Discrete Probability Distributions PDF

Chapter 5 Discrete Probability Distributions A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 1  Objectives In this chapter, you learn: The properties of a probability distr...

Chapter 5 Discrete Probability Distributions A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 1  Objectives In this chapter, you learn: The properties of a probability distribution. How to calculate the expected value and variance of a probability distribution. How to calculate probabilities from binomial and Poisson distributions. How to use the binomial and Poisson distributions to solve business problems. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 2 Random Variables A random variable is a numerical description of the outcome of an experiment. – Formally, a random variable is a function that assigns a real number to each element of a sample space. E.g., sum of dice, time for computer repair (numerical outcomes); 1 if consumer likes a product and 0 if if consumer dislikes it (categorical outcomes). – Usually denoted by capital italic letters, such as X or Y. A discrete random variable is one for which the number of possible outcomes can be counted. A continuous random variable has outcomes over one or more continuous intervals of real numbers. 2025-IS6051 | Probability | S. Seppälä 3 Based on: Evans, 2021, Chapter 5.  Definitions Discrete variables produce outcomes that come from a counting process (e.g. number of classes you are taking). Continuous variables produce outcomes that come from a measurement (e.g. your annual salary, or your weight). A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 4 Probability Distributions A probability distribution is a characterization of the possible values that a random variable may assume along with the probability of assuming these values. – It can be either discrete or continuous depending on the nature of the random variable. We may develop a probability distribution using any one of the three perspectives of probability: classical, relative frequency, and subjective. 2025-IS6051 | Probability | S. Seppälä 5 Based on: Evans, 2021, Chapter 5.  Types Of Variables Types Of Variables Ch. 5 Discrete Continuous Ch. 6 Variable Variable A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 6  Discrete Variables Can only assume a countable number of values. Examples: Roll a die twice Let X be the number of times 4 occurs (then X could be 0, 1, or 2 times). Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5). A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 7  Probability Distribution For A Discrete Variable A probability distribution for a discrete variable is a mutually exclusive list of all possible numerical outcomes for that variable and a probability of occurrence associated with each outcome. Interruptions Per Day In Probability Computer Network 0 0.35 1 0.25 2 0.20 3 0.10 4 0.05 5 0.05 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 8 Probability Distributions Are Often Represented Graphically P(X) 0.4 0.3 0.2 0.1 0 1 2 3 4 5 X A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 9 Expected Value Of Discrete Variables, Measuring Center Expected Value (or mean) of a discrete variable (Weighted Average): N  E(X)  x i P ( X  x i ) i 1 Interruptions Per Day In Probability Computer Network (xi) P(X = xi) xiP(X = xi) 0 0.35 (0)(0.35) = 0.00 1 0.25 (1)(0.25) = 0.25 2 0.20 (2)(0.20) = 0.40 3 0.10 (3)(0.10) = 0.30 4 0.05 (4)(0.05) = 0.20 5 0.05 (5)(0.05) = 0.25 1.00 μ = E(X) = 1.40 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 10  Discrete Variables: Measuring Dispersion Variance of a discrete variable. N σ 2  [x i  E(X)]2 P(X  x i ) i 1 Standard Deviation of a discrete variable. N σ  σ2   i [x  i 1 E(X)]2 P(X  x i ) where: E(X) = Expected value of the discrete variable X xi = the ith outcome of X P(X=xi) = Probability of the ith occurrence of X A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 11 Discrete Variables: Measuring Dispersion (continued) N σ  [x i 1 i 2  E(X)] P(X  x i ) Interruptions Per Day In Computer Probability Network (xi) P(X = xi) [xi – E(X)]2 [xi – E(X)]2P(X = xi) 0 0.35 (0 – 1.4)2 = 1.96 (1.96)(0.35) = 0.686 1 0.25 (1 – 1.4)2 = 0.16 (0.16)(0.25) = 0.040 2 0.20 (2 – 1.4)2 = 0.36 (0.36)(0.20) = 0.072 3 0.10 (3 – 1.4)2 = 2.56 (2.56)(0.10) = 0.256 4 0.05 (4 – 1.4)2 = 6.76 (6.76)(0.05) = 0.338 5 0.05 (5 – 1.4)2 = 12.96 (12.96)(0.05) = 0.648 σ2 = 2.04, σ = 1.4283 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 12  Probability Distributions Probability Distributions Ch. 5 Discrete Continuous Ch. 6 Probability Probability Distributions Distributions Binomial Normal Poisson A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 13 Binomial Distribution In some cases, a mathematical expression or model can be used to calculate the probability of a value, or outcome, for a variable of interest. For discrete variables, such mathematical models are also known as probability distribution functions. One such function that can be used in many business situations is the binomial distribution. Use the binomial distribution when the discrete variable is the number of events of interest in a sample of n observations. 2025-IS6051 | Probability | S. Seppälä 14 Based on: Levine et al., 2019, p. 242.  Binomial Probability Distribution A fixed number of observations, n. e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse. Each observation is classified into one of two mutually exclusive & collectively exhaustive categories. e.g., head or tail in each toss of a coin; defective or not defective light bulb. The probability of being classified as the event of interest, π, is constant from observation to observation. Probability of getting a tail is the same each time we toss the coin. Since the two categories are mutually exclusive and collectively exhaustive, when the probability of the event of interest is π, the probability of the event of interest not occurring is 1 – π. The value of any observation is independent of the value of any other observation. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 15  Possible Applications for the Binomial Distribution A manufacturing plant labels items as either defective or acceptable. A firm bidding for contracts will either get a contract or not. A marketing research firm receives survey responses of “yes I will buy” or “no I will not.” New job applicants either accept the offer or reject it. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 16  The Binomial Distribution Counting Techniques Suppose the event of interest is obtaining heads on the toss of a fair coin. You are to toss the coin three times. In how many ways can you get two heads? Possible ways: HHT, HTH, THH, so there are three ways you can get two heads. This situation is fairly simple. We need to be able to count the number of ways for more complicated situations. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 17  Counting Techniques Rule of Combinations The number of combinations of selecting x objects out of n objects is: n! n Cx  x! (n  x)! where: n! =(n)(n - 1)(n - 2)... (2)(1) x! = (X)(X - 1)(X - 2)... (2)(1) 0! = 1 (by definition) A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 18  Counting Techniques Rule of Combinations How many possible 3 scoop combinations could you create at an ice cream parlor if you have 31 flavors to select from and no flavor can be used more than once in the 3 scoops? The total choices is n = 31, and we select X = 3. 31! 31! 31  30  29  28! 31 C 3    31  5  29 4,495 3!(31  3)! 3!28! 3  2  1  28! A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 19  Binomial Distribution Formula n! x n- x P(X=x |n,π) = π (1-π) x! (n - x )! P(X=x|n,π) = probability that X = x events of interest, given n and π Example: Flip a coin four times, let x = # heads: x = number of “events of interest” in sample, (x = 0, 1, 2,..., n) n=4 π = 0.5 n = sample size (number of trials or observations) 1 - π = (1 - 0.5) = 0.5 π = probability of “event of interest” X = 0, 1, 2, 3, 4 1 – π = probability of not having an event of interest A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 20 Example: Calculating a Binomial Probability What is the probability of one success in five observations if the probability of an event of interest is 0.1? x = 1, n = 5, and π = 0.1 n! P(X 1 | 5,0.1)   x (1   ) n  x x! (n  x)! 5!  (0.1)1 (1  0.1) 5 1 1!(5  1)!  (5)(0.1)(0.9) 4  0.32805 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 21 The Binomial Distribution Example Suppose the probability of an invoice payment being late is 0.10. What is the probability of 1 late invoice payment in a group of 4 invoices? x = 1, n = 4, and π = 0.10 n! P(X 1 | 4, 0.10)   x (1  ) n  x x! (n  x)! 4!  (0.10)1 (1  0.10) 4 1 1!(4  1)!  (4)(0.10)( 0.729)  0.2916 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 22 Excel, JMP, & Minitab Can Be Used To Calculate Binomial Probabilities A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 23  The Binomial Distribution Shape The shape of the Here, n = 5 and π = 0.1. binomial distribution P(X=x|5, 0.1) depends on the values.6 of π and n..4 When π ≠ 0.5, both π.2 0 and n affect the 0 1 2 3 4 5 x skewness of the distribution. Here, n = 5 and π = 0.5. Whenever π = 0.5, the binomial distribution is P(X=x|5, 0.5).6 symmetrical, regardless.4 of how large or small the.2 value of n. 0 0 1 2 3 4 5 x A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 24 Binomial Distribution Characteristics Mean: μ E(X) n Variance and Standard Deviation: 2 σ n (1 -  ) σ  n (1 -  ) Where n = sample size π = probability of the event of interest for any trial (1 – π) = probability of no event of interest for any trial A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 25  The Binomial Distribution Characteristics Examples P(X=x|5, 0.1) μ nπ (5)(.1) 0.5.6.4.2 σ  nπ (1 - π )  (5)(.1)(1 .1) 0  0.6708 0 1 2 3 4 5 x P(X=x|5, 0.5) μ nπ (5)(.5) 2.5.6.4 σ  nπ (1 - π )  (5)(.5)(1 .5).2 0 1.118 0 1 2 3 4 5 x A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 26 Computing Binomial Probabilities for Service at a Fast-Food Restaurant (1) Accuracy in taking orders at a drive-through window is important for fast-food chains. Peri-odically, QSR Magazine publishes “The Drive-Thru Performance Study: Order Accuracy” that measures the percentage of orders that are filled correctly. In a recent month, the percentage of orders filled correctly at Wendy’s was approximately 86.9%. Suppose that you go to the drive-through window at Wendy’s and place an order. Two friends of yours independently place orders at the drive-through window at the same Wendy’s. What are the probabilities that all three, that none of the three, and that at least two of the three orders will be filled correctly? What are the mean and standard deviation of the binomial distribution for the number of orders filled correctly? 2025-IS6051 | Probability | S. Seppälä 27 Based on: Levine et al., 2019, p. 247-248. Computing Binomial Probabilities for Service at a Fast-Food Restaurant (2) Because there are three orders and the probability of a correct order is 0.869, n = 3, and π = 0.869, using Equation (5.5) on page 243, The mean number of orders filled correctly in a sample of three orders is 2.607, and the standard deviation is 0.5844. The probability that all three orders are filled correctly is 0.6562, or 65.62%. The probability that none of the orders are filled correctly is 0.0022 (0.22%). The probability that at least two orders are filled correctly is 0.9530 (95.30%). 2025-IS6051 | Probability | S. Seppälä 28 Based on: Levine et al., 2019, p. 247-248.  The Poisson Distribution Definitions You use the Poisson distribution when you are interested in the number of times an event occurs in a given area of opportunity. An area of opportunity is a continuous unit or interval of time, volume, or such area in which more than one occurrence of an event can occur. The number of scratches in a car’s paint. The number of mosquito bites on a person. The number of computer crashes in a day A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 29  The Poisson Distribution Apply the Poisson Distribution when: You are interested in counting the number of times a particular event occurs in a given area of opportunity. An area of opportunity is defined by time, length, surface area, and so forth. The probability that an event occurs in a given area of opportunity is the same for all the areas of opportunity. The number of events that occur in one area of opportunity is independent of the number of events that occur in any other area of opportunity. The probability that two or more events will occur in an area of opportunity approaches zero as the area of opportunity becomes smaller. The average number of events per unit is  (lambda). A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 30  Poisson Distribution Formula The mathematical expression for the Poisson distribution for computing the probability of X = x events, given that events are expected.  x e  P( X  x |  )  x! where: x = number of events in an area of opportunity  = expected number of events e = base of the natural logarithm system (2.71828...) A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 31 Poisson Distribution Characteristics Mean: μ λ Variance and Standard Deviation: σ 2 λ σ λ where  = expected number of events. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 32 Applying the Poisson distribution Consider a study that seeks to examine the number of customers arriving during the lunch hour at a specific bank branch in the central business district of a large city. If the research problem is stated as the number of customers who arrive each minute, does this study match the four properties needed to use the Poisson distribution? First, the event of interest is the arrival of a customer, and the given area of opportunity is defined as a one-minute interval. Will zero customers arrive, one customer arrive, two customers arrive, and so on? Second, a reasonable assumption is that the probability that a customer arrives during a particular one-minute interval is the same as the probability for all the other one-minute intervals. Third, the arrival of one customer in any one-minute interval has no effect on (is independent of) the arrival of any other customer in any other one-minute interval. Finally, the probability that two or more customers will arrive in a given time period approaches zero as the time interval becomes small. For example, the probability is virtually zero that two customers will arrive in a time interval of 0.01 second. Therefore, using the Poisson distribution to determine probabilities involving the number of customers arriving at the bank in a one-minute time interval during the lunch hour is appropriate. 2025-IS6051 | Probability | S. Seppälä 33 Based on: Levine et al., 2019, p. 249. The Poisson Distribution Example (1) The mean number of customers who arrive per minute at a bank during the noon-to-1pm hour is 3.0. What is the probability that 2 customers arrive in a given minute? x = 2, λ = 3.0 e    x e  3.0 3.0 2 P(X 2|3.0)   x! 2! 9  2.71828 3 ( 2 ) 0.2240 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 34 The Poisson Distribution Example (2) 2025-IS6051 | Probability | S. Seppälä 35 Based on: Levine et al., 2019, p. 250. Excel & Minitab & JMP Can Automate Poisson Probability Calculations The mean number of customers who arrive per minute at a bank during the noon-to-1pm hour is 3.0. λ = 3.0 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 36  Graph of Poisson Probabilities Graphically:  = 0.50 = X 0.50 0 0.6065 1 0.3033 2 0.0758 3 0.0126 4 0.0016 5 0.0002 6 0.0000 7 0.0000 P(X = 2 | =0.50) = 0.0758 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 37  Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameter :  = 0.50  = 3.00 A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 38  Chapter Summary In this chapter we covered: The properties of a probability distribution. Computing the expected value and variance of a probability distribution. Computing probabilities from binomial and Poisson distributions. Using the binomial and Poisson distributions to solve business problems. A LWA Y S L E A R N I N G Copyright © 2020 Pearson Education Ltd. All Rights Slide 39