 # Slide Slide 1 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Lecture Slides Elementary Statistics Tenth Edition and the.

## Presentation on theme: "Slide Slide 1 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Lecture Slides Elementary Statistics Tenth Edition and the."— Presentation transcript:

Slide Slide 1 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Lecture Slides Elementary Statistics Tenth Edition and the Triola Statistics Series by Mario F. Triola

Slide Slide 2 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Chapter 5 Probability Distributions 5-1 Overview 5-2 Random Variables 5-3 Binomial Probability Distributions 5-4 Mean, Variance and Standard Deviation for the Binomial Distribution 5-5 The Poisson Distribution

Slide Slide 3 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Created by Tom Wegleitner, Centreville, Virginia Section 5-1 Overview

Slide Slide 4 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Overview This chapter will deal with the construction of discrete probability distributions by combining the methods of descriptive statistics presented in Chapter 2 and 3 and those of probability presented in Chapter 4. Probability Distributions will describe what will probably happen instead of what actually did happen.

Slide Slide 5 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Combining Descriptive Methods and Probabilities In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect.

Slide Slide 6 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Created by Tom Wegleitner, Centreville, Virginia Section 5-2 Random Variables

Slide Slide 7 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept This section introduces the important concept of a probability distribution, which gives the probability for each value of a variable that is determined by chance. Give consideration to distinguishing between outcomes that are likely to occur by chance and outcomes that are “unusual” in the sense they are not likely to occur by chance.

Slide Slide 8 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definitions  Random variable a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure  Probability distribution a description that gives the probability for each value of the random variable; often expressed in the format of a graph, table, or formula

Slide Slide 9 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definitions  Discrete random variable either a finite number of values or countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they result from a counting process  Continuous random variable infinitely many values, and those values can be associated with measurements on a continuous scale in such a way that there are no gaps or interruptions

Slide Slide 10 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Graphs The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities.

Slide Slide 12 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Requirements for Probability Distribution P ( x ) = 1 where x assumes all possible values.  0  P ( x )  1 for every individual value of x.

Slide Slide 13 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Mean, Variance and Standard Deviation of a Probability Distribution µ =  [x P(x)] Mean  2 =  [ (x – µ) 2 P(x )] Variance  2 = [  x 2 P ( x )] – µ 2 Variance (shortcut )  =  [ x 2 P ( x )] – µ 2 Standard Deviation

Slide Slide 14 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Roundoff Rule for µ,  and  2 Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ,  and  2 to one decimal place.

Slide Slide 15 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Identifying Unusual Results Range Rule of Thumb According to the range rule of thumb, most values should lie within 2 standard deviations of the mean. We can therefore identify “unusual” values by determining if they lie outside these limits: Maximum usual value = μ + 2σ Minimum usual value = μ – 2σ

Slide Slide 16 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Identifying Unusual Results Probabilities Rare Event Rule If, under a given assumption (such as the assumption that a coin is fair), the probability of a particular observed event (such as 992 heads in 1000 tosses of a coin) is extremely small, we conclude that the assumption is probably not correct.  Unusually high: x successes among n trials is an unusually high number of successes if P(x or more) ≤ 0.05.  Unusually low: x successes among n trials is an unusually low number of successes if P(x or fewer) ≤ 0.05.

Slide Slide 17 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definition E =  [x P(x)] The expected value of a discrete random variable (with many trials) is denoted by E, and it represents the average value of the outcomes. It is obtained by finding the value of  [x P(x)].

Slide Slide 18 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Number of Radios Sold at Sound City in a Week Let x be the random variable of the number of radios sold per week –x has values x = 0, 1, 2, 3, 4, 5 Given: Frequency distribution of sales history over past 100 weeks –Let f be the number of weeks (of the past 100) during which x number of radios were sold # Radios, x FrequencyRelative Frequency 0f(0) =33/100 = 0.03 1f(1) =2020/100 = 0.20 2f(2) =500.50 3f(3) =200.20 4f(4) =50.05 5f(5) = 20.02 1001.00

Slide Slide 19 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example We can interpret the relative frequencies as probabilities –So for any value x, f(x)/n = p(x) –Assuming that sales remain stable over time Number of Radios Sold at Sound City in a Week Radios, x Probability, p(x) 0p(0) = 0.03 1p(1) = 0.20 2p(2) = 0.50 3p(3) = 0.20 4p(4) = 0.05 5p(5) = 0.02 1.00

Slide Slide 20 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example What is the chance that two radios will be sold in a week? –p(x = 2) = 0.50

Slide Slide 21 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example Continued What is the chance that fewer than 2 radios will be sold in a week? –p(x < 2) = p(x = 0 or x = 1) = p(x = 0) + p(x = 1) = 0.03 + 0.20 = 0.23 What is the chance that three or more radios will be sold in a week? –p(x ≥ 3) = p(x = 3, 4, or 5) = p(x = 3) + p(x = 4) + p(x = 5) = 0.20 + 0.05 + 0.02 = 0.27 Using the addition rule for the mutually exclusive values of the random variable.

Slide Slide 22 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Number of Radios Sold at Sound City in a Week How many radios should be expected to be sold in a week? –Let’s calculate the expected value of the number of radios sold,  X Radios, x Probability, p(x) x p(x) 0p(0) = 0.030  0.03 = 0.00 1p(1) = 0.201  0.20 = 0.20 2p(2) = 0.502  0.50 = 1.00 3p(3) = 0.203  0.20 = 0.60 4p(4) = 0.054  0.05 = 0.20 5p(5) = 0.025  0.02 = 0.10 1.00 2.10 On average, we expect to sell 2.1 radios per week

Slide Slide 23 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Number of Radios Sold at Sound City in a Week Let us calculate the variance and standard deviation of the number of radios sold at Sound City in a week Variance Radios, xProbability, p(x) (x -  X ) 2 p(x) 0p(0) = 0.03(0 – 2.1) 2 (0.03) = 0.1323 1p(1) = 0.20(1 – 2.1) 2 (0.20) = 0.2420 2p(2) = 0.50(2 – 2.1) 2 (0.50) = 0.0050 3p(3) = 0.20(3 – 2.1) 2 (0.20) = 0.1620 4p(4) = 0.05(4 – 2.1) 2 (0.05) = 0.1805 5p(5) = 0.02(5 – 2.1) 2 (0.02) = 0.1682 1.00 0.8900 Standard deviation

Slide Slide 24 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  Combining methods of descriptive statistics with probability.  Probability histograms.  Requirements for a probability distribution.  Mean, variance and standard deviation of a probability distribution.  Random variables and probability distributions.  Identifying unusual results.  Expected value.

Slide Slide 25 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Created by Tom Wegleitner, Centreville, Virginia Section 5-3 Binomial Probability Distributions

Slide Slide 26 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept This section presents a basic definition of a binomial distribution along with notation, and it presents methods for finding probability values. Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died.

Slide Slide 27 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definitions A binomial probability distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure). 4. The probability of a success remains the same in all trials.

Slide Slide 28 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Notation for Binomial Probability Distributions S and F (success and failure) denote two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so P(S) = p(p = probability of success) P(F) = 1 – p = q(q = probability of failure)

Slide Slide 29 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Notation (cont) n denotes the number of fixed trials. x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. p denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. P(x) denotes the probability of getting exactly x successes among the n trials.

Slide Slide 30 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Important Hints  Be sure that x and p both refer to the same category being called a success.  When sampling without replacement, consider events to be independent if n < 0.05N.

Slide Slide 31 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Methods for Finding Probabilities We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution.

Slide Slide 32 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Method 1: Using the Binomial Probability Formula P(x) = p x q n-x ( n – x )! x ! n !n ! for x = 0, 1, 2,..., n where n = number of trials x = number of successes among n trials p = probability of success in any one trial q = probability of failure in any one trial (q = 1 – p)

Slide Slide 33 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Method 2: Using Table A-1 in Appendix A Part of Table A-1 is shown below. With n = 4 and p = 0.2 in the binomial distribution, the probabilities of 0, 1, 2, 3, and 4 successes are 0.410, 0.410, 0.154, 0.026, and 0.002 respectively.

Slide Slide 34 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Method 2: Using Table A-1 in Appendix A Part of Table A-1 is shown below. With n = 12 and p = 0.80 in the binomial distribution, the probabilities of 4, 5, 6, and 7 successes are 0.001, 0.003, 0.016, and 0.053 respectively.

Slide Slide 35 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities. Method 3: Using Technology Minitab STATDISK

Slide Slide 36 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Method 3: Using Technology STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities. Excel TI-83 Plus calculator

Slide Slide 37 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Strategy for Finding Binomial Probabilities  Use computer software or a TI-83 Plus calculator if available.  If neither software nor the TI-83 Plus calculator is available, use Table A-1, if possible.  If neither software nor the TI-83 Plus calculator is available and the probabilities can’t be found using Table A-1, use the binomial probability formula.

Slide Slide 38 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Incidence of Nausea after Treatment x = number of patients who will experience nausea following treatment with Phe-Mycin antibiotic out of the 4 patients tested Let us find the probability that at least 3 of the 4 patients treated will experience nausea We set x = 3, n = 4, p = 0.1, so q = 1 – p = 1 – 0.1 = 0.9 Then: Using the addition rule for the mutually exclusive values of the binomial random variable with a binomial table

Slide Slide 39 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Binomial Probability Table (see table in the book) values of p (.05 to.50) Table 4.7(a) for n = 4, with x = 2 and p = 0.1 P(x = 2) = 0.0486 p = 0.1

Slide Slide 40 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Rationale for the Binomial Probability Formula P(x) = p x q n-x n !n ! ( n – x )! x ! The number of outcomes with exactly x successes among n trials

Slide Slide 41 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Binomial Probability Formula P(x) = p x q n-x n !n ! ( n – x )! x ! Number of outcomes with exactly x successes among n trials The probability of x successes among n trials for any one particular order

Slide Slide 42 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example Incidence of Nausea after Treatment x = number of patients who will experience nausea following treatment with Phe-Mycin out of the 4 patients tested Let us find the probability that 2 of the 4 patients treated will experience nausea Given: n = 4, p = 0.1, with x = 2 Then: q = 1 – p = 1 – 0.1 = 0.9 and

Slide Slide 43 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Binomial Distribution n = 4, p = 0.1

Slide Slide 44 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Several Binomial Distributions with different n and p parameters

Slide Slide 45 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  The definition of the binomial probability distribution.  Important hints.  Three computational methods.  Rationale for the formula.  Notation.

Slide Slide 46 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Created by Tom Wegleitner, Centreville, Virginia Section 5-4 Mean, Variance, and Standard Deviation for the Binomial Distribution

Slide Slide 47 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept In this section we consider important characteristics of a binomial distribution including center, variation and distribution. That is, we will present methods for finding its mean, variance and standard deviation. As before, the objective is not to simply find those values, but to interpret them and understand them.

Slide Slide 48 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. For Any Discrete Probability Distribution: Formulas Mean µ =  [ x P ( x )] Variance  2  = [  x 2 P ( x ) ] – µ 2 Std. Dev  = [  x 2 P(x) ] – µ 2

Slide Slide 49 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Binomial Distribution: Formulas Std. Dev.  = n p q Mean µ = n p Variance  2  = n p q Where n = number of fixed trials p = probability of success in one of the n trials q = probability of failure in one of the n trials (Note: You could use both sets of formulas to obtain)

Slide Slide 50 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. This scenario is a binomial distribution where: n = 14 p = 0.5 q = 0.5 Example Find the mean and standard deviation for the number of girls in groups of 14 births.

Slide Slide 51 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. This scenario is a binomial distribution where n = 14 p = 0.5 q = 0.5 Using the binomial distribution formulas: µ = (14)(0.5) = 7 girls  = (14)(0.5)(0.5) = 1.9 girls (rounded) Example (cont)

Slide Slide 52 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Interpretation of Results Maximum usual values = µ + 2  Minimum usual values = µ – 2  It is especially important to interpret results. The range rule of thumb suggests that values are unusual if they lie outside of these limits:

Slide Slide 53 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. For this binomial distribution, µ = 50 girls (n.p)   = 5 girls (n.p.q) ½ µ + 2  = 50 + 2(5) = 60 µ - 2  = 50 - 2(5) = 40 The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result. Example Determine whether 68 girls among 100 babies could easily occur by chance.

Slide Slide 54 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  Mean,variance and standard deviation formulas for the any discrete probability distribution.  Interpreting results.  Mean,variance and standard deviation formulas for the binomial probability distribution.

Slide Slide 55 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Created by Tom Wegleitner, Centreville, Virginia Section 5-5 The Poisson Distribution

Slide Slide 56 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept The Poisson distribution is important because it is often used for describing the behavior of rare events (with small probabilities).

Slide Slide 57 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definition The Poisson distribution is a discrete probability distribution that applies to occurrences of some event over a specified interval. The random variable x is the number of occurrences of the event in an interval. The interval can be time, distance, area, volume, or some similar unit. P(x) = where e  2.71828 µ x e -µ x!x! Formula

Slide Slide 58 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Poisson Distribution Requirements  The random variable x is the number of occurrences of an event over some interval.  The occurrences must be random.  The occurrences must be independent of each other.  The occurrences must be uniformly distributed over the interval being used. Parameters  The mean is µ.  The standard deviation is  = µ.

Slide Slide 59 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Difference from a Binomial Distribution The Poisson distribution differs from the binomial distribution in these fundamental ways:  The binomial distribution is affected by the sample size n and the probability p, whereas the Poisson distribution is affected only by the mean μ.  In a binomial distribution the possible values of the random variable x are 0, 1,... n, but a Poisson distribution has possible x values of 0, 1,..., with no upper limit.

Slide Slide 60 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example World War II Bombs In analyzing hits by V-1 buzz bombs in World War II, South London was subdivided into 576 regions, each with an area of 0.25 km 2. A total of 535 bombs hit the combined area of 576 regions If a region is randomly selected, find the probability that it was hit exactly twice. The Poisson distribution applies because we are dealing with occurrences of an event (bomb hits) over some interval (a region with area of 0.25 km 2 ).

Slide Slide 61 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example The mean number of hits per region is number of bomb hits number of regions The probability of a particular region being hit exactly twice is P(2) = 0.170. Where x=2 Number of occurrences)

Slide Slide 62 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: ATC Center Errors An air traffic control (ATC) center has been averaging 20.8 errors per year and lately has been making 3 errors per week. Is this unusual enough to justify an investigation? Then:  = 0.4 errors per week Because there are 52 weeks per year,  for a week is:  = (20.8 errors/year) / (52 weeks/year) = 0.4 errors/week Let x be the number of errors made by the ATC center during one week Given:  year  = 20.8 errors per year

Slide Slide 63 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: ATC Center Errors Continued Find the probability that 3 errors (x =3) will occur in a week –Want p(x = 3) when  = 0.4 Find the probability that no errors (x = 0) will occur in a week –Want p(x = 0) when  = 0.4

Slide Slide 64 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Poisson Probability Table , Mean number of Occurrences  =0.4

Slide Slide 66 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Poisson Distribution  = 0.4

Slide Slide 68 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Back to Example – Find  In the ATC center situation, 28.0 errors occurred on average per year mean  = 0.4 errors/week Because there are 52 weeks per year,  for a week is  = (20.8 errors/year) / (52 weeks/year) = 0.4 errors/week standard deviation  = 0.6325 errors/week. Because  = √ 0.4 = 0.6325 Assume that the number x of errors during any span of time follows a Poisson distribution for that time span Per week, the parameters of the Poisson distribution are:

Slide Slide 69 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Poisson as Approximation to Binomial Rule of Thumb  n  100  np  10 The Poisson distribution is sometimes used to approximate the binomial distribution when n is large and p is small.

Slide Slide 70 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Poisson as Approximation to Binomial - μ Value for μ  = n p  n  100  np  10

Slide Slide 71 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed:  Definition of the Poisson distribution.  Difference between a Poisson distribution and a binomial distribution.  Poisson approximation to the binomial.  Requirements for the Poisson distribution.