Probability Distributions

Chapter 5 Probability Distributions Recall that one of the objectives of statistics is to make generalizations regarding a specified population. And these generalizations are always subject to uncertainties due to the limited information that can be obtained from sample observations. One of the ways to deal with this problem is through the study of probability theories. Probability theory provides a way to construct a model that theoretically describes the behavior of a population that is associated with the statistical experiment involved. 5.1 Random Variables We recall that a statistical experiment is yields random outcomes. And there are 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 36 possible outcomes. If we are just interested in the number of heads in the outcome of the toss, then we are only considering once characteristic of the outcome of the experiment. And since the outcomes can vary from sample to sample we may consider this characteristic our variable. Thus we define what we mean by a random variable. 59

CHAPTER 5. PROBABILITY DISTRIBUTIONS 60 A random variable is defined to be a function whose value is a real number determined 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 of the random variable are represented by a small letter. Example For instance, in an experiment of tossing a coin thrice and we are only concerned with the outcome of the number of heads occurring in the experiment we may associate the number 0,1,2, and 3 to the number of head that may occur in a particular outcome. To represent these values we may want to use a variable, a random variable. Random since we are not definite about the values 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 the outcome 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 HHH x 0 1 2 3 We recall that the sample space of a given experiment is the set of all possible outcomes. And so if we define our random variable based on that sample space we can categorize a random variable in the following manner: Discrete and Continuous. Discrete Sample Space - If a sample space contains a finite number of possibilities or an unending sequence with as many elements as there whole numbers, it is called a discrete sample space. Continuous Sample Space - If a sample space contains an infinite number of possibilities equal to a number of points on a line segment, it is called a continuous sample space. Remark A Discrete Random Variable is a random variable which is defined on a discrete sample space while a Continuous Random Variable is a random variable defined on a continuous sample space.

CHAPTER 5. PROBABILITY DISTRIBUTIONS 61 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 Discrete Probability Distribution- A table or a formula listing all possible values that a discrete random variable can assume, along with the associated probabilities, is called a discrete probabilities distribution. Example In an experiment of tossing a coin three times the following sample space is obtained: S = HHH,HHT,HTH,THH,HTT,THT,TTH,TTT. We define the random variable X, as the number of head in an outcome. We summarize the result of the experiment and identify the values of our random variable as well as the associated probability with each value of the random variable. 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

CHAPTER 5. PROBABILITY DISTRIBUTIONS 62 Example Find the probability distribution given a random variable X defined as the sum of the numbers when a pair of dice is tossed. The following diagram illustrates all possible outcomes when a pair of dice is tossed and the associated value of the random variable X. x 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 5 4 3 2 1 f(x) 36 36 36 36 36 36 36 36 36 36 36 Usually it is convenient to represent all the probabilities associated with each value of a random variable by a formula. And such formulas are called probability functions or probability distributions. Probability Density Function - The function with values f(x) is called a probability density function for a continuous random variable x if the total area under its curve and above the x axis is equal to 1 and if the area under the curve between any two ordinates a and b gives the probability that the random variable xbetween aand b.

CHAPTER 5. PROBABILITY DISTRIBUTIONS 63 5.2 Mean and Variance of Discrete Random Variables Given a discrete random variable X with a probability distribution f(x),the mean or expected value of the random variable X is given by µ = E[X] = n x i f(x i ) i=1 And the variance of X is given by σ 2 = Var[X] = E [ (X µ) 2] = n (x i µ) 2 f(x i ) i=1 Example Find the mean and variance of H, where H is a random variable which represents the number of automobiles that are used for official business purpose on any given workday by a certain company. The probability distribution for H is as follows: Var[X] = µ = E[X] = H 1 2 3 f(h) 0.3 0.4 0.3 n x i f(x i ) = (1)(0.3)+(2)(0.4)+(3)(0.3) = 2 i=1 n (x i µ) 2 f(x i ) = (1 2) 2 (0.3)+(2 2) 2 (0.4)+(3 2) 2 (0.3) = 0.6 i=1

CHAPTER 5. PROBABILITY DISTRIBUTIONS 64 Example A shipment of 7 television sets contains 2 defectives were delivered by Tan Electronic Company at a certain mall in Manila. If Dusit Hotel makes a random purchase of 3 of the sets. If G is the number of defective sets purchased by the hotel, find the mean and variance of G. Solution: First we have to determine the probabilities associated with each value 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 probabilities where the g = 0,1, and 2. f(0) is the probability that the hotel purchases 3 television sets where none of the set is defective. f(1) is the probability that 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 following computation: f(g) = n, where n is the number of outcomes in the event where the hotel N purchases g defective television sets and N is the total number of ways of selecting 3 television sets out of the 7 sets that were shipped; so we have,n = 7 C 3 = 35. Suppose g = 0. Since in this event we are selecting none of the defective and 3 out of the 5 television sets that are not defective we have the following computation; n = 5 C 3 2 C 0 = 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 the 5 television sets that are not defective we have the following computation; n = 5 C 2 2 C 1 = 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 the 5 television sets that are not defective we have the following computation; n = 5 C 1 2 C 2 = 5. Thus we have f(2) = 5. 35 g 0 1 2 f(g) 10 35 Now, computation of the mean and variance of G gives us µ = 6 7 and σ2 = 0.4081653265 respectively. 20 35 5 35

CHAPTER 5. PROBABILITY DISTRIBUTIONS 65 5.3 Properties of the Mean and Variance of a Random Variable Theorem If a and b are consants, then µ ax+b = E[aX +b] = ae[x]+b = aµ X +b and σ 2 ax+b = a 2 σ 2 X. The above theorem can ge generalized so that we have a function of a random variable say, g(x) instead of a linear expression of the random variable X. Theorem Suppose we have a function g(x) of a discrete random variable X, we have i=0 E[g(X)] = x i f(x i ). We note that for a countable sample space, the limits of the summation notation should be up to infinity. n Exercises (1) In an experiment of selecting 3 persons to form a committee from a set of 4 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 at random from a collection consisting of 5 jazz records, 2 classical records, and 3 polka records. (3) Acoinistossedthreetimes. LetYbetherandomvariablethatrepresents the number of tails. Find the probability distribution of Y. Find the mean and variance of the probability distribution of the random variable Y.

CHAPTER 5. PROBABILITY DISTRIBUTIONS 66 (4) In an experiment of tossing a dice first and then tossing a coin, where the coin is tossed once if the dice resulted in an even number and twice if the dice resulted to an odd number. Find the probability distribution of the random variable Y, where Y represents the number or heads in the outcome. (5) A stockroom clerk returns three safety helmets at random to three steel mill employees who had previously checked them. If Smith, Jones, and Brown, in that order, receive one of the three hats, list the sample space points for the possible orders of returning the helmets, and find the value m of the random variable M that represents the number of correct matches. (6) A shipment of 8 similar microcomputers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives. (7) An attendant at a car wash is paid according to the number of cars that pass through. Suppose the probabilities are 1 12, 1 12, 1 4, 1 4, 1 6 and 1 6, respectively, and that the attendant receives $7, $9, $11, $13, $15, or $17 between 4:00PM and 5:00PM on any sunny Friday. Find the attendant s expected earnings for this particular period. (8) By investing in a particular stock, a person can make a profit in one year of $4000 with probability 0.3 or take a loss of $1000 with probability 0.7. What is this person s expected gain?

Chapter 6 Discrete Probability Distributions Probability distributions describe the behavior of our random variable and this is presented either in a tabular form like the probability histogram or in a tabular form. And often we just need to generalize and summarize how to describe the distribution of the random variable. This is obtained by representing the probability distribution by means of a mathematical function or formula. And in practice, we only need a handful of important discrete probability distributions that would describe most random variable that can be encountered in real world applications. Some of the discrete probability distributions are the following: Discrete Uniform, Binomial, Negative Binomial, Geometric, Hypergeometric, and Poisson distribution. 6.1 Discrete Uniform Probability Distributions The simplest of all disrete probability distribution is one where the values of the random variable have equal probabilities. If the random variable X assumes the values x 1,x 2,...,x k with equal probabilities, then the discrete uniform distribution is given by f(x;k) = 1 k,x = x 1,x 2,...,x k. 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 bulb from this box. We now have a uniform distribution with x {40,60,75,100} each with equal probabilities and hence, f(x;4) = 1 4 x = 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. Therefore, we have a uniform distribution, with 6 f(x;6) = 1 6, x = 1,2,3,4,5,6. 6.2 Binomial Probability Distributions A binomial experiment or sometimes referred to as the Bernoulli Process has the 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 a failure 3. The probability of a success, denoted by p, remains constant from trial to trial. 4. The repeated trials are independent. Usually if the first 3 conditions are already met, the last condition is presumably a forgone conclusion.

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 69 The number x of success of a random variable X in n trials of a binomials experiment is called a binomial random variable. If a binomial experiment can result in a success with probability p and the failure with the probability q = 1 p, then the probability distribution of the binomial random variable X, as the number of success in n independent trials is b(x;n;p) = ( n x ) p x q n x for x = 0,1,2...,n. The mean and variance of the binomial random variable are given by µ = np and σ 2 = npq.

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 70 Example Find the probability of obtaining exactly three 2 s if an ordinary dice is tossed 5 times. Solution: Suppose X is the random variable representing the number of 2 s occurring in tossing a dice 5 times. Check if the conditions of the binomial experiment are satisfied. 1. The experiment consists of n repeated trials: There are 5 repeated trials of tossing a dice 2. Each trial results in an outcome that may be classified as a success or a failure: The outcome can be classified as a success when the 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 constant from trial to trial: The probability of a success on each of the 5 trials is 1 6 and the probability of failure is 5 6. 4. The repeated trials are independent: We conclude that the trials are independent from one another since the result of the first toss does not affect of the result of the next toss. From this we see that the experiment satisfies the conditions of a binomial experiment. Thus the random variable X is a binomial random variable with n = 5; p = 1 6 ; q = 5 ; and x = 3. 6 Thus the probability of obtaining exactly three 2 s if an ordinary dice is tossed 5 times is given by b(x;n;p) = ( n x ) ( 5 p x q n x = 3 )( ) 3 ( ) 2 1 5 = 0.032 6 6

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 71 6.3 Hypergeometric Probability Distributions 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. A random variable defined as the number of successes in a Hypergeometric experiment is called a Hypergeometric Random Variable. If the population of size contains k items labeled as success and N k items labeled as failures then the probability distribution of the Hypergeometric random 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 ( ) for x = 0,1,2,...,n N n The mean and variance of the hypergeometric random variable are given µ = nk ( )( )( ) N n nk N k N and σ2 = N 1 N N

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 72 6.4 Negative Binomial Probability Distributions Consider an experiment in which the properties are the same as those listed for a binomial experiment, with the exception that the trials will be repeated until a fixed number of successes occur. From this, instead of finding the probability of x number of successess in n trials where n is fixed, we are now interested in the probability that the k th succcess to occurr on the x th trial. This kind of experiments are called negative binomial experiments. If repeated independent trials can result in a success with probability p and a failure with probability q = 1 p, then the probability distribution of the random variable X, the number of the trial on which the k th success occurs is given by Example b (x;k,p) = ( x 1 k 1 ) p k q x k x = k,k +1,k +2,... Find the probability that a person tossing three coins will get either all heads or 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 where k = 1, we have a probability distribution for the number of trials required for a single/first success. This is called the geometric distribution. If repeated independent trials can result in a success with probability p and a failure with probability q = 1 p, then the probability distribution of the random variable X, the number of the trial on which the first success occurs is given by g(x;p) = pq x 1 x = 1,2,3,...

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 73 Example Find the probability that a person flipping a balanced coins requires 4 tosses to get a head. g(4; 1 2 ) = 1 16. Example In a certain manufacturing process it is known that, on the average 1 in every 100 items is defective. What is the probability that the fifth item inspected is the 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 A poisson experiment has the following properties: 1. The number of outcomes occurring in one time interval or a specified region is independent of the number of outcomes that occur in any other disjoint time interval or region space. 2. The probability that a single outcome will occur during a very short time interval or in a small region is proportional to the length of time interval or the size of the region and does not depend on the outcomes occurring outside this time interval or region. 3. The probability that more than one outcome will occur in a very short time interval or a small region is very small and can be assumed to be negligible.

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 74 The number X of success in a poisson experiment is called a poisson random variable. The probability distribution of a poisson random variable X representing the number of outcomes occurring in the given time interval or specified region is p(x;µ) = e µ µ x x! for x = 0,1,2,... Whereµistheaveragenumberofoutcomesoccurringinthegiventimeinterval or specified region and e = 2.71828... is the natural number. Example The average number of days school is closed due to floods during the rainy season in a city in Pampanga is 4. What is the probability that the schools in this particular city in Pampanga will close for 6 days during a winter? Solution: p(x = 6;µ = 4) = e 4 4 6 0.1042 6!

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 75 Exercises (1) On the average, the intersection of Taft Avenue and Buendia results in 3 traffic accidents per month. What is the probability that in any given month 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 probability that he gets exactly 1 shoot in his next 5 times attempt to shoot the ball? (3) A multiple-choice quiz has 10 questions, each with 4 possible answers of which only one is correct. What is the probability that sheers guess work yields from 3 to 6 correct answers? (4) If probability that a patient recovers from a leukemia is 0.4. And if 15 people are known to have contracted this disease, what is the probability that (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 is given as the reason for 55% of all thefts. What is the probability that exactly 2 of the next 4 theft cases-reported to MMDA resulted from the need for money to buy drugs? (6) A homeowner plants 5 bulbs selected at random from a box containing 5 rose bulbs and 4 sampaguita bulbs. What is the probability that he planted 2 sampaguita bulbs and 3 rose bulbs? (7) A professor in biology gave a multiple choice quiz with 10 items, each with 5 possible answers and only one of which is correct. (a) Whatistheprobabilitythatastudenttookthetestmymerelyguessing and got a score of 5?

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 76 (b) What is the probability that merely guessing the answers from the test would yield a score of 4 to 8? (c) What is the probability that merely guessing the answers from the test would yield a score of at least 5? (8) What is the probability that a waiter will refuse to serve alcoholic drinks to only 2 minors if he randomly checks the Identification cards of 5 students from among the 10 students where 4 of which are not of legal age? (9) The average number of patients arriving at the emergency room of Philippine General Hospital (PGH) on Monday nights between 9:00 pm up to 12:00 midnight is 5. If we assume that the patients arrive at random and independently, what is the probability that less than 5 patients arrive at the emergency room of PGH on a Monday night from 9:00 pm to 12:00 midnight? (10) A box contains10 red marbles and 15 blue marbles and 5 marbles are selected 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 Mindanao 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 typographical 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 their backyard is 0.35. Find the probability that 4 out of the 10 randomly selected houses has a Pomelo tree in their backyard.

CHAPTER 6. DISCRETE PROBABILITY DISTRIBUTIONS 77 (14) Batanes is hit by 8 storms per year on the average. What is the probability 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 repair in 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 warranty repair? (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 batch of chocolate chip cookies that has just been baked. If the production process 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 7 Continuous Probability Distributions A continuous random variable has a probability of zero of assuming exactly any of its values. And due to the nature of the random variable, we cannot enumerate all of its possible values. Thus when we consider continuous random variables and their probabilities, we only look at probabilities of the random variable have a value in a specified interval. However, we will only consider one type of continuous random variable, the Normal random variable and its associated probability distribution. 7.1 Normal Probability Distributions The normal distribution is one of the most important continuous distribution in the entire field of statistics. And the graph of this distribution is called the 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 is a 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 we proceed in either direction away from the mean. (The graph approaches the x-axis but the graph will never intersect the x-axis). 5. The total area under the curve and above the horizontal axis is equal to 1. A continuous random variable X having the bell-shaped distribution is called a normal random variable. The mathematical equation for the probability distribution of the normal random variable depends on two parameters µ and σ ; its mean and standard deviation. Thus we denote the probability density of X by N(x;µ;σ). If X is a normal random variable with mean µ and variance σ 2, then the equation of the normal curve is N(x;µ;σ) = 1 2πσ e 1 2( x µ σ ) 2 for < x <. Remark It is difficult to compute for the probabilities of a normal random variable using the above formula. However, another way of calculating such probabilities is through the transformation of a normal random variable to its corresponding standard normal random variable. By transforming a normal random variable to a standard normal random variable we can now determine probabilities of the said random variable. Thus we define the standard normal random variable and its distribution.

CHAPTER 7. CONTINUOUS PROBABILITY DISTRIBUTIONS 80 The distribution of a normal random variable with mean µ = 0 and standard deviation σ = 1 is called a standard normal distribution. In order to transform a normal random variable to a standard normal one, we use the following formula: X µ Z=. σ By using the table for the standard normal random variable, we can now determine the probability of any normal random variable by transforming the given random variable to its corresponding standard normal random variable. Example Given a normally distributed random variable X with mean 18 and standard deviation of 2.5, find 1. P (X < 15). Solution: 2. P (17 < X < 21). 15 18 = P (Z < 1.2) = 0.1151 (a) P (X < 15) = P Z < 2.5 Refer to the standard normal table: 17 18 21 18 (b) P (17 < X < 21) = P <Z< 2.5 2.5 P (Z < 1.2) P ( 0.4) = 0.8849 0.3446 = 0.5403 = P (0.4 < Z < 1.2) =

CHAPTER 7. CONTINUOUS PROBABILITY DISTRIBUTIONS 81 Example An electrical firm manufacturers light bulbs that have a length of life that is normally distributed with mean equal to 800 hours and standard deviation of 40 hours. Find the probability that the bulb burns between 778 and 834 hours. Solution: The distribution of the light bulbs is illustrated by the figure below: The z values corresponding to x 1 = 778 and x 2 = 834 are Hence, z 1 = 778 800 40 z 2 = 834 800 40 = 0.55, = 0.85. P(778 < X < 834) = P( 0.55 < Z < 0.85) = P(Z < 0.85) P(Z < 0.55) = 0.8023 0.2912 = 0.5111

CHAPTER 7. CONTINUOUS PROBABILITY DISTRIBUTIONS 82 Exercises (1) Given a normally distributed random variable X with mean 18 and standard 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 a standard deviation of 0.5 years. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years. (3) An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 834 hours. (4) If the average height of miniature poodles is 30 centimeters, with a standard deviation of 4.1 cm, what percentage of miniature poodles exceeds 35 cm in height, assuming that the height follows a normal distribution and can be measured to any desired degree of accuracy? (5) The quality grade-point averages of 300 college freshmen follow approximatelyanormaldistributionwithameanof2.1andastandarddeviation of 0.8. How many of these freshmen would you expect to have a score between 2.5 and 3.5 inclusive if the point averages are computed to the nearest tenth? (6) A set of final examination grades in an introductory statistics course was found to be normally distributed, with a mean of 73 and a variance of 64. (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 what grade? (7) Plastic bags used for packaging produce re manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 poundspersquareinchandastandarddeviationof1.5poundspersquare inch.

CHAPTER 7. CONTINUOUS PROBABILITY DISTRIBUTIONS 83 (a) What proportion of the bags produced have a mean breaking strength of between 5 and 5.5 pounds per square inch? (b) What is the probability that a randomly selected bag will have a mean breaking strength of at least 6 pounds per square inch? (c) What percentage of the bags have a mean breaking strength of less than 4.17 pound per square inch? (d) Between what two values symmetrically distributed around the mean will 95% of the breaking strengths fall? (8) If we know that the length of time it takes a college student to find a parking spot in the university parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the 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 8 Estimation of Parameters Recall that one of the objectives of statistics is to make inferences concerning a population. And these inferences are based only in partial information regarding the population, since the information of the statistics is based on the 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 associated with the statistic involved in our inference. Another concern regarding inference based on sample information is the factor of how the samples are taken and how large the sample size is so that meaningful interpretations can be drawn from the sample. This concern is addressed in specialized study of statistics, Sampling Theory, which is beyond the scope of our study in this course. But an overview of terms and concepts 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. However, there are random samples thus producing different values for a certain statistic. Since statistic varies from sample to sample then we can say that a statistic is also a random variable. Recall that we can construct a probability distribution for a random variable 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 The sampling distribution of a statistic is the probability distribution for the possible values of the statistic that results when random samples of size n are repeatedly drawn from the population. Example A population consists of N = 5 numbers :1, 2, 3, 4, and 5. If a random sample of 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 and σ 2 = 2. Since there are only 5 distinct and equally likely elements in our population the probability that one will occur is the same for all elements in the population, that is, P(x) = 1. Since we are only choosing 3 from the 5 population there are only 5 C 3 = 10 different possible samples and they are as follows: No. Sample Sample Mean x 1 1,2,3 2 2 1,2,4 2.333 3 1,2,5 2.667 4 1,3,4 2.667 5 1,3,5 3 6 1,4,5 3.333 7 2,3,4 3 8 2,3,5 3.333 9 2,4,5 3.667 10 3,4,5 4 Thus the sampling distribution of the sample mean is x 2 2.333 2.667 3 3.333 3.667 4 f( x) 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 the value 3 and a variance of 1. But if we increase our sample size say n = 4 and 3 compute for the sampling distribution of x again, we will still get a mean of 3 but a variance of 0.125.

CHAPTER 8. ESTIMATION OF PARAMETERS 86 Remark We can notice that µ x the mean of the sample means is equal to the population mean, and the variance σ 2 x or the standard deviation σ x will decrease as our sample size increases. If all possible random samples of size n are drawn, without replacement, from a finite population of size N with mean µ and standard deviation σ, then the sampling distribution of the sample mean will be approximately normally distributed and the mean and standard deviation is given by The factor N n N 1 µ x = µ and σ x = σ n N n N 1. is called the finite correction factor. For large or infinite populations, this correction factor will be approximately equal to 1. Hence σ x = σ n The above notion regarding the sampling distribution of the sample mean gives us the foundation of the next theorem; the central limit theorem. The central limit theorem states that in general situations and condition sums and means of samples of random observations that are drawn from a population of any distribution tends to possess, approximately, a bell shaped distribution in repeated sampling. And thus the distribution can be assumed approximately normal. One of the significance of the central limit theorem is that it explains why some of the observations in the real world tends to possess an approximately a normal distribution. To illustrate this significance, consider the weight of a person. Weight can be affected by many factors whether environmental or genetics for instance, family lineage such as the parents weights. Another factor can be the physical activities of the person. All this possibilities may really affect the weight of a person but the central limit theorem together with other theorems applicable to the normal distribution provides an explanation of this events. Another significance of the central limit theorem and probably the most important attribute is its application to statistical inference. Many statistical estimators that are used to make inferences about a population have parameters that are sums and averages of sample observations.

CHAPTER 8. ESTIMATION OF PARAMETERS 87 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 µ x and σ x = σ n where µ and sigma are the mean and standard deviation of the population, respectively. Thus, z = ( x µ x ) is a value of a standard normal random variable Z. n σ x = σ Remark 1. If samples are taken from a population having a normal distribution, then the sampling distribution of the sample mean will have a normal distribution no matter what n is. 2. If samples are taken from a population which is not normally distributed, then the sampling distribution of the sample mean will have an approximate normal distribution only for large samples, that is, when n 30 3. The standard deviation of the sampling distribution of x,σ x, is called the standard error of the sample mean. 8.2 Sampling Procedures The need for sampling arises from the fact that information that is needed cannot be obtained easily if the size of the population is large. And it is impractical to observe each element in a population. The next obvious question is: How large should the sample size be so that meaningful interpretations can be drawn from them? But to answer this question, another problem arises. The design of how the information is collected needs to be considered so that the appropriate sample size can be determined. All this notions and concepts are beyond the scope of this book. But this section introduces the

CHAPTER 8. ESTIMATION OF PARAMETERS 88 concepts in sampling theory. So far the discussions of sampling distributions from the previous chapters have been restricted and assumed simple random samples but there are other ways of obtaining samples from a population. And these other sampling methods offers a more precise and efficient in such a way that more information regarding the population is attained at a lower cost than simple random sampling. We can classify sampling methods in to two types: probability sampling and non-probability sampling. The difference between them is that in probability sampling, every unit has a chance of being selected. This is not true for non-probability sampling; every item in a population does not have an equal chance of being selected. The following methods of sampling falls under the probability sampling. In simple random sampling, each member of a population has an equal chance of being selected. One of the necessary elements to perform a simple random 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 observed element is either replaced in the population or not. In the sampling with replacement, there would be a possibility that the sampled observation may be selected twice or more. Usually, the simple random sampling approach is conducted without replacement because it is more convenient and gives more precise results. For simplicity and scoping of our discussions, when we discuss simple random sampling, we will refer to sampling without replacement. 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. Another property of the simple random sample is its independence. That is the selection of one element has no influence on the selection of other elements. However, a completely unbiased and independent sample is very hard to find in the real world.

CHAPTER 8. ESTIMATION OF PARAMETERS 89 In stratified sampling, the population is divided into homogeneous and non overlapping groups called strata, and then independent random samples are selected from each stratum. This means that elements in a certain strata, elements have characteristics in common. To achieve the assumption of homogeneity, stratification must be done in such a way that elements in a stratum have some characteristics in common and those characteristic has in some way related to the characteristic in study. Any of the sampling methods can be used to sample within each stratum. The sampling method can vary from one stratum to another. If simple random sampling is used to select the sample within each stratum, the sample design is called stratified simple random sampling. An advantage of using stratified sampling is that it can make the sampling strategy more efficient. For example, if every person in a certain group had the same salary, then a sample of one individual would be enough to get a precise estimate of the average salary. Creating strata where units share similar characteristics is one factor for efficient selection of samples. You would only need to select minimal sample from this group since information captured in a strata would somehow describe the whole strata. Then you could combine this information from each stratum to get a precise estimate for the entire population. But using a simple random sampling approach in the whole population without stratification, the sample would need to be larger than the total of all stratum samples to get an estimate of total income with the same level of precision. Stratification would entail varying sizes in each stratum, and thus consideration on the sample size in each stratum must be given. A procedure called proportional allocation is implemented such that sample sizes are proportional to the size of different strata. Sample Sizes for Proportional Allocation: If a population of size N is divided into k strata, N 1,N 2,...,N k and then the sample size for the i th stratum is given by, ( ) nni n i = N Where n is the total sample size.

CHAPTER 8. ESTIMATION OF PARAMETERS 90 Systematic random sampling means that there is a gap, or interval, between each selected unit in the sample. This type of sampling method is also called Interval Sampling. In order to select a systematic sample, you need to follow these steps: 1. Number the units on your population from 1 to N (where N is the population size). 2. Determine the sampling interval (k) by dividing the number of units in the population by the desired sample size. 3. Selectanumberbetweenoneandk atrandom. Thisnumberiscalledthe random start and would be the first number included in your sample. 4. Select every k th unit after that first number. In cluster sampling, the population is divided into clusters and then a random sample of the clusters is selected. The selected clusters may be completely sampled or a random sample may be obtained from the selected clusters. For instance, a large company like San Miguel Food Corporation has 30 plants located throughout the Philippines. In order to access a new total quality plan, the 30 plants are considered clusters and five of the plants are randomly selected. All of the quality control personnel at the five selected plants 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, a simple random sample is performed so that every house in the selected block is surveyed. Note. 1. Most statistical methods depend on the independence and lack of bias of the simple random sample. The result discussed in this book applies to simple random sample ONLY. And for other sampling procedures, theresultswillhavetobemodifiedandthatdiscussionsareoutofscope

CHAPTER 8. ESTIMATION OF PARAMETERS 91 in this book. Reference to specialized sampling textbooks is encouraged to explore those concepts. 2. We must also take note that without the assumption of randomization in obtaining the samples, results and inferences that will be obtained from the data cannot be dependable. No matter how sophisticated the modification is in the statistical analysis, if the assumption of randomness is not present, interpretations in the sample will be doubtful. The significance of randomness statistically guarantees the accuracy of a survey. 8.3 Estimating the Population Mean Procedures and formulas used in estimating values of unknown population parameters that are based on information provided in a sample data are based on the theory of sampling distributions and the methods used to collect these sample. The sampling distributions allow us to associates specific levels of confidence with each statistical inference. And thus enabling us to quantify how much confidence we place in a sample statistic correctly estimating the population parameter. An estimator is a rule, usually expressed as a formula that tells us how to calculate an estimate based on information in the sample. We can classify estimators into two, point estimators and interval estimators. 1. Point estimation - Based on sample data, a single number is calculated to estimate the population parameter. The rule or formula that describes this calculation is called the point estimator, and the resulting number is called the point estimate. 2. Interval estimation - Based on sample data, two numbers are calculated to form an interval within which the parameter is expected to lie. The rule or formula that describes this calculation is called the interval estimator, and the resulting pair of numbers is called an interval estimate or confidence interval.

CHAPTER 8. ESTIMATION OF PARAMETERS 92 A. Point Estimation (1) The best point estimate for the population mean,µ, is the sample mean, x. (2) The point estimator x is unbiased with standard error given by SE = σ n. (3) The margin of error of the point estimate, x, is given by x±1.96se. (4) If σ is unknown and n 30, the sample standard deviation s can be used to approximate σ. B. Interval estimation To construct and interval estimate for the population mean, we consider two cases. One case is when the standard deviation of the population is known or unknown by the sample size is large enough, that is, n 30. The other case is when the standard deviation 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: x±zα 2 ( ) σ n where: x = sample mean Zα = z-score with an area of α to the right 2 2 n = sample size σ = population standard deviation (b) CASE 2: If σ unknown and n 30, a (1 α)100% confidence interval for a population mean,µ is given by: x±tα 2 ( s n ) where: x = sample mean tα = critical t-value with an area of α 2 2 n = sample size s = sample standard deviation to the right and a degree of freedom n 1

CHAPTER 8. ESTIMATION OF PARAMETERS 93 Remark (1) If x is used as an estimate of µ, we ( can ) then( be (1 α)100% confident σ that the error will not exceed Zα 2 n or tα s 2 n ). (2) If x is used as an estimate of µ, we can then be (1 α)100% confident that the error will not exceed a specified amount e when the sample size ( Zα2 ) σ 2 is n =. e Exercises (1) A scientist interested in monitoring chemical contaminants in food, and thereby the accumulation of contaminants in human diets, selected a random sample of n = 50 male adults. It was found that the average daily intake of dairy products was 756 grams per day with a standard deviation of 35 grams per day. Construct a 95% confidence interval for the mean daily intake of dairy products for men. (2) A random sample of 12 female students in a certain dormitory showed an average weekly expenditure of P400 for snack foods, with a standard deviation of P12.50. Construct a 90% confidence interval for the average amount 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 the mean content of all such containers, assuming an approximate normal distribution for container contents. (4) The mean and standard deviation for the quality grade point averages of a random sample of 36 college seniors are calculated to be 2.6 and 0.3 respectively. Find the 95% and 99% confidence intervals for the mean of the entire senior class. (5) The following data were collected based from a sample in an experiment: n = 64, x = 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?

CHAPTER 8. ESTIMATION OF PARAMETERS 94 (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 length of the call is recorded, at the end of each call. A random sample of 9 reports yields a mean length of call of 1.2 minutes. Construct a 95% confidence interval for the mean length of call for the whole telephone answering service company if it is known that the population is normally distributed with a standard deviation of 0.6 minutes. (7) A random sample of 10 chocolate bars has an average of 230 calories with a standard deviation of 15 calories. Assuming that the distribution of the calories is approximately normal. (a) Construct a 99% confidence mean calories content of this chocolate bar. (b) How large a sample is needed if we wish to be 99% confident that our sample mean will be within 5 calories of the true mean?

Chapter 9 Statistical Tables and Formulas 95

CHAPTER 9. STATISTICAL TABLES AND FORMULAS 96 Standard Normal Table 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