DEPARTMENT OF QUANTITATIVE METHODS INFORMATION SYSTEMS Introduction to

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120 Chapter 5 Discrete random variables and their probability distribution

Chapter 5: Random Variables 2 Random variables # of PC’s owned Frequency Relative Frequency 0 120 . 12 1 180 . 18 2 470 . 47 3 230 . 23 N = 1000 Sum = 1. 000 Let x denote the number of PCs owned by a family. Then x can take any of the four possible values (0, 1, 2, and 3). Ø A random variable (RV) is a variable whose value is determined by the outcome of a random experiment. Ø

Chapter 5: Random Variables 3 Discrete random variable A random variable that assumes countable values is called a discrete random variable. Ø Ø Examples of discrete RVs: Ø Number of cars sold at a dealership during a week Ø Number of houses in a certain block Ø Number of fish caught on a fishing trip Ø Number of costumers in a bank at any given day Continuous random variable A random variable that can assume any value contained in one or more intervals is called a continuous random variable. Ø

Chapter 5: Random Variables 4 Ø Examples of continuous RVs: Ø Height of a person Ø Time taken to complete a test Ø Weight of a fish Ø Price of a car Example: Classify each of the following RVs as discrete or continuous. Ø The number of new accounts opened at a bank during a week Ø The time taken to run a marathon Ø The price of a meal in fast food restaurant Ø The score of a football game Ø The weight of a parcel

Chapter 5: Probability Distribution of a Discrete RV 5 The probability distribution of a discrete RV lists all the possible value that the RV can assume and their corresponding probabilities. Ø Example: Write the probability distribution of the number of PCs owned by a family. # of PC’s owned Frequency Relative Frequency 0 120 . 12 1 180 . 18 2 470 . 47 3 230 . 23 N = 1000 Sum = 1. 000 X P(x) 0 . 12 1 . 18 2 . 47 3 . 23 P(x) = 1. 000

Chapter 5: Probability Distribution of a Discrete RV 6 The following two characteristics must hold for any discrete probability distribution: Ø Ø Ø The probability assigned to each value of a RV x lies in the range 0 to 1; that is 0 P(x) 1 for each x. The sum of the probabilities assigned to all possible values of x is equal to 1. 0; that is P(x) = 1. Example: Each of the following tables lists certain values of x and their probabilities. Determine whether or not each table represents a valid probability distribution. x P(x) 0 . 08 0 . 25 4 . 2 1 . 11 1 . 34 5 . 3 2 . 39 2 . 28 6 . 6 3 . 27 3 . 13 8 -. 1

Chapter 5: Probability Distribution of a Discrete RV 7 Example: The following table lists the probability distribution of a discrete RV x. x P(x) 0 1 2 3 4 5 6 . 11. 19. 28. 15. 12. 09. 06 a) P(x = 3) b) P(x 2) c) P(x 4) d) P(1 x 4) e) Probability that x assumes a value less than 4 f) Probability that x assumes a value greater than 2 g) Probability that x assumes a value in the interval 2 to 5

Chapter 5: Probability Distribution of a Discrete RV 8 Example: For the following table x 1 2 3 4 5 f 8 20 24 16 12 a) Construct a probability distribution table. Draw a graph of the probability distribution. 0. 35 x f P(x) 1 8 . 1 0. 25 2 20 . 25 0. 2 3 24 . 3 4 16 . 2 5 12 . 15 0. 3 0. 15 0. 1 0. 05 0 1 b) Find the following probabilities i. P(x = 3) ii. P(x < 4) iii. P(x 3) 2 3 4 5 iv. P(2 x 4)

Chapter 5: Mean of a discrete RV 9 Mean of a discrete RV The mean -or expected value E(x)- of a discrete RV is the value that you would expect to observe on average if the experiment is repeated again and again Ø Ø It is denoted by Illustration: Let us toss two fair coins, and let x denote the number of heads observed. We should have the following probability distribution table Ø x 0 P(x) 1/4 1 2 1/4 Suppose we repeat the experiment a large number of times, say n =4, 000. We should expect to have approximately

Chapter 5: Mean of a discrete RV 10 1 thousand zeros, 2 thousand ones, and 1 thousand twos. Then the average value of x would equal Similarly, if we use the , we would have

Chapter 5: Mean of a discrete RV 11 Example: Recall “ number of PC’s owned by a family” example. Find the mean number of PCs owned by a family. x P(x) 0 . 12 1 . 18 2 . 47 3 . 23 We need to find x. p(x) for each value of x and then add them up together Thus, on average, we expect to see 1. 81 PC owned by a family!

Chapter 5: Mean of a discrete RV 12 Example: In a lottery conducted to benefit the local fire company, 8000 tickets are to be sold at $5 each. The prize is a $12, 000. If you purchase two tickets, what is your expected gain?

Chapter 5: Mean of a discrete RV 13 Example: Determine the annual premium for a $1000 insurance policy covering an event that over a long period of time, has occurred at the rate of 2 times in 100. Let x : the yearly financial gain to the insurance company resulting from the sale of the policy C : unknown premium Calculate the value of C such that the expected gain E(x) will equal to zero so that the company can add the administrative costs and profit.

Chapter 5: Mean of a discrete RV 14 Example: You can insure a $50, 000 diamond for its total value by paying a premium of D dollars. If the probability of theft in a given year is estimated to be. 01, what premium should the insurance company charge if it wants the expected gain to equal $1000.

Chapter 5: Standard Deviation of a Discrete RV 15 Ø Ø The standard deviation of a discrete RV x, denoted by , measures the spread of its probability distribution. A higher value of indicates that x can assume values over a larger range about the mean. While, a smaller value indicates that most of the values that can x assume are clustered closely around the mean. The standard deviation can be found using the following formula: Hence, the variance 2 can be obtained by squaring its standard deviation .

Chapter 5: Standard Deviation of a Discrete RV 16 Example: Recall “ number of PC’s owned by a family” example. Find the standard deviation of PCs owned by a family.

Chapter 5: Standard Deviation of a Discrete RV 17 Example: An electronic store sells a particular model of a computer notebook. There are only four notebooks in stocks, and the manager wonders what today’s demand for this particular model will be. She learns from marketing department that the probability distribution for x, the daily demand for the laptop, is as shown in the table. x 0 1 2 3 4 5 P(x) . 10 . 40 . 20 . 15 . 10 . 05 Find the mean, variance, and the standard deviation of x. Is it likely that five or more costumers will want to buy a laptop? Solution:

Chapter 5: Standard Deviation of a Discrete RV 18 Example: A farmer will earn a profit of $30, 000 in case of heavy rain next year, $60, 000 in case of moderate rain, and $15, 000 in case of little rain. A meteorologist forecasts that the probability is. 35 for heavy rain, . 40 for moderate rain, and. 25 for little rain next year. Let x be the RV that represents next year’s profits in thousands of dollars for this farmer. Write the probability distribution of x. Find the mean, variance, and the standard deviation of x.

Chapter 5: Standard Deviation of a Discrete RV 19 Example: An instant lottery ticket costs $2. Out of a total of 10, 000 tickets printed for this lottery, 1000 tickets contain a prize of $5 each. 100 tickets contain a prize of $10 each, 5 tickets contain a prize of $1000 each, and 1 ticket has the prize of $5000. Let x be the RV that denotes the net amount player wins by playing this lottery. Write the probability distribution of x. Determine the mean and the standard deviation of x.

Chapter 5: Standard Deviation of a Discrete RV 20 Example: Based in its analysis of future demand for its products, the financial department a company has determined that there is a. 17 probability that the company will lose $1. 2 million during the next year, a. 21 probability that it will lose $. 7 million, a. 37 probability that it will make a profit of $0. 9 million, and a. 25 probability that it will make a profit of $2. 3 million. a) Let x be a RV that denotes the profit earned by this company during the next year. Write the probability distribution of x.

Chapter 5: Standard Deviation of a Discrete RV 21 b) Find the mean and standard deviation of the probability of part a.

Chapter 5: The Binomial Probability Distribution 22 The most widely used discrete probability distribution. Ø Ø It is applied to find the probability that an event will occur x times in n repetitions of an experiment (under certain conditions). Suppose the probability that a VCR is defective at a factory is. 05. We can apply the binomial probability distribution to find the odds of getting exactly one defective VCR out three VCRs.

Chapter 5: The Binomial Probability Distribution 23 Binomial experiment Ø An experiment that satisfies the following four conditions is called a binomial experiment. Ø Ø There are n identical trials Ø Each trail has only two possible outcomes. Ø The probabilities of the two outcomes remain constant. Ø The trials are independent. A success does not mean that the corresponding outcome is considered favorable. Similarly, a failure doesn't necessarily refer to unfavorable outcome.

Chapter 5: The Binomial Probability Distribution 24 Example: Consider the experiment consisting of 10 tosses of a coin. Determine whether or not it is a binomial experiment. Solution:

Chapter 5: The Binomial Probability Distribution 25 The binomial probability distribution and its formula. Ø The RV x that represents the number of successes in n trials for a binomial experiment is called a binomial RV. Ø Ø The probability distribution of x is called the binomial distribution. Consider the VCRs example. Let x be the number of defective VCRs in a sample of 3, x can assume any of the values 0, 1, 2, 3.

Chapter 5: The Binomial Probability Distribution 26 Ø For a binomial experiment, the probability of exactly x successes in n trials is given by the binomial formula where n = total number of trials, p = probability of success, q = probability of failure 1 – p, x = number of successes in n trials, n – x = number of failures in n trials

Chapter 5: The Binomial Probability Distribution 27 Example: Find P(2) for a binomial RV with n = 5 and p =0. 1 Solution:

Chapter 5: The Binomial Probability Distribution 28 Example: Over a long period of time it has been observed that a given sniper can hit a target on a single trial with a probability =. 8. Suppose he fires four shots at the target. Solution: a) What is the probability that he will hit the target exactly two times? b) What is the probability that he will hit the target at least once?

Chapter 5: The Binomial Probability Distribution 29 Example: Five percent of all VCRs manufactured by a large factory are defective. A quality control inspector selects three VCRs from the production line. What is the probability that exactly one of these three VCRs is defective Solution:

Chapter 5: The Binomial Probability Distribution 30 Example: At the Express Delivery Services (EDS), providing high-quality service to its customers is the top priority of the management. The company guarantees a refund of all charges if a package is not delivered at its destination by the specified time. It is known from the past that despite all efforts, 2% of the packages mailed through this company does not arrive on time. A corporation mailed 10 packages through EDS. Solution: a) Find the probability that exactly 1 of these 10 packages will not arrive on time b) Find the probability that at most 1 of these 10 packages will not arrive on time.

Chapter 5: The Binomial Probability Distribution 31 Example: According to the U. S. Bureau of Labor Statistics, 56% of mothers with children under 6 years of age work outside their homes. A random sample of 3 mothers with children under 6 years of age is selected. Let x denote the number of mothers who work outside their homes. Write the probability distribution of x and draw a graph of the probability distribution. Solution:

Chapter 5: The Binomial Probability Distribution 32 Using the table of binomial probabilities Ø We can use the tables of binomial probabilities (Table IV) given in Appendix C to calculate the required probabilities. P n x . 05 . 1 . 2 … . 95 1 0 . 9500 . 9000 . 8000 … . 0500 1 . 0500 . 1000 . 2000 … . 9500 0 . 9025 . 8100 . 6400 … . 0025 1 . 0950 . 1800 . 3200 … . 0950 2 . 0025 . 0100 . 0400 … . 9025 0 . 8574 . 7290 . 5120 … . 0001 1 . 1354 . 2430 . 3840 … . 0071 2 . 0071 . 0270 . 0960 … . 1354 3 . 0001 . 0010 . 0080 … . 8574 2 3

Chapter 5: The Binomial Probability Distribution 33 Example: Based on data from A peter D. Hart Research Associates’ poll on consumer buying habits and attitudes, It was estimated that 5% of American shoppers are status shoppers, that is, shoppers who love to buy designer labels. A random sample of eight American shoppers is selected. Using Binomial Table, answer the following: Solution: a) Find the probability that exactly 3 shoppers are status shoppers. b) Find the probability that at most 2 shoppers are status shoppers.

Chapter 5: The Binomial Probability Distribution 34 c) Find the probability that at least 3 shoppers are status shoppers. d) Find the probability that 1 to 3 shoppers are status shoppers. e) Let x be the number of status shoppers, Write the probability distribution of x and draw a graph of this probability.

Chapter 5: The Binomial Probability Distribution 35 Probability of success and the shape of the binomial distribution For any number of trials n: Ø Ø The binomial probability distribution is symmetric if p =. 5 n p n x . 05 . 1 … . 5 … . 95 4 0 . 8145 . 6561 … . 0625 … . 0001 . 0000 1 . 1715 . 2916 … . 2500 … . 0036 . 0005 2 . 0135 . 0486 … . 3750 … . 0486 . 0135 3 . 0005 . 0036 … . 2500 … . 2916 . 1715 4 . 0000 . 0001 … . 0625 … . 6561 . 8145

Chapter 5: The Binomial Probability Distribution 36 Ø The binomial probability distribution is skewed to the right if p is less than. 5 p n x . 05 . 1 … . 5 … . 95 4 0 . 8145 . 6561 … . 0625 … . 0001 . 0000 1 . 1715 . 2916 … . 2500 … . 0036 . 0005 2 . 0135 . 0486 … . 3750 … . 0486 . 0135 3 . 0005 . 0036 … . 2500 … . 2916 . 1715 4 . 0000 . 0001 … . 0625 … . 6561 . 8145

Chapter 5: The Binomial Probability Distribution 37 Ø The binomial probability distribution is skewed to the left if p is greater than. 5 p n x . 05 . 1 … . 5 … . 95 4 0 . 8145 . 6561 … . 0625 … . 0001 . 0000 1 . 1715 . 2916 … . 2500 … . 0036 . 0005 2 . 0135 . 0486 … . 3750 … . 0486 . 0135 3 . 0005 . 0036 … . 2500 … . 2916 . 1715 4 . 0000 . 0001 … . 0625 … . 6561 . 8145

Chapter 5: The Binomial Probability Distribution 38 Mean and standard deviation of the binomial distribution Ø Although we can still use the mean and standard deviations formulas learned in 5. 3 and 5. 4, it is more convenient and simpler to use the following formulas once the RV x is known to be a binomial RV

Chapter 5: The Binomial Probability Distribution 39 Example: Refer to the mothers with children under 6 years of age example, find the mean and the standard deviation of the probability distribution. Solution:

Chapter 5: The Binomial Probability Distribution 40 Example: Let x be a discrete RV that posses a binomial distribution. Using the binomial formula, find the following probabilities. a) P(x = 5) for n = 8 and p =. 6 b) P(x = 3) for n = 4 and p =. 3 c) P(x = 2) for n = 6 and p =. 2 Verify your answer by using Table of Binomial Probabilities.

Chapter 5: The Binomial Probability Distribution 41 Example: Let x be a discrete RV that posses a binomial distribution. a) Table of Binomial Probabilities, write the probability distribution for x for n = 7 and p =. 3 and graph it. b) What are the mean and the standard deviation of the probability distribution developed in part a?

Chapter 5: The Binomial Probability Distribution 42 Ø An experiment that satisfies the following four conditions is called a binomial experiment. Ø There are n identical trials Ø Each trail has only two possible outcomes. Ø The probabilities of the two outcomes remain constant. Ø The trials are independent.

Chapter 5: The Hypergeometric Probability Distribution 43 Ø Ø We learned that one of the conditions required to apply the binomial distribution is that the trials are independent so the probability of the two outcomes remain constant. What if the probability of the outcomes is not constant? In such cases we replace the binomial distribution by the hypergeometric probability distribution. Such a case occurs when a sample is drawn without replacement from a finite population.

Chapter 5: The Hypergeometric Probability Distribution 44 Hypergeometric 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) x = number of successes in n trials n – x = number of failures in n trials Ø The probability of x successes in n trials is given by The mean and the variance are given by

Chapter 5: The Hypergeometric Probability Distribution 45 Example: Dawn corporation has 12 employees who hold managerial positions. Of them, 7 are female and 5 are male. The company is planning to send 3 of these 12 managers to a conference. If 3 mangers are randomly selected out of 12, a) find the probability that all 3 of them are female

Chapter 5: The Hypergeometric Probability Distribution 46 b) find the probability that at most 1 of them is a female

Chapter 5: The Hypergeometric Probability Distribution 47 Example: A case of soda has 12 bottles, 3 of which contain diet soda. A sample of 4 bottles is randomly selected from the case a) find the probability distribution of x, the number of diet sodas in the sample

Chapter 5: The Hypergeometric Probability Distribution 48 b) what are the mean and variance of x?

Chapter 5: The Hypergeometric Probability Distribution 49 Example: GESCO Insurance company has prepared a final list of 8 candidates for 2 positions. Of the 8 candidates, 5 are business majors and 3 are engineers. If the company manager decides to select randomly two candidates from this list, find the probability that a) both candidates are business majors b) neither of the two candidates is a business major c) at most one of the candidates is a business major

Chapter 5: The Poisson Probability Distribution 50 Ø Ø Ø The Poisson distribution is another discrete probability distribution that has numerous practical applications. It provides a good model for data that represent the number of occurrence of a specified event in a given unit of time or space. Here are some examples of experiments for which the RV x can modeled by the Poisson RV: Ø Ø The number of phone calls received by an operator during a day The number of customer arrivals at checkout counter during an hour Ø The number of bacteria per a cm 3 of a fluid Ø The number of machine breakdowns during a given day Ø The number of traffic accidents at a given time period

Chapter 5: The Poisson Probability Distribution 51 Ø The Poisson probability distribution is applied to experiments with random and independent occurrences Ø Ø Ø The occurrences are random in the sense they do not follow any pattern and, hence, the are unpredictable. Independence of occurrences means that the occurrence or nonoccurrence of an event does not influence the successive occurrence of that next event. The occurrences are always considered with respect to an interval. The interval may be a time interval, a space interval, or a volume interval. If the average number of occurrences ( ) for a given interval is known, then by using the Poisson probability we can compute the probability of a certain number of occurrences x in that interval.

Chapter 5: The Poisson Probability Distribution 52 Ø Ø Ø The following three conditions must be satisfied to apply the Poisson probability distribution Ø x is a discrete RV. Ø The occurrences are random. Ø The occurrences are independent. The Poisson probability formula is given by where (pronounced lambda) is the mean number of occurrences in that interval and the value of e is approximately 2. 71828.

Chapter 5: The Poisson Probability Distribution 53 Ø Ø The mean number of occurrences, denoted by , is called the parameter of the Poisson probability distribution or the Poisson parameter. Remember: The interval of and x must be of the same length. If they are not, the mean should be redefined to make them equal. For instance, if was given in hours and we were asked to find the event in minutes, needs to be divided by 60 to have both x and in the same interval length

Chapter 5: The Poisson Probability Distribution 54 Example: The automatic teller machine (ATM) installed outside Mansfield Savings and Loan is used on average by five costumers per hour. The bank closed this ATM for one hour for repairs. What is the probability that during that hour eight customers came to use this ATM? Solution:

Chapter 5: The Poisson Probability Distribution 55 Example: The average number of traffic accidents on a certain section of highway is two per week. assume that the number of accidents follows a Poisson distribution with = 2. 1. Find the probability of no accidents on this section of highway during a 1 -week period. 2. Find the probability of at most three accidents on this section of highway during a 2 -week period.

Chapter 5: The Poisson Probability Distribution 56 Using the table of Poisson probabilities Ø The probabilities for Poisson distribution can also be found using Poisson Probability Table

Chapter 5: The Poisson Probability Distribution 57 Example: On average, two new accounts are opened per day at an Imperial Savings Bank branches. Using Table VI of Appendix C, find the probability that on a given day the number of new accounts opened at this bank will be a) Exactly 6 b) At most 3 c) At least 7

Chapter 5: The Poisson Probability Distribution 58 Mean and standard deviation of the Poisson probability distribution Ø For the Poisson distribution = 2 = Example: An auto salesperson sells an average of. 9 cars per day. Let x be the number of cars sold by this salesperson on any given day. Using the Poisson probability table, write the probability distribution of x, draw the probability distribution, and find the mean and standard deviation of x