# Chapter 5 - Chapter 5 Chapter 5 Discrete Probability...

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Unformatted text preview: Chapter 5 Chapter 5 Discrete Probability Distributions 0. A random variable is a numerical description of the outcome of an experiment. Random Variables 1. A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes. 2. A discrete random variable may assume either a finite number of values or an infinite sequence of values. 3. A continuous random variable may assume any numerical value in an interval or collection of intervals. Example: JSL Appliances 4. Discrete random variable with a finite number of values Let x = number of TV sets sold at the store in one day where x can take on 5 values (0, 1, 2, 3, 4) 5. Discrete random variable with an infinite sequence of values Let x = number of customers arriving in one day where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive. Discrete Probability Distributions values of the random variable. 6. The probability distribution for a random variable describes how probabilities are distributed over the 7. The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable. 8. The required conditions for a discrete probability function are: 9. We can describe a discrete probability distribution with a table, graph, or equation. Example: JSL Appliances d Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below right) was developed. f(x) > 0 Σ f(x) = 1 Graphical Representation of the Probability Distribution 10. The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula. Discrete Uniform Probability Distribution Discrete Uniform Probability Distribution 11. The discrete uniform probability function is where: n = the number of values the random variable may assume 12. Note that the values of the random variable are equally likely. Expected Value and Variance f(x) = 1/n 13. The expected value, or mean, of a random variable is a measure of its central location. 0. Expected value of a discrete random variable: E(x) = µ = Σ xf(x) 14. The variance summarizes the variability in the values of a random variable. 1. Variance of a discrete random variable: Var(x) = σ 2 = Σ (x ­ µ )2f(x) 15. The standard deviation, σ, is defined as the positive square root of the variance. Example: JSL Appliances 16. Expected Value of a Discrete Random Variable x 0 1 2 3 4 f(x) .40 .25 .20 .05 .10 xf(x) .00 .25 .40 .15 .40 E(x) = 1.20 The expected number of TV sets sold in a day is 1.2 Example: JSL Appliances 17. Variance and Standard Deviation of a Discrete Random Variable x 0 1 2 3 4 x ­ µ ­1.2 ­0.2 0.8 1.8 2.8 (x ­ µ )2 1.44 0.04 0.64 3.24 7.84 f(x) .40 .25 .20 .05 .10 (x ­ µ )2f(x) .576 .010 .128 .162 .784 1.660 = σ 2 The variance of daily sales is 1.66 TV sets squared. The standard deviation of sales is 1.2884 TV sets. The standard deviation of sales is 1.2884 TV sets. Binomial Probability Distribution 18. Properties of a Binomial Experiment 2. The experiment consists of a sequence of n identical trials. 3. Two outcomes, success and failure, are possible on each trial. 4. The probability of a success, denoted by p, does not change from trial to trial. 5. The trials are independent. Example: Evans Electronics 19. Binomial Probability Distribution Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year? Let: p = .10, n = 3, x = 1 Binomial Probability Distribution 20. Binomial Probability Function f ( x) = n! ( n −x ) p x (1 − p ) x!( n − x )! where: f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial Example: Evans Electronics 21. Using the Binomial Probability Function n! ( n −x ) p x (1 − p ) x!( n − x )! 3! ( 3 −1 ) f (1 ) = 0 .11 (1 − 0 .1 ) 1!( 3 − 1 )! f ( x) = f (1 ) = 3 ∗ 2 ∗1 (2 ) 0 .1( 0 .9 ) =(3)(0.1)(0.81)=0.243 2! Using the Tables of Binomial Probabilities (A-12 textbook) p .30 .3430 .4410 .1890 .0270 n 3 x 0 1 2 3 .10 .7290 .2430 .0270 .0010 .15 .6141 .3251 .0574 .0034 .20 .5120 .3840 .0960 .0080 .25 .4219 .4219 .1406 .0156 .35 .2746 .4436 .2389 .0429 .40 .2160 .4320 .2880 .0640 .45 .1664 .4084 .3341 .0911 .50 .1250 .3750 .3750 .1250 Using a Tree Diagram Binomial Probability Distribution s Expected Value E(x) = µ = np s Variance Var(x) = σ2 = np(1 ­ p) s Standard Deviation SD(x) = σ = np(1 − p) Example: Evans Electronics s Binomial Probability Distribution • Expected Value E(x) = µ = 3(.1) = .3 employees out of 3 • Variance Var(x) = σ 2 = 3(.1)(.9) = .27 • Standard Deviation SD( x ) = σ = 3 (.1)(.9 ) = .52 employees Poisson Probability Distribution s Properties of a Poisson Experiment • The probability of an occurrence is the same for any two intervals of equal length. • The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval. Poisson Probability Function f (x) = where: µ x e −µ x! f(x) = probability of x occurrences in an interval µ = mean number of occurrences in an interval e = 2.71828 Example: Mercy Hospital s Using the Poisson Probability Function Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening? µ = 6/hour = 3/half­hour, x = 4 f (4 ) = Using the Tables of Poisson Probabilities (A­19 textbook) x 0 1 2 3 4 5 6 7 8 2.1 .1225 .2572 .2700 .1890 .0992 .0417 .0146 .0044 .0011 2.2 .1108 .2438 .2681 .1966 .1082 .0476 .0174 .0055 .0015 2.3 .1003 .2306 .2652 .2033 .1169 .0538 .0206 .0068 .0019 2.4 .0907 .2177 .2613 .2090 .1254 .0602 .0241 .0083 .0025 2.5 .0821 .2052 .2565 .2138 .1336 ..0668 .0278 .0099 .0031 3 4 (2 .71828 ) −3 = .1680 4! µ 2.6 .0743 .1931 .2510 .2176 .1414 .0735 .0319 .0118 .0038 2.7 .0672 .1815 .2450 .2205 .1488 .0804 .0362 .0139 .0047 2.8 .0608 .1703 .2384 .2225 .1557 .0872 .0407 .0163 .0057 2.9 .0550 .1596 .2314 .2237 .1622 .0940 .0455 .0188 .0068 3.0 .0498 .1494 .2240 .2240 .1680 .1008 .0504 .0216 .0081 Hypergeometric Probability Distribution With the hypergeometric distribution, the trials are not independent, and the probability of success changes from trial to trial. Hypergeometric Probability Function s s The hypergeometric distribution is closely related to the binomial distribution. r N − r x n − x f (x) = N n where: f(x) = probability of x successes in n trials Example: Neveready s n = number of trials N = number of elements in the population r = number of elements in the population labeled success Hypergeometric Probability Distribution Bob Neveready has removed two dead batteries from a flashlight and inadvertently mingled them with the two good batteries he intended as replacements. The four batteries look identical. Bob now randomly selects two of the four batteries. What is the probability he selects the two good batteries? Using Hypergeometric Probability Distribution: where: r N − r 2 2 2! 2! x n − x 2 0 2!0! 0!2! 1 = = = = .167 f ( x ) = 6 N 4 4! n 2 2!2! x = 2 = number of good batteries selected n = 2 = number of batteries selected N = 4 = number of batteries in total r = 2 = number of good batteries in total Exercises: 1. The following represents the probability distribution for the daily demand of microcomputers at a local store. Demand 0 1 2 3 4 Probability 0.1 0.2 0.3 0.2 0.2 a. What is the expected daily demand? b. What is the probability of having a demand for at least two microcomputers? 2. The student body of a large university consists of 60% female students. A random sample of 8 students is selected. (Use Binomial Probability Function) a. What is the probability that among the students in the sample exactly two are female? b. What is the probability that among the students in the sample at least 7 are female? c. What is the probability that among the students in the sample at least 6 are male? 3. Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. SEATWORK a. What is the probability that the sample contains 2 female voters? b. What is the probability that there are no females in the sample? 4. AMR is a computer consulting firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown below. Number of New Clients 0 1 2 3 Probability 0.05 0.10 0.15 0.35 4 5 6 0.20 0.10 0.05 a. What is the expected number of new clients per month? b. What is the variance? c. What is the standard deviation? 5 The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages (x) in the city has the following probability distribution. SEATWORK x 0 1 2 3 f(x) 0.80 0.15 0.04 0.01 What is the mean and the standard deviation for the number of electrical outages (respectively)? 6. Seventy percent of the students applying to a university are accepted. Using the binomial probability tables, what is the probability that among the next 18 applicants a. At least 6 will be accepted? b. Exactly 10 will be accepted? c. Exactly 5 will be rejected? d. Fifteen or more will be accepted? e. Determine the expected number of acceptances f. Compute the standard deviation. 7. A salesperson contacts eight potential customers per day. From past experience, we know that the probability of a potential customer making a purchase is .10. SEATWORK a. What is the probability the salesperson will make exactly two sales in a day? b. What is the probability the salesperson will make at least two sales in a day? c. What percentage of days will the salesperson not make a sale? d. What is the expected number of sales per day? Homework: p. 202 #43 (Due Tomorrow) ...
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