TYPES OF RANDOM VARIABLES DISRETE RANDOM VARIABLES AND THEIR PROBABILITY DISTRIBUTIONS We distinguish between two types of random variables: Discrete random variables ontinuous random variables Discrete Random Variable Definition A random variable that assumes finite or countable values is called a discrete random variable Eamples 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 4 PROBABLITY DISTRIBUTION OF A DISRETE RANDOM VARIABLE Definition The probability distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities Two haracteristics of a Probability Distribution tion The probability distribution of a discrete random variable possesses the following two characteristics: P () for each value of P () = 5 6
Eample : the frequency and relative frequency distributions of the number of vehicles owned by families Eample : the frequency and relative frequency distributions of the number of vehicles owned by families (continued) 7 Number of Relative Frequency Vehicles Owned Frequency / = 5 47/ = 5 85/ = 45 4 47 85 49 6 49/ = 45 6/ = 8 N = Sum = Now let be the number of vehicles owned by a randomly selected family Write the probability distribution of 8 Number of Vehicles Owned Probability ) 5 5 45 45 4 8 ) = Graphical presentation of the probability distribution from previous Eample Eample : Each of the following tables lists certain values of and their probabilities ),45,4,5,,5,,5,,5 4 an you determine whether or not each table represents a valid probability distribution? a) ) b) ) c) ) 8 5 7 7 9 4 4 8 7 5 8 9 5-9 Answer Key Eample a) NO: the respective probabilities DO NOT sum up to, though all are in the range of [,]; b) YES: both characteristics ti are satisfied; The following table lists the probability distribution of the number of breakdowns per week for a machine based on past data c) NO: the probability bilit of an event (when takes value equal to 9) cannot be negative, though they sum up to Breakdowns per week Probability 5 5
haracteristics: Probability p() or f() function and cumulative distribution function F() Eample (continued) a) Present this probability distribution graphically Number in tossing fair die once Sum of two numbers in tossing two fair dice once b) Find the probability that the number of breakdowns for this machine during a given week is: i eactly ii to iii more than iv at most 4 Eample (continued) Let denote the number of breakdowns for this machine during a given week; the following table lists the probability distribution of ) 5 5 ) = a) Eample (continued) ),4,5,,5, 5,5,,5 5 6 Eample (concluded) b) i P (eactly breakdowns) = P ( = ) = 5 ii P ( to breakdowns) = P ( ) = P ( = ) + P ( = ) + P ( = ) = 5+ + 5= 7 iii P (more then breakdown) = P ( > ) = P ( =)+P P ( =) = 5 + = 65 Eample According to a survey, 6% of all students at a large university suffer from math aniety Two students are randomly selected from this university Let denote the number of students in this sample who suffer from math aniety Your task is to develop the probability bilit distribution of iv P (at most one breakdown) = P ( ) = P ( = ) + P ( = ) = 5 + = 5 7 8
Eample ontinued: Tree Diagram Eample (continued) Let us define the following two events: N = the student selected does not suffer from math aniety M = the student selected suffers from math aniety P ( = ) = NN) = 6 P ( = ) = NM or MN) = NM) + MN) = 4 + 4 = 48 P ( = ) = MM) = 6 9 Eample (concluded) Probability distribution of the number of students with math aniety in a sample of two students is summarized in the following table: ) 6 48 6 ) = MEAN OF A DISRETE RANDOM VARIABLE The mean of a discrete variable is the value that is epected to occur per repetition, on average, if an eperiment is repeated a large number of times It is denoted by µ and calculated as μ = P () The mean of a discrete random variable is also called its epected value and is denoted by E (); that is, μ = E () = P () Eample 4: omputing the Mean Recall the table from Eample, where represents the number of breakdowns for a machine during a given week, and P () is the probability of the corresponding value of Your task is to find the mean number (or the epected value) of breakdowns per week for this machine Eample 4: omputing the Mean (concluded) ) ) 5 (5) = () = 5 (5) = 7 () = 9 ) = 8 Thus the mean is The mean is μ = P () = 8 4
STANDARD DEVIATION OF A DISRETE RANDOM VARIABLE The standard deviation of a discrete random variable measures the spread of its probability distribution and is computed as P ( ) Eample 5 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 denote the number of defective computer parts in a shipment of 4 The following table gives the probability distribution of ompute the standard d deviation of 4 5 5 6 ) 8 7 Eample 5: omputing the Standard Deviation ) ) ² ²) 4 5 8 6 9 4 4 ) = 5 4 9 6 5 7 6 5 8 4 5 P 4 5 defective computer parts in P 77 (5 ) 45 defective computer parts 4 ²) = 77 8 Eample 6 Loraine orporation is planning to market a new makeup product According to the analysis made by the financial department of the company, it will earn an annual profit of $45 million if this product has high sales and an annual profit of $ million if the sales are mediocre, and it will lose $ million a year if the sales are low The probabilities of these three scenarios are, 5 and 7 respectively Thus: a) Let be the profits (in millions of dollars) earned per annum by the company from this product Write the probability bilit distribution ib ti of b) alculate the mean and the standard deviation of Eample 6 (continued) Eample 6 (concluded): omputations to Find the Mean and Standard Deviation 9 a) The following table lists the probability distribution of ) 45 5-7 7 ) ) ² ²) 45-5 7 44 6-9 5 44 44 59 648 744 899 P P ) = 66 ²) = 87 $66 $4 million million 87 (66)
FATORIALS AND OMBINATIONS Factorials Factorials ombinations Using the Table of ombinations Definition The symbol n!, read as n factorial, represents the product of all the integers from n to In other words, n! = n(n - )(n )(n ) By definition,! = Eample 7 Solution Evaluate the following: a) 7! b)! c) ( 4)! d) (5 5)! a) 7! = 7 6 5 4 = 54 b)! = 9 8 7 6 5 4 =,68,8 c) ( 4)! = 8! = 8 7 6 5 4 = 4, d) (5 5)! =! = 4 ombinations ombinations cont Definition ombinations give the number of ways elements can be selected from n elements The notation used to denote the total number of combinations is n which is read as the number of combinations of n elements selected at a time = n n denotes the total number of elements the number of combinations of n elements selected at a time denotes the number of elements selected per selection 5 6
ombinations (continued) Eample 8 The number of combinations for selecting from n distinct elements is given by the formula n n!!( n )! An ice cream parlor has si flavors of ice cream Kristen wants to buy two flavors of ice cream If she randomly selects two flavors out of si, how many combinations are there? 7 8 n = 6 = 6!!(6 )! Answer Key 6! 65 4 5!4! 4 6 Eample 9 Three members of a jury will be randomly selected from five people How many different combinations are possible? Thus, there are 5 ways for Kristin to select two ice cream flavors out of si 9 4 Solution 5! 5!!(5 )!!! 6 5 Eample : Using the Table of ombinations Marv & Sons advertised to hire a financial analyst The company has received applications from candidates who seem to be equally qualified The company manager has decided to call only of these candidates for an interview i If she randomly selects candidates from the, how many total selections are possible? 4 4
Eample : Determining the Value of THE BINOMIAL PROBABILITY DISTRIBUTION n = X = n 45 The Binomial Eperiment The Binomial Probability Distribution and binomial Formula Using the Table of Binomial Probabilities Probability of Success and the Shape of the Binomial Distribution 4 44 45 The Binomial Probability Distribution and Binomial Formula For a binomial eperiment, the probability of eactly successes in n trials is given by the binomial formula where n p q n - ) n p q n n p = total number of trials = probability bilit of success = p = probability of failure = number of successes in n trials = number of failures in n trials q n 46 Binomial Distribution Number of Successes in a of n Observations (Trials) # Reds in 5 Spins of Roulette Wheel # Defective Items in a Batch of 5 Items # orrect on a Question Eam # ustomers Who Purchase Out of ustomers Who Enter Store # of Bush-heney supporters in survey of people Binomial Distribution-How tion to find it Sequence of n Identical Trials Each Trial Has Outcomes Success (Desired/specified Outcome) or Failure onstant Trial Probability 4 Trials Are Independent 5 # of successes in n trials is a binomial random variable??? Binomial??? Pick 6 students from this class Each flips a coin ount # of heads Pick 6 students from this class X= # of st year students selected Random digit dialing of numbers # of Bush-heney supporters Random digit dialing of numbers Sum of ages of respondents 47 48
Eample Tree diagram for selecting three VRs Five percent of all VRs manufactured by a large electronics company are defective A quality control inspector randomly selects three VRs from the production line What is the probability that eactly one of these three VRs are defective? 49 5 Eample : Answer Key EampleAns Eample : Answer erke Key (continued) Let D = a selected VR is defective G = a selected VR is good P (DGG ) = P (D )P (G )P (G ) = (5)(95)(95) = 45 P (GDG ) = P (G )P (D )P (G ) = (95)(5)(95) = 45 P (GGD ) = P (G )P (G )P (D ) = (95)(95)(5) = 45 Therefore, P ( VR is defective in ) = P (DGG or GDG or GGD ) = P (DGG ) + P (GDG ) + P (GGD ) = 45 + 45 + 45 = 5 5 5 EampleAns Eample : Answer erke Key (continued) EampleAns Eample : Answer erke Key (concluded) n = total number of trials = VRs = number of successes = number of defective VRs = n =- = p = P (success) = 5 q = P (failure) = p = 95 Therefore, the probability of selecting eactly one defective VR P ( ) (5) (95) ()(5)(95) 54 The probability 54 is slightly different from the earlier calculation 5 because of rounding 5 54
Eample Eample 55 At the Epress House Delivery Service, providing high-quality service to customers is the top priority of fthe 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, % of the packages mailed through this company do not arrive at their destinations within the specified time Suppose a corporation mails packages through Epress House Delivery Service on a certain day 56 a) Find the probability that eactly of these packages will not arrive at its destination within the specified time b) Find the probability that at most of these packages will not arrive at its destination within the specified time Eample : Answer Key EampleAns Eample : Answer erke Key (continued) n = total number of packages mailed = p = P (success) = q = P (failure) = = 98 a) ) = number of successes = n = number of failures = = 9 () (98) 9! () (98)!( )! ()()(874776) 667 9 57 58 Eample : Answer Key (concluded) Eample b) ) ) ) () (98) () (98) ()()(8778) ()()(874776) 87 667 988 9 According to an Allstate Survey, 56% of Baby Boomers have car loans and are making payments on these loans (USA TODAY, October 8, ) Assume that this result holds true for the current population p of all Baby Boomers Let denote the number in a random sample of three Baby Boomers who are making payments on their car loans Write the probability distribution of and draw a bar graph for this probability distribution 59 6
Eample : Answer Key Eample : Answer Key n = total Baby boomers in the sample = p = P (a Baby Boomer is making car loan payments) = 56 q = P (a Baby Boomer is not making car loan payments) = - 56 = 44 ) (56) (44) ()()(8584) 85 ) (56) (44) ()(56)(96) 5 ) (56) (44) ()(6)(44) 44 ) (56) (44) ()(7566)() 756 6 6 The following table represents the probability distribution ib ti of P () 85 5 44 756 Here we have a bar graph of the probability distribution of ),45 4,4,5,,5,,5,,5 6 6 64 Eample 4: Using the Table of Binomial Probabilities Eample 4 65 According to a study of college students by Harvard University s School of Public health, 9% of those included in the study abstained from drinking (USA TODAY, April, ) Suppose that t of all current college students t in the United States, % abstain from drinking A random sample of si college students is selected 66 Using Table IV of Appendi, answer the following a) Find the probability that eactly three college students in this sample abstain from drinking b) Find the probability that at most two college students in this sample abstain from drinking c) Find the probability that at least three college students in this sample abstain from drinking d) Find the probability that one to three college students in this sample abstain from drinking e) Let be the number of college students in this sample who abstain from drinking Write the probability distribution of and draw a bar graph for this probability distribution
Determining P ( = ) for n = 6 and p = p = p n 5 95 n = 6 6 75 54 6 54 9 5 984 458 = 46 89 4 54 5 5 5 6 75 67 P ( = ) = 89 Eample 4: Answers a) P ( = ) = 89 b) P (at most ) = P ( or or ) = P ( =)+P P ( =)+P P ( =) = 6 + 9 + 458 = 9 c) P (at least ) = or 4 or 5 or 6) = P ( = ) + P ( = 4) + P ( =5) + P ( = 6) = 89 + 54 + 5 + = 989 d) P ( to ) = P ( = ) + P ( = ) + P ( = ) = 9 + 458 + 89 = 79 68 Probability Distribution of for n = 6 and p= Bar graph for the probability distribution of 4 5 6 ) 6 9 458 89 54 5 ),45,4,5,,5,,5,,5 4 5 6 69 7 Probability of Success and the Shape of the Binomial Distribution Bar graph from the probability distribution from previous table The binomial probability distribution is symmetric if p = 5 ),4 Probability Distribution of for n = 4 and p = 5 7 ) 65 5 75 5 4 65 7,,, 4
Probability of Success and the Shape of the Binomial Distribution cont Bar graph for the probability distribution for the previous table The binomial probability distribution is skewed to the right if p is less than 5 ),5 7 Probability Distribution of for n = 4 and p = ) 4 46 646 756 4 8 74,4,,, 4 Probability of Success and the Shape of the Binomial Distribution cont The binomial probability distribution is skewed to the left if p is greater than 5 Bar graph for the probability distribution from previous table ),5 4,4 75 Probability Distribution of for n =4 and p = 8 ) 6 56 56 496 4 496 76,,, 4 Mean and Standard Deviation of the Binomial Distributiontion The mean and standard deviation of a binomial distribution are np and npq where n is the total number of trails, p is the probability of success, and q is the probability of failure General rules for mean (epected values) If X and Y are random variables, a and b are constants: ) Ea ( ) a ) EaX ( ) aex ( ) EaX ( b) aex ( ) b ) E ( X Y ) E ( X ) E ( Y ) EaX ( by) aex ( ) bey ( ) 4) E( XY) E( X) E( Y) if X and Y are independent random variables 77 78
Rules for Variance If X and Y are random variables, a and b are constants, the variance V(X), V(Y) have: V( a) V( ax) a V( X) ( ) ( ) V ax b a V X 4 V( ax by ) av( X ) bv( Y ) if X and Y are independent Eample 5 In a Martiz poll of adult drivers conducted in July, 45% said that they often or sometimes eat or drink while driving i (USATODAY TODAY, October, ) Assume that this result is true for the current population p of all adult drivers A sample of 4 adult drivers is selected Let be the number of drivers in this sample who often or sometimes eat or drink while driving Find the mean and standard deviation of the probability distribution of 79 8 Eample 5: Answer Key n = 4 p = 45, and q = 55 np 4(45) 45) 8 npq (4)( 4)(45)(55) 45)( 55) 46 THE HYPERGEOMETRI PROBABILITY DISTRIBUTION Let 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) = number of successes in n trials n = number of failures in n trials 8 8 THE HYPERGEOMETRI PROBABILITY DISTRIBUTION HYPERGEOMETRI vs BINOMIAL The probability of successes in n trials is given by number of ways successes can be selected from a total of r successes in the population ) r N r N n n number of ways n failures can be selected from a total of N r failures in the population The hypergeometric distribution is closely related to the binomial distribution However, for the hypergeometric distribution: the trials are not independent, and the probability bilit of success changes from trial to trial 8 number of ways a sample of size n can be selected from a population of size N 84
Hypergeometric limit distribution Hypergeometric limit distribution onsider a hypergeometric distribution with n trials and let p = (r/n) denote the probability of a success on the first trial If the population size is large, the term (N n)/(n ) approaches The epected value and variance can be written E() = np and Var() = np( p) Note that these are the epressions for the epected value and variance of a binomial distribution onsider a hypergeometric distribution with n trials and let p = (r/n) denote the probability of a success on the first trial If the population size is large, the term (N n)/(n ) approaches The epected ed value and variance a can be written E() = np and Var() = np( p)?? Binomial?? When the population size is large, a hypergeometric distribution can be approimated by a binomial distribution with n trials and a probability of success p = (r/n) 85 86 Eample 6 Eample 6: Solution Brown Manufacturing makes auto parts that are sold to auto dealers Last week the company shipped 5 auto parts to a dealer Later on, it found out that five of those parts were defective By the time the company manager contacted t the dealer, four auto parts from that shipment have already been sold What is the probability that three of those four parts were good parts and one was defective? r N r n 5 P ( ) N n! 5!!( )!!(5 )! 5! 4!(5 4)! 5 (4)(5) 456,65 Thus, the probability that three of the four parts sold are good and one is defective is 456 4 87 88 Eample 7 Eample 7: Answers Dawn orporation has employees who hold managerial positions Of them, seven are female and five are male The company is planning to send of these managers to a conference If managers are randomly selected out of, a) Find the probability that all of them are female b) Find the probability that at most of them is a female (a) r N r n 5 ) N n (5)() 7 59 Thus, the probability that all three of managers selected re female ae is 59 89 9
Eample 7: Answers THE POISSON PROBABILITY DISTRIBUTION (b) r N r n 5 ) P ( ) r N N r N n n n 7 ()() 7 5 455 (7)() 8 ) ) ) 455 8 67 Using the Table of Poisson probabilities Mean and Standard Deviation of the Poisson Probability Distribution 9 9 THE POISSON PROBABILITY DISTRIBUTION (continued) Eamples onditions to Apply the Poisson Probability Distribution The following three conditions must be satisfied to apply the Poisson probability distribution is a discrete random variable The occurrences are random The occurrences are independent The number of accidents that occur on a given highway during a one-week period The number of customers entering a grocery store during a one hour interval The number of television i sets sold at a department t store during a given week 9 94 THE POISSON PROBABILITY DISTRIBUTION (continued) Poisson Probability Distribution Formula According to the Poisson probability bilit distribution, ib ti the probability of occurrences in an interval is 95 P ( ) e! where is the mean number of occurrences in that interval and the value of e is approimately 788 96 Poisson distribution-characteristicstion Poisson variables is used to count the number of successes within a specified time or region Eg number of customers arriving at a bank teller in the net 5 minutes Poisson variables possess the following properties: The number of successes occurred in any interval is independent of the number of successes occurred in any other interval The probability that a success will occur in an interval is the same for all intervals of equal size and is proportional to the size of the interval The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller
Poisson distribution-link tion to binomial Poisson variables can be used to approimate Binomial variables when: p in binomial is small (less than 5) and n is large (larger than ), then we can use np to approimate binomial distribution Eample-using Poisson instead of binomial Probability of a defective screw p = Probability bilit of more than defects in a lot of screws? Binomial distribution: = np = ()() = Since p <<, can use Poisson distribution to approimate solution j X ) F() f ( j ) e e! j j j! X ) X ) 8% 997!! 97 98 Eample 8 Eample 8: Solution On average, a household receives 95 telemarketing phone calls per week Using the Poisson distribution formula, find the probability that a randomly selected household receives eactly si telemarketing phone calls during a given week e (95) 6 e 95 6)! 6! (75,9896)(7485) 7 764 99 Eample 9 Solution to Eample 9 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 net month this machine will have ( a) () e (9)(497877) P ( ) 4! a) eactly two breakdowns b) at most one breakdown ( b) () e () e P ( ) P ( )!! ()(497877) ()(497877) 498 494 99
Eample Eample : Answer Key ynthia s Mail Order ompany provides free eamination of its products for seven days If not completely satisfied, a customer can return the product within that period and get a full refund According to past records of the company, an average of of every products sold by this company are returned for a refund Using the Poisson probability distribution formula, find the probability that eactly 6 of the 4 products sold by this company on a given day will be returned for a refund = 8 =6 e 6)! 6 (8) e 6! 8 (6,44)(546) 7 4 Eample : Using the Table of Poisson Probabilities Portion of Table of Poisson Probabilities for = 5 On average, two new accounts are opened per day at an Imperial Saving Bank branch Using the Poisson table, find the probability that on a given day the number of new accounts opened at this bank will be a) eactly 6 b) at most c) at least 7 6 5 77 77 4 5 6 7 8 9 84 9 6 4 9 = = 6 P ( = 6) Eample : Answers Mean and Standard Deviation of the Poisson Probability bilit Distribution ib ti 7 a) P ( = 6) = b) P (at most ) =P ( = ) + P ( = ) + P ( = ) + P ( = ) =5 +77 + 77 + 84 = 857 c) P (at least 7) = P ( = 7) + P ( = 8) + P ( = 9) = 4 + 9 + = 45 8
Eample Eample : Answers Probability bilit Distribution ib ti of for = 9 9 An auto salesperson sells an average of 9 car per day Let be the number of cars sold by this salesperson on any given day Using the Poisson probability distribution table, a) Write the probability distribution of b) Draw a graph of the probability distribution c) Find the mean, variance, and standard deviation 4 5 6 P () 466 659 647 494 Eample : Answers (continued) Bar graph for the probability distribution of previous table Eample : Answers (concluded) ),45,4,5,,5,,5, 5,5 4 5 6 9 9 car 9 949 car