random variable, we can nowdetermine the probability of any normal random variable by transforming thegiven random variable to its corresponding standard normal random variable.Example

Given a normally distributed random variable X with mean 18 and standarddeviation of 2.5, find

1. P (X < 15). 2. P (17 < X < 21).

Solution:

(a) P (X < 15) = P

(

Z <15− 18

2.5

)

= P (Z < −1.2) = 0.1151

Refer to the standard normal table:

(b) P (17 < X < 21) = P

(

17− 18

2.5< Z <

21− 18

2.5

)

= P (0.4 < Z < 1.2) =

P (Z < 1.2)− P (−0.4) = 0.8849− 0.3446 = 0.5403


Example

An electrical firm manufacturers light bulbs that have a length of life that isnormally distributed with mean equal to 800 hours and standard deviationof 40 hours. Find the probability that the bulb burns between 778 and 834hours.Solution:

The distribution of the light bulbs is illustrated by the figure below:

The z values corresponding to x1 = 778 and x2 = 834 are

z1 =778− 800

40= −0.55,

z2 =834− 800

40= 0.85.

Hence,

P (778 < X < 834) = P (−0.55 < Z < 0.85)= P (Z < 0.85)− P (Z < −0.55)= 0.8023− 0.2912= 0.5111


Exercises

(1) Given a normally distributed random variable X with mean 18 and stan-dard deviation of 2.5, find the value of k such that

(2) A certain type of storage battery last on the average 3.0 years, with astandard deviation of 0.5 years. Assuming that the battery lives arenormally distributed, find the probability that a given battery will lastless than 2.3 years.

(3) An electrical firm manufactures light bulbs that have a length of life thatis normally distributed with mean equal to 800 hours and a standarddeviation of 40 hours. Find the probability that a bulb burns between778 and 834 hours.

(4) If the average height of miniature poodles is 30 centimeters, with a stan-dard deviation of 4.1 cm, what percentage of miniature poodles exceeds35 cm in height, assuming that the height follows a normal distributionand can be measured to any desired degree of accuracy?

(5) The quality grade-point averages of 300 college freshmen follow approxi-mately a normal distribution with a mean of 2.1 and a standard deviationof 0.8. How many of these freshmen would you expect to have a scorebetween 2.5 and 3.5 inclusive if the point averages are computed to thenearest tenth?

(6) A set of final examination grades in an introductory statistics course wasfound to be normally distributed, with a mean of 73 and a variance of64.

(a) What is the probability of getting a grade of 91 or less in this exam?

(b) What percentage of students scored between 81 and 89?

(c) Only 5% of the students taking the test scored higher than whatgrade?

(7) Plastic bags used for packaging produce re manufactured so that thebreaking strength of the bag is normally distributed with a mean of 5pounds per square inch and a standard deviation of 1.5 pounds per squareinch.


(a) What proportion of the bags produced have a mean breaking strengthof between 5 and 5.5 pounds per square inch?

(b) What is the probability that a randomly selected bag will have amean breaking strength of at least 6 pounds per square inch?

(c) What percentage of the bags have a mean breaking strength of lessthan 4.17 pound per square inch?

(d) Between what two values symmetrically distributed around the meanwill 95% of the breaking strengths fall?

(8) If we know that the length of time it takes a college student to find aparking spot in the university parking lot follows a normal distributionwith a mean of 3.5 minutes and a standard deviation of 1 minute, findthe probability that if we select 36 randomly selected college students,the average time it would take for them to find a parking spot is

(a) less than 3.2 minutes?

(b) between 3.4 and 3.7 minutes?

(c) more than 3.8 minutes?

Chapter 8Estimation of Parameters

Recall that one of the objectives of statistics is to make inferences concern-ing a population. And these inferences are based only in partial informationregarding the population, since the information of the statistics is based onthe sample. And the value of our statistics may vary from sample to sample.Because of this, we need to understand first the variations that are associatedwith the statistic involved in our inference.

Another concern regarding inference based on sample information is thefactor of how the samples are taken and how large the sample size is so thatmeaningful interpretations can be drawn from the sample. This concern isaddressed in specialized study of statistics, Sampling Theory, which is be-yond the scope of our study in this course. But an overview of terms andconcepts in sampling theory are discussed in section 6.2.

8.1 Sampling and Sampling Distribution

A statistic is a numerical descriptive measure derived from a sample. How-ever, there are random samples thus producing different values for a certainstatistic. Since statistic varies from sample to sample then we can say thata statistic is also a random variable.

Recall that we can construct a probability distribution for a random vari-able hence probability distribution for a statistic can also be constructed.We call the probability distribution of a statistic a sampling distribution.

84

CHAPTER 8. ESTIMATION OF PARAMETERS 85

Definition

The sampling distribution of a statistic is the probability distributionfor the possible values of the statistic that results when random samples ofsize n are repeatedly drawn from the population.

Example

A population consists of N = 5 numbers :1, 2, 3, 4, and 5. If a random sampleof size n = 3 is selected, find the sampling distributions for the sample mean.Solution:

Computation of the population mean and variance will give us µ = 3 andσ2 = 2. Since there are only 5 distinct and equally likely elements in ourpopulation the probability that one will occur is the same for all elements in

the population, that is, P (x) =1

5. Since we are only choosing 3 from the

population there are only 5C3 = 10 different possible samples and they are asfollows:

No. Sample Sample Mean x̄

1 1,2,3 22 1,2,4 2.3333 1,2,5 2.6674 1,3,4 2.6675 1,3,5 36 1,4,5 3.3337 2,3,4 38 2,3,5 3.3339 2,4,5 3.66710 3,4,5 4

Thus the sampling distribution of the sample mean is

x̄ 2 2.333 2.667 3 3.333 3.667 4f(x̄) 0.01 0.01 0.02 0.02 0.02 0.01 0.01

Notice that if we take the average of all the sample means we will get thevalue 3 and a variance of 1

3. But if we increase our sample size say n = 4 and

compute for the sampling distribution of x̄ again, we will still get a mean of 3but a variance of 0.125.


Remark

We can notice that µx̄ the mean of the sample means is equal to the popula-tion mean, and the variance σ2

x̄ or the standard deviation σx̄will decrease asour sample size increases.If all possible random samples of size n are drawn, without replacement, froma finite population of size N with mean µ and standard deviation σ, then thesampling distribution of the sample mean will be approximately normallydistributed and the mean and standard deviation is given by

µx̄ = µ and σx̄ =σ√n

√

N − n

N − 1.

The factor√

N−nN−1

is called the finite correction factor. For large or

infinite populations , this correction factor will be approximately equal to 1.Hence σx̄ = σ√

n

The above notion regarding the sampling distribution of the sample meangives us the foundation of the next theorem; the central limit theorem. Thecentral limit theorem states that in general situations and condition sums andmeans of samples of random observations that are drawn from a populationof any distribution tends to possess, approximately, a bell shaped distributionin repeated sampling. And thus the distribution can be assumed approxi-mately normal.

One of the significance of the central limit theorem is that it explains whysome of the observations in the real world tends to possess an approximatelya normal distribution. To illustrate this significance, consider the weight ofa person. Weight can be affected by many factors whether environmental orgenetics for instance, family lineage such as the parents weights. Anotherfactor can be the physical activities of the person. All this possibilities mayreally affect the weight of a person but the central limit theorem togetherwith other theorems applicable to the normal distribution provides an expla-nation of this events.

Another significance of the central limit theorem and probably the mostimportant attribute is its application to statistical inference. Many statis-tical estimators that are used to make inferences about a population haveparameters that are sums and averages of sample observations.


Theorem

If random samples of size n are taken from a non-normal population, then

as n becomes large, the sampling distribution of the sample mean becomes

approximately normally distributed with mean and standard deviation

µx̄ and σx̄ =σ√n

where µ and sigma are the mean and standard deviation of the population,

respectively.

Thus, z =x̄− µx̄

(

σx̄ = σ√n

) is a value of a standard normal random variable Z.

Remark

1. If samples are taken from a population having a normal distribution,then the sampling distribution of the sample mean will have a normaldistribution no matter what n is.

2. If samples are taken from a population which is not normally dis-tributed, then the sampling distribution of the sample mean will havean approximate normal distribution only for large samples, that is,when n ≥ 30

3. The standard deviation of the sampling distribution of x̄ ,σx̄ , is calledthe standard error of the sample mean .

8.2 Sampling Procedures

The need for sampling arises from the fact that information that is neededcannot be obtained easily if the size of the population is large. And it is im-practical to observe each element in a population. The next obvious questionis: How large should the sample size be so that meaningful interpretationscan be drawn from them? But to answer this question, another problemarises. The design of how the information is collected needs to be consideredso that the appropriate sample size can be determined. All this notions andconcepts are beyond the scope of this book. But this section introduces the


concepts in sampling theory.So far the discussions of sampling distributions from the previous chap-

ters have been restricted and assumed simple random samples but there areother ways of obtaining samples from a population. And these other sam-pling methods offers a more precise and efficient in such a way that moreinformation regarding the population is attained at a lower cost than simplerandom sampling.

We can classify sampling methods in to two types: probability sam-

pling and non-probability sampling. The difference between them is thatin probability sampling, every unit has a ’chance’ of being selected. This isnot true for non-probability sampling; every item in a population does nothave an equal chance of being selected. The following methods of samplingfalls under the probability sampling.

Definition

In simple random sampling, each member of a population has an equalchance of being selected. One of the necessary elements to perform a simplerandom sample is a list of all of the elements in the population.

Simple random sampling can be performed with or without replacement.This means that whenever a element in the population is selected the ob-served element is either replaced in the population or not. In the samplingwith replacement, there would be a possibility that the sampled observa-tion may be selected twice or more. Usually, the simple random samplingapproach is conducted without replacement because it is more convenientand gives more precise results. For simplicity and scoping of our discussions,when we discuss simple random sampling, we will refer to sampling withoutreplacement.

One of the properties of the simple random sample is that it is unbiased.This means that each element has the same chance of being selected. An-other property of the simple random sample is its independence. That is theselection of one element has no influence on the selection of other elements.However, a completely unbiased and independent sample is very hard to findin the real world.


Definition

In stratified sampling, the population is divided into homogeneous and nonoverlapping groups called strata, and then independent random samples areselected from each stratum. This means that elements in a certain strata,elements have characteristics in common. To achieve the assumption of homo-geneity, stratification must be done in such a way that elements in a stratumhave some characteristics in common and those characteristic has in some wayrelated to the characteristic in study. Any of the sampling methods can beused to sample within each stratum. The sampling method can vary from onestratum to another. If simple random sampling is used to select the samplewithin each stratum, the sample design is called stratified simple random

sampling.

An advantage of using stratified sampling is that it can make the sam-pling strategy more efficient. For example, if every person in a certain grouphad the same salary, then a sample of one individual would be enough to geta precise estimate of the average salary. Creating strata where units sharesimilar characteristics is one factor for efficient selection of samples. Youwould only need to select minimal sample from this group since informationcaptured in a strata would somehow describe the whole strata. Then youcould combine this information from each stratum to get a precise estimatefor the entire population. But using a simple random sampling approachin the whole population without stratification, the sample would need to belarger than the total of all stratum samples to get an estimate of total incomewith the same level of precision.

Stratification would entail varying sizes in each stratum, and thus con-sideration on the sample size in each stratum must be given. A procedurecalled proportional allocation is implemented such that sample sizes areproportional to the size of different strata.

Sample Sizes for Proportional Allocation: If a population of size N isdivided into k strata, N1, N2, . . . , Nk and then the sample size for the ith

stratum is given by ,

ni =

(

nNi

N

)

Where n is the total sample size.


Definition

Systematic random sampling means that there is a gap, or interval,between each selected unit in the sample. This type of sampling method isalso called Interval Sampling. In order to select a systematic sample, youneed to follow these steps:

1. Number the units on your population from 1 to N (where N is thepopulation size).

2. Determine the sampling interval (k) by dividing the number of units inthe population by the desired sample size.

3. Select a number between one and k at random. This number is called therandom start and would be the first number included in your sample.

4. Select every kth unit after that first number.

Definition

In cluster sampling, the population is divided into clusters and then a randomsample of the clusters is selected. The selected clusters may be completelysampled or a random sample may be obtained from the selected clusters.

For instance, a large company like San Miguel Food Corporation has 30plants located throughout the Philippines. In order to access a new totalquality plan, the 30 plants are considered clusters and five of the plants arerandomly selected. All of the quality control personnel at the five selectedplants are asked to evaluate the total quality plan.

Another example might be the national survey during election periods.A city can be divided in to blocks and then from the clusters of blocks, asimple random sample is performed so that every house in the selected blockis surveyed.

Note.

1. Most statistical methods depend on the independence and lack of biasof the simple random sample. The result discussed in this book appliesto simple random sample ONLY. And for other sampling procedures,the results will have to be modified and that discussions are out of scope


in this book. Reference to specialized sampling textbooks is encouragedto explore those concepts.

2. We must also take note that without the assumption of randomizationin obtaining the samples, results and inferences that will be obtainedfrom the data cannot be dependable. No matter how sophisticated themodification is in the statistical analysis, if the assumption of random-ness is not present, interpretations in the sample will be doubtful. Thesignificance of randomness statistically guarantees the accuracy of asurvey.

8.3 Estimating the Population Mean

Procedures and formulas used in estimating values of unknown populationparameters that are based on information provided in a sample data are basedon the theory of sampling distributions and the methods used to collect thesesample. The sampling distributions allow us to associates specific levels ofconfidence with each statistical inference. And thus enabling us to quantifyhow much confidence we place in a sample statistic correctly estimating thepopulation parameter.Definition

An estimator is a rule, usually expressed as a formula that tells us how tocalculate an estimate based on information in the sample.

We can classify estimators into two, point estimators and interval esti-mators.

1. Point estimation - Based on sample data, a single number is cal-culated to estimate the population parameter. The rule or formulathat describes this calculation is called the point estimator, and theresulting number is called the point estimate.

2. Interval estimation - Based on sample data, two numbers are calcu-lated to form an interval within which the parameter is expected to lie.The rule or formula that describes this calculation is called the inter-val estimator, and the resulting pair of numbers is called an interval

estimate or confidence interval.


A. Point Estimation

(1) The best point estimate for the population mean,µ , is the samplemean,x̄ .

(2) The point estimator x̄ is unbiased with standard error given by SE =σ√n.

(3) The margin of error of the point estimate,x̄ , is given byx̄± 1.96SE.

(4) If σ is unknown and n ≥ 30, the sample standard deviation s can beused to approximate σ.

B. Interval estimation To construct and interval estimate for the pop-ulation mean, we consider two cases. One case is when the standarddeviation of the population is known or unknown by the sample size islarge enough, that is, n ≥ 30. The other case is when the standarddeviation is not known and the sample size is less than 30.

(a) CASE 1: If σ is known or σ unknown but n ≥ 30, a (1 − α)100%confidence interval for a population mean,µ is given by:

x̄± Zα

2

(

σ√n

)

where:x̄ = sample meanZα

2= z-score with an area of α

2to the right

n = sample sizeσ = population standard deviation

(b) CASE 2: If σ unknown and n ≥ 30, a (1 − α)100% confidenceinterval for a population mean,µ is given by:

x̄± tα

2

(

s√n

)

where:x̄ = sample meantα

2= critical t-value with an area of α

2to the right and a degree of freedom n− 1

n = sample sizes = sample standard deviation


Remark

(1) If x̄ is used as an estimate of µ , we can then be (1− α)100% confident

that the error will not exceed Zα

2

(

σ√n

)

or tα

2

(

s√n

)

.

(2) If x̄ is used as an estimate of µ , we can then be (1− α)100% confidentthat the error will not exceed a specified amount e when the sample size

is n =(

Zα

2σ

e

)2

.

Exercises

(1) A scientist interested in monitoring chemical contaminants in food, andthereby the accumulation of contaminants in human diets, selected arandom sample of n = 50 male adults. It was found that the averagedaily intake of dairy products was 756 grams per day with a standarddeviation of 35 grams per day. Construct a 95% confidence interval forthe mean daily intake of dairy products for men.

(2) A random sample of 12 female students in a certain dormitory showedan average weekly expenditure of P400 for snack foods, with a standarddeviation of P12.50. Construct a 90% confidence interval for the averageamount spent on snack foods by female students living in this dormitory,assuming the expenditures to be normally distributed.

(3) The contents of 7 similar containers of sulfuric acid are 9.8, 10.2, 10.4,9.8, 10.0, 10.2, and 9.6 liters. Find a 95% confidence interval for themean content of all such containers, assuming an approximate normaldistribution for container contents.

(4) The mean and standard deviation for the quality grade point averagesof a random sample of 36 college seniors are calculated to be 2.6 and 0.3respectively. Find the 95% and 99% confidence intervals for the mean ofthe entire senior class.

(5) The following data were collected based from a sample in an experiment:n = 64,x̄ = 22.5 and s = 3.4.

(a) What is the point estimate of µ?

(b) What is the margin of error associated with the point estimate of µ?


(c) Construct a 99% confidence interval for µ.

(d) What is the maximum error of the estimate for (c)?

(6) A telephone answering service completes a report in which the lengthof the call is recorded, at the end of each call. A random sample of 9reports yields a mean length of call of 1.2 minutes. Construct a 95%confidence interval for the mean length of call for the whole telephoneanswering service company if it is known that the population is normallydistributed with a standard deviation of 0.6 minutes.

(7) A random sample of 10 chocolate bars has an average of 230 calorieswith a standard deviation of 15 calories. Assuming that the distributionof the calories is approximately normal.

(a) Construct a 99% confidence mean calories content of this chocolatebar.

(b) How large a sample is needed if we wish to be 99% confident thatour sample mean will be within 5 calories of the true mean?

Chapter 9Statistical Tables and Formulas

95

CHAPTER 9. STATISTICAL TABLES AND FORMULAS 96

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

-3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002

-3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003

-3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005

-3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007

-3 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010

-2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014

-2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019

-2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026

-2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036

-2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048

-2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064

-2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084

-2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110

-2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143

-2 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183

-1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233

-1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294

-1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367

-1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455

-1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559

-1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681

-1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823

-1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985

-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170

-1 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379

-0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611

-0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867

-0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148

-0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451

-0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

-0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121

-0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483

-0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859

-0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247

0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641

Standard Normal Table


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852

0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890

2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916

2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952

2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964

2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974

2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981

2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993

3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995

3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997

3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998

Standard Normal Table

Bibliography

[1] R. Walpole Introduction to Statistics. Pearson Education South Asia PteLtd.2004.

[2] R. Walpole, R. Myers, K. Ye , and S. Myers Probability and Statistics for

Engineers and Scientists. Pearson Education International.2007.

[3] L. Stephens Schaums’s Outline of Theory and Problems in Beginning

Statistics. The McGraw-Hill Companies, Inc.1998.

[4] L. Kazmier Schaums’s Easy Outlines: Business Statistics. The McGraw-Hill Companies, Inc.2003.

[5] L. Gonick and W. Smith Cartoon Guide to Statistics. HarperCollins Pub-lisher, 1993.

[6] A. Graham Developing Thinking in Statistics Paul Chapman Publishing2006.

[7] R. Khazanie Elementary Statistics: In a World of Applications. GoodyearPublishing Inc., 1979.

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