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Prepared by Lloyd R. Jaisingh

A PowerPoint Presentation Package to Accompany. Applied Statistics in Business & Economics, 4 th edition David P. Doane and Lori E. Seward. Prepared by Lloyd R. Jaisingh. Chapter Contents 6.1 Discrete Distributions 6.2 Uniform Distribution 6.3 Bernoulli Distribution

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Prepared by Lloyd R. Jaisingh

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  1. A PowerPoint Presentation Package to Accompany Applied Statistics in Business & Economics, 4th edition David P. Doane and Lori E. Seward Prepared by Lloyd R. Jaisingh

  2. Chapter Contents 6.1 Discrete Distributions 6.2 Uniform Distribution 6.3 Bernoulli Distribution 6.4 Binomial Distribution 6.5 Poisson Distribution 6.6 Hypergeometric Distribution 6.7Geometric Distribution (Optional) 6.8Transformations of Random Variables (Optional) Discrete Probability Distributions Chapter 6

  3. Chapter Learning Objectives (LO’s) LO6-1:Define a discrete random variable. LO6-2: Solve problems using expected value and variance. LO6-3:Define probability distribution, PDF, and CDF. LO6-4:Know the mean and variance of a uniform discrete model. LO6-5:Find binomial probabilities using tables, formulas, or Excel. Discrete Probability Distributions Chapter 6

  4. Chapter Learning Objectives (LO’s) LO6-6: Find Poisson probabilities using tables, formulas, or Excel. LO6-7: Use the Poisson approximation to the binomial (optional). LO6-8: Find hypergeometric probabilities using Excel. LO6-9: Calculate geometric probabilities (optional). LO6-10: Select an appropriate discrete model from problem context. LO6-11: Apply rules for transformations of random variables (optional). Discrete Probability Distributions Chapter 6

  5. 6.1 Discrete Distributions LO6-1 Chapter 6 LO6-1: Define a discrete random variable. Random Variables • A random variableis a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment. • Nomenclature:- Upper case letters are used to representrandom variables(e.g., X, Y).- Lower case letters are used to represent values of the random variable (e.g., x, y). • A discrete random variablehas a countable number of distinct values.

  6. 6.1 Discrete Distributions LO6-1 Chapter 6 Probability Distributions • A discrete probability distribution assigns a probability to each value of a discrete random variable X. • To be a valid probability distribution, the following must be satisfied.

  7. 6.1 Discrete Distributions LO6-1 Chapter 6 Example: Coin Flips Table 6.1) When a coin is flipped 3 times, the sample space will be S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. If X is the number of heads, then X is a random variable whose probability distribution is as follows:

  8. 6.1 Discrete Distributions LO6-1 Chapter 6 Example: Coin Flips Note also that a discrete probability distribution is defined only at specific points on the X-axis. Note that the values of X need not be equally likely. However, they must sum to unity.

  9. 6.1 Discrete Distributions LO6-2 Chapter 6 LO6-2: Solve problems using expected value and variance. Expected Value • The expected value E(X) of a discrete random variable is the sum of all X-values weighted by their respective probabilities. • If there are n distinct values of X, then • E(X) is a measure of central tendency.

  10. 6.1 Discrete Distributions LO6-2 Chapter 6 Example: Service Calls E(X) = μ =0(.05) + 1(.10) + 2(.30) + 3(.25) + 4(.20) + 5(.10) = 2.75

  11. However, the mean is still the balancing point, or fulcrum. m = 2.75 6.1 Discrete Distributions LO6-2 Chapter 6 Example: Service Calls This particular probability distribution is not symmetric around the mean m = 2.75. E(X) is an average and it does not have to be an observable point.

  12. 6.1 Discrete Distributions LO LO6-2 Chapter 6 Variance and Standard Deviation • If there are n distinct values of X, then the variance of a discrete random variable is: • The variance is a weighted average of the dispersion about the mean, denoted either as s2 or V(X). It is a measure of variability. • The standard deviation is the square root of the variance and is denoted s.

  13. 6.1 Discrete Distributions LO6-2 Chapter 6 Example: Bed and Breakfast

  14. 6.1 Discrete Distributions LO6-2 Chapter 6 Example: Bed and Breakfast The histogram shows that the distribution is skewed to the left and bimodal. The mode is 7 rooms rented but the average is only 4.71 room rentals. s = 2.06 indicates considerable variation around m.

  15. 6.1 Discrete Distributions LO6-3 Chapter 6 LO6-3: Define probability distribution, PDF, and CDF. What is a PDF or CDF? • A probability distribution function (PDF) is a mathematical function that shows the probability of each X-value. • A cumulative distribution function (CDF) is a mathematical function that shows the cumulative sum of probabilities, adding from the smallest to the largest X-value, gradually approaching unity.

  16. Illustrative PDF(Probability Density Function) Cumulative CDF(Cumulative Density Function) 6.1 Discrete Distributions LO6-3 Chapter 6 What is a PDF or CDF? Consider the following illustrative histograms: CDF = P(X ≤ x) PDF= P(X = x)

  17. 6.2 Uniform Distribution LO6-4 Chapter 6 LO6-4: Know the mean and variance of a uniform discrete model. Characteristics of the Uniform Discrete Distribution • The uniform distribution describes a random variable with a finite number of integer values from a to b (the only two parameters). • Each value of the random variable is equally likely to occur. • Consider the following characteristics of the uniform discrete distribution:

  18. 6.2 Uniform Distribution LO6-4 Chapter 6 Characteristics of the Uniform Discrete Distribution

  19. PDF for one die CDF for one die 6.2 Uniform Distribution LO6-4 Chapter 6 Example: Rolling a Die • The number of dots on the roll of a die form a uniform random variable with six equally likely integer values: 1, 2, 3, 4, 5, 6 • What is the probability of getting any of these on the roll of a die?

  20. 6.2 Uniform Distribution LO6-4 Chapter 6 Example: Rolling a Die • The PDF for all x is: • Calculate the mean as: • Calculate the standard deviation as:

  21. 6.3 Bernoulli Distribution Chapter 6 Bernoulli Experiments • A random experiment with only 2 outcomes is a Bernoulli experiment. • One outcome is arbitrarily labeled a “success” (denoted X = 1) and the other a “failure” (denoted X = 0). p is the P(success), 1 – p is the P(failure). • “Success” is defined as the less likely outcome so that p < .5 for convenience. • Note that P(0) + P(1) = (1 – p) + p = 1 and 0 ≤ p ≤ 1. • The expected value (mean) and variance of a Bernoulli experiment is calculated as: E(X) =  and V(X) = (1 - )

  22. 6.4 Binomial Distribution LO6-5 Chapter 6 LO6-5: Find binomial probabilities using tables, formulas, or Excel. Characteristics of the Binomial Distribution • The binomial distribution arises when a Bernoulli experiment is repeated n times. • Each trial is independent so the probability of success p remains constant on each trial. • In a binomial experiment, we are interested in X = number of successes in n trials. So, X = X1 + X2 + ... + Xn • The probability of a particular number of successes P(X) is determined by parameters n and p.

  23. 6.4 Binomial Distribution LO6-5 Chapter 6 Characteristics of the Binomial Distribution

  24. 6.4 Binomial Distribution LO6-5 Chapter 6 Example: Quick Oil Change Shop • It is important to quick oil change shops to ensure that a car’s service time is not considered “late” by the customer. • Service times are defined as either late or not late. • X is the number of cars that are late out of the total number of cars serviced. • Assumptions: - cars are independent of each other - probability of a late car is consistent

  25. 6.4 Binomial Distribution LO6-5 Chapter 6 Example: Quick Oil Change Shop • What is the probability that exactly 2 of the next n = 12 cars serviced are late (P(X = 2))? • P(car is late) = p = .10 • P(car is not late) = 1 - p = .90

  26. 6.4 Binomial Distribution LO6-5 Chapter 6 Application: Uninsured Patients • On average, 20% of the emergency room patients at Greenwood General Hospital lack health insurance. • In a random sample of 4 patients, what is the probability that at least 2 will be uninsured? • X = number of uninsured patients (“success”) 20% or .20 • P(uninsured) = p = • P(insured) = 1 – p = 1 – .20 = .80 • n = 4 patients • The range is X = 0, 1, 2, 3, 4 patients.

  27. 6.4 Binomial Distribution LO6-5 Chapter 6 Compound Events • Individual probabilities can be added to obtain any desired event probability. • For example, the probability that the sample of 4 patients will contain at least 2 uninsured patients is • HINT: What inequality means “at least?” P(X  2) = P(2) + P(3) + P(4) = .1536 + .0256 + .0016 = .1808

  28. 6.4 Binomial Distribution LO6-5 Chapter 6 Compound Events • What is the probability that fewer than 2 patients have insurance? • HINT: What inequality means “fewer than?” P(X< 2) = P(0) + P(1) = .4096 + .4096 = .8192 • What is the probability that no more than2patients have insurance? • HINT: What inequality means “no more than?” P(X 2) = P(0) + P(1) + P(2) = .4096 + .4096 + .1536 = .9728

  29. 6.4 Binomial Distribution LO6-5 Chapter 6 Compound Events It is helpful to sketch a diagram:

  30. 6.5 Poisson Distribution LO6-6 Chapter 6 LO6-6: Poisson Events Distributed over Time. • Called the model of arrivals, most Poisson applications model arrivals per unit of time. • The events occur randomly and independently over a continuum of time or space: Each dot (•) is an occurrence of the event of interest.

  31. 6.5 Poisson Distribution LO6-6 Chapter 6 • Let X = the number of events per unit of time. • X is a random variable that depends on when the unit of time is observed. • For example, we could get X = 3 or X = 1 or X = 5 events, depending on where the randomly chosen unit of time happens to fall. The Poisson model’s only parameter is l (Greek letter “lambda”) wherel represents the mean number of events per unit of time or space.

  32. 6.5 Poisson Distribution LO6-6 Chapter 6 Characteristics of the Poisson Distribution

  33. 6.5 Poisson Distribution LO6-6 Chapter 6 Example: Credit Union Customers • On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the Oxnard University Credit Union at a mean rate of 1.7 customers per minute. • Find the PDF, mean and standard deviation: Note the unit for the mean and standard deviation is customers/minute.

  34. 6.5 Poisson Distribution LO6-6 Chapter 6 Compound Events • Cumulative probabilities can be evaluated by summing individual X probabilities. • What is the probability that two or fewer customers will arrive in a given minute? P(X 2) = P(0) + P(1) + P(2) = .1827 + .3106 + .2640 = .7573 • What is the probability of at least three customers (the complimentary event)? P(X 3) = 1 - P(X  2) = 1 - .7573 =.2427

  35. 6.5 Poisson Distribution LO6-7 Chapter 6 LO6-7: Use the Poisson approximation to the binomial (optional). The Poisson distribution may be used to approximate a binomial by setting = n. This approximation is helpful when the binomial calculation is difficult (e.g., when n is large). The general rule for a good approximation is that n should be “large” and should be “small.” A common rule of thumb says the approximation is that n should be “large” and  should be “small.” A common rule of thumb says the approximation is adequate if n  20 and   .05.

  36. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 LO6-8: Find hypergeometric probabilities using Excel. Characteristics of the Hypergeometric Distribution • The hypergeometric distribution is similar to the binomial distribution. • However, unlike the binomial, sampling is without replacement from a finite population of N items. • The hypergeometric distribution may be skewed right or left and is symmetric only if the proportion of successes in the population is 50%.

  37. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Characteristics of the Hypergeometric Distribution

  38. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Example: Damaged iPods • In a shipment of 10 iPods, 2 were damaged and 8 are good. • The receiving department at Best Buy tests a sample of 3 iPods at random to see if they are defective. • Let the random variable X be the number of damaged iPods in the sample. N = 10 (number of iPods in the shipment) n = 3 (sample size drawn from the shipment) s = 2 (number of damaged iPods in the shipment (“successes” in population)) N–s = 8 (number of non-damaged iPods in the shipment) x = number of damaged iPods in the sample (“successes” in sample) n–x = number of non-damaged iPods in the sample

  39. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Example: Damaged iPods • This is not a binomial problem because p is not constant. • What is the probability of getting a damaged iPod on the first draw from the sample? p1 = 2/10 • Now, what is the probability of getting a damaged iPod on the second draw? p2 = 1/9 (if the first iPod was damaged) or = 2/9 (if the first iPod was undamaged) • What about on the third draw? p3 = 0/8 or = 1/8 or = 2/8 depending on what happened in the first two draws.

  40. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Using the Hypergeometric Formula Since there are only 2 damaged iPods in the population, the only possible values of x are 0, 1, and 2. Here are the probabilities:

  41. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Using Software: Excel Since the hypergeometric formula and tables are tedious and impractical, use Excel’s hypergeometric function to find probabilities. Figure 6.27

  42. 6.6 Hypergeometric Distribution LO6-8 Chapter 6 Binomial Approximation to the Hypergeometric • Both the binomial and hypergeometric involve samples of size n and treat X as the number of successes. • The binomial samples with replacement while the hypergeometric samples without replacement.

  43. 6.7 Geometric Distribution LO6-9 Chapter 6 LO6-9: Calculate Geometric probabilities. Characteristics of the Geometric Distribution • The geometric distribution describes the number of Bernoulli trials until the first success. • X is the number of trials until the first success. • X ranges from {1, 2, . . .} since we must have at least one trial to obtain the first success. However, the number of trials is not fixed. • p is the constant probability of a success on each trial.

  44. 6.7 Geometric Distribution LO6-9 Chapter 6 Characteristics of the Geometric Distribution Example: Telefund Calling • At Faber University, 15% of the alumni make a donation or pledge during the annual telefund. • What is the probability that the first donation will occur on the 7th call?

  45. 6.7 Geometric Distribution LO6-9 Chapter 6 Example: Telefund Calling • What is p? p = .15 • The PDF is: P(x) = p(1–p)x1 P(7) = .15(1–.15)71 = .15(.85)6 = .0566 • What are the mean and standard deviation of this distribution?

  46. 6.8 Transformation of Random Variables LO6-11 Chapter 6 LO6-11: Apply rules for transformations of random variables. Linear Transformations • A linear transformation of a random variable X is performed by adding a constant or multiplying by a constant. NOTE: LO6-10 (not presented)relates to the selection of the appropriate discrete distribution. Please refer to the text.

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