Chapter 5: Discrete Random Variables and Their Probability Distributions 5.1 Random Variables 5.2 Probability Distribution of a Discrete Random Variable 5.3 Mean and Standard Deviation of a Discrete Random Variable 5.4 The Binomial Probability Distribution 5.5 The Hypergeometric Probability Distribution 5.6 The Poisson Probability Distribution Introduction We discussed concepts and rules of probability in chapter 4. It helps in solving simple problem, but not complicated ones such as finding probability of getting at least 5 heads in 10 tosses of a fair coin. We want to study the probability mathematically, so we assign numerical values to experimental outcomes and define random variables. Study the probability characteristic of random variables the topic of chapters 5 & 6 5-1 5-2 5.1 Random Variables Continuous Random Variable A random variable is a variable whose value is determined by the outcome of a random experiment Discrete Random Variable A random variable that assumes countable values is called a discrete random variable Examples of discrete random variables The number of cars sold at a dealership during a given month The number of houses in a certain block The number of fish caught on a fishing trip The number of complaints received at the office of an airline on a given day The number of customers who visit a bank during any given hour The number of heads obtained in three tosses of a coin A random variable that can assume any value contained in one or more intervals is called a continuous random variable Examples of continuous random variables The length of a room The time taken to commute from home to work The amount of milk in a gallon (note that we do not expect a gallon to contain exactly one gallon of milk but either slightly more or slightly less than one gallon) The weight of a letter The price of a house 5-3 5-4 1
5.2 Probability Distribution of a Discrete Random Variable The probability distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities Two Characteristics of a Probability Distribution 0 P(x) 1 for each value of x ΣP(x) = 1 Example of tossing 2 coins, X = # of heads Example 5-1 Recall the frequency and relative frequency distributions of the number of vehicles owned by families given in Table 5.1. That table is reproduced below as Table 5.2. Let x be the number of vehicles owned by a randomly selected family. Write the probability distribution of x. X P 5-5 5-6 Example 5-2 Example 5-3 Each of the following tables lists certain values of x and their probabilities. Determine whether or not each table represents a valid probability distribution The following table lists the probability distribution of the number of breakdowns per week (x) for a machine based on past data a) No, since the sum of all probabilities is not equal to 1.0. b) Yes c) No since one of the probabilities is negative. Present this probability distribution graphically. Find the probability that the number of breakdowns for this machine during a given week is exactly 2 0 to 2 more than 1 at most 1 5-7 5-8 2
Example 5-3: Solution Example 5-4 Graphical presentation Finding probabilities P(exactly 2 breakdowns) = P(x = 2) =.35 P(0 to 2 breakdowns) = P(0 x 2) = P(x = 0) + P(x = 1) + P(x = 2) =.15 +.20 +.35 =.70 P(more then 1 breakdown) = P(x > 1) = P(x = 2) + P(x = 3) =.35 +.30 =.65 P(at most one breakdown) = P(x 1) = P(x = 0) + P(x = 1) =.15 +.20 =.35 According to a survey, 60% of all students at a large university suffer from math anxiety. Two students are randomly selected from this university. Let x denote the number of students in this sample who suffer from math anxiety. Develop the probability distribution of x. Let us define the following two events: N = the student selected does not suffer from math anxiety M = the student selected suffers from math anxiety Then P(x = 0) = P(NN) =.16 P(x = 1) = P(NM or MN) = P(NM) + P(MN) =.24 +.24 =.48 P(x = 2) = P(MM) =.36 5-9 5-10 5.3 Mean and Standard Deviation of a Discrete Random Variable The mean of a discrete variable x is the value that is expected to occur per repetition, on average, if an experiment is repeated a large number of times. It is denoted by µ and calculated as µ = Σ x P(x) The mean of a discrete random variable x is also called its expected value and is denoted by E(x); that is, E(x) = Σ x P(x) Example of tossing a coin twice, x = # of heads Example 5-5 Recall Example 5-3 of Section 5-2. The probability distribution Table 5.4 from that example is reproduced on the next slide. In this table, x represents the number of breakdowns for a machine during a given week, and P(x) is the probability of the corresponding value of x. Find the mean number of breakdown per week for this machine. X 0 1 2 P 1/4 2/4 ¼ xp 0 2/4 2/4 µ = Σ x P(x) = 1 The mean is µ = Σx P(x) = 1.80 5-11 5-12 3
Standard Deviation of a Discrete Random Variable The standard deviation of a discrete random variable, denoted by σ, measures the spread of its probability distribution. A higher value for the standard deviation of a discrete random variable indicates that x can assume values over a larger range about the mean. A smaller value for the standard deviation indicates that most of the values that x can assume are clustered closely about the mean. Definition of variance - 2 Deviation Standard deviation = square root of variance Book s definition of σ Interpretation of the Standard Deviation same way as Section 3.4 of Chapter 3 2 2 x P( x) Example 5-6 Baier s Electronics manufactures computer parts that are supplied to many computer companies. Despite the fact that two quality control inspectors at Baier s Electronics check every part for defects before it is shipped to another company, a few defective parts do pass through these inspections undetected. Let x denote the number of defective computer parts in a shipment of 400. The following table gives the probability distribution of x. Compute the standard deviation of x. 5-13 5-14 Book s Method to Find the Standard Deviation Recommended Method Computations to Find the Mean and Standard Deviation 2 = x 2 P(x) - 2 = 7.7 2.5 2 = 1.45 => = X P(x) xp(x) (X-) 2 = (X - ) 2 (X-) 2 P(x) 0 0.02 0 6.25 0.125 1 0.2 0.2 2.25 0.45 2 0.3 0.6 0.25 0.075 3 0.3 0.9 0.25 0.075 4 0.1 0.4 2.25 0.225 5 0.08 0.4 6.25 0.5 1 2.5 2 = 1.45 σ 1.45 5-15 5-16 4
Easy Example The following table gives the probability distribution for a random Variable X, the number of DVDs that were returned late in a local Blockbuster per week. X 0 1 2 3 4 P 0.45 0.3? 0.1 0.05 1. Find the probability that one or two DVDs were returned late. 2. Find the probability that at least one DVD was returned late. 3. Find X's mean 4. Find X's variance 2 5. Find X's standard deviation Why not use Excel? 5.4 The Binomial Probability Distribution Binomial Experiment: an experiment satisfying five conditions There are fixed n identical trials Each trial has only two outcomes, success & failure Probability of success p remains constant for each trial The trials are independent X = the number of successes in n trials X is called a binomial random variable and its distribution called BPD X ~ B(n, p) Examples of binomial experiment Tossing a coin 10 times Example 5-8 Example 5-9: 5% of all DVD players made by a large electronics company are defective and 3 DVD players are randomly selected Rolling a die (trial, not experiment, has 2 outcomes) Random guess for answers of a multiple-choice test/quiz 5-17 5-18 Notation and Formula Example 5-10 Notation n = total number of trials p = probability of success q = 1 p = probability of failure x = number of successes in n trials n x =number of failures in n trials Formula for X ~ B(n, p) x n x P( x) C p q, x = 0, 1,, n n x For a problem, first check if it is binomial experiment by using the five conditions If answer is a yes, then identify n, p, x. Use formula to obtain binomial probability distribution Five percent of all DVD players manufactured by a large electronics company are defective. A quality control inspector randomly selects three DVD player from the production line. What is the probability that exactly one of these three DVD players is defective? 5-19 5-20 5
Example 5-18: Solution Let D = a selected DVD player is defective P(D)=.05 G = a selected DVD player is good P(G)=.95 P(DGG) = P(D)P(G)P(G) = (.05)(.95)(.95) =.0451 P(GDG) = P(G)P(D)P(G) = (.95)(.05)(.95) =.0451 P(GGD) = P(G)P(G)P(D) = (.95)(.95)(.05) =.0451 P(1 DVD player in 3 is defective) = P(DGG or GDG or GGD) = P(DGG)+P(GDG)+P(GGD) =.0451 +.0451 +.0451 =.1353 Formula way n = 3, p = 0.05, q = 0.95, x = 1 Example 5 11 At the Express House Delivery Service, providing highquality service to customers is the top priority of the management. The company guarantees a refund of all charges if a package it is delivering does not arrive at its destination by the specified time. It is known from past data that despite all efforts, 2% of the packages mailed through this company do not arrive at their destinations within the specified time. Suppose a corporation mails 10 packages through Express House Delivery Service on a certain day. Find the probability that exactly one of these 10 packages will not arrive at its destination within the specified time. Find the probability that at most one of these 10 packages will not arrive at its destination within the specified time. 5-21 5-22 Example 5 12 In a Pew Research Center nationwide telephone survey conducted in March through April 2011, 74% of college graduates said that college provided them intellectual growth (Time, May 30, 2011). Assume that this result holds true for the current population of college graduates. Let x denote the number in a random sample of three college graduates who hold this opinion. Write the probability distribution of x and draw a bar graph for this probability distribution. 0 3 Px ( 0) 3C0(.74) (.26) (1)(1)(.017576).0176 1 2 Px ( 1) 3C1(.74) (.26) (3)(.74)(.0676).1501 2 1 Px ( 2) 3C2(.74) (.26) (3)(.5476)(.26).4271 3 0 Px ( 3) C(.74) (.26) (1)(.405224)(1).4052 3 3 Automatic Way to Find Binomial Probabilities Table I in Appendix C, the table of binomial probabilities. List the probabilities of x for n = 1 to n = 25. List the probabilities of x for selected values of p Hence the table is very limited Using Excel Binom.dist(number_s, trials, probability_s, cumulative) = binomdist(x, n, p, cumulative) Cumulative = 0, false for cumulative, it gives P(x) Cumulative = 1, true for cumulative, it gives cumulative probabilities from 0 to x, i.e., sum of P(0) through P(x). Calculator TI 83: 2 nd => DISTR => 0 (A) binompdf(n, p, x): gives P(x). binomcdf(n, p, x): gives cumulative probabilities from 0 to x, i.e., binomcdf(n, p, x) = P(0) + P(1) + + P(x). 5-23 5-24 6
Example 5 13 In an NPD Group survey of adults, 30% of 50-year-old or older (let us call them 50-plus) adult Americans said that they would be willing to pay more for healthier options at restaurants (USA TODAY, 2011). Suppose this result holds true for the current population of 50-plus adult Americans. A random sample of six 50-plus adult Americans is selected. Answer the following. Find the probability that exactly 3 persons in this sample hold the said opinion. Find the probability that at most two persons in this sample hold the said opinion. Find the probability that at least three persons in this sample hold the said opinion. Find the probability that one to three persons in this sample hold the said opinion. Let x be the number of persons in this sample who hold the said opinion. Write the probability distribution of x, and draw a bar graph for this probability distribution. 5-25 Determining P(x = 3) for n = 6 and p =.30 Table way Excel way In a cell of Excel type in binom.dist(3, 6, 0.3, 0) and hit enter key P(x = 3) = binom.dist(3, 6, 0.3, 0) = 0.18522 5-26 Probability of Success and the Shape of the Binomial Distribution The binomial probability distribution is symmetric if p =.50 For any n, it gives a symmetric bell-shape The binomial probability distribution is skewed to the right if p is less than.50. The binomial probability distribution is skewed to the left if p is greater than.50. For large n, any p, it gives rough bell-shaped Using Excel to show such feature Mean and Standard Deviation of the Binomial Distribution General formula Mean Variance Special formula = n p 2 = n p q Examples Examples from the book Tossing a fair coin twice, X = # of heads Tossing a fair coin 10 times, X = # of heads Finding any kind of probability Example 5-14 at page 236 5-27 5-28 7
5.5 The Hypergeometric Probability Distribution Notations N = total number of elements in the population r = number of successes in the population N r = number of failures in the population n = number of trials (sample size) x = number of successes in n trials n x = number of failures in n trials The probability of x successes in n trials is given by rcx NrCnx Px ( ) C N 5-29 n Example 5-15 Brown Manufacturing makes auto parts that are sold to auto dealers. Last week the company shipped 25 auto parts to a dealer. Later, it found out that 5 of those parts were defective. By the time the company manager contacted the dealer, 4 auto parts from that shipment had already been sold. What is the probability that 3 of those 4 parts were good parts and 1 was defective? : N = 25, r = 20, N r = 5, n = 4, x = 3, n x = 1 20! 5! r Cx C N r nx 20C3 5C1 3!(20 3)! 1!(5 1)! P( x 3) C 25! N n 25C4 4!(25 4)! (1140)(5).4506 12,650 5-30 5.6 The Poisson Probability Distribution The following three conditions must be satisfied to apply the Poisson probability distribution. x is a discrete random variable. The occurrences are random. The occurrences are independent. Examples of Poisson Probability Distribution The number of accidents that occur on a given highway during a 1- week period. The number of customers entering a grocery store during a 1 hour interval. The number of television sets sold at a department store during a given week. Poisson Probability Distribution Formula According to the Poisson probability distribution, the probability of x occurrences in an interval is x e P ( x) x! where λ (pronounced lambda) is the mean number of occurrences in that interval and the value of e is approximately 2.71828. Mean and Standard Deviation 2 5-31 5-32 8
Example 5-17 On average, a household receives 9.5 telemarketing phone calls per week. Using the Poisson distribution formula, find the probability that a randomly selected household receives exactly 6 telemarketing phone calls during a given week. x 6 9.5 e (9.5) e Formula way P( x 6) x! 6! Excel way: POISSON.DIST(6, 9.5, 0) = (735,091.8906)(.00007485) 720 0.0764 Example 5-18 A washing machine in a laundromat breaks down an average of three times per month. Using the Poisson probability distribution formula, find the probability that during the next month this machine will have exactly two breakdowns at most one breakdown Formula way Excel way Calculator way TI-84 calculator way poissonpdf(9.5, 6) = 0.07642 5-33 5-34 Example 5-20 Example 5-21 On average, two new accounts are opened per day at an Imperial Saving Bank branch. Using Table III of Appendix C, find the probability that on a given day the number of new accounts opened at this bank will be exactly 6 at most 3 at least 7 Three ways An auto salesperson sells an average of.9 car per day. Let x be the number of cars sold by this salesperson on any given day. Find the mean, variance, and standard deviation. s.9 2.9 car.9.949 car 5-35 5-36 9