3 Probability Distributions This chapter will deal the construction of probability distribution. By combining the methods of descriptive statistics in Chapter 2 and those of probability presented in Chapter 3. Probability Distributions will describe what will probably happen instead of what actually did happen.
4 Combining Descriptive Methods and Probabilities In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect.
5 Definitions A random variable is a variable (typically represented by X) that has a single numerical value, determined by chance, for each outcome of a procedure. A probability distribution is a graph, table, or formula that gives the probability for each value of the random variable.
6 2 Types of Random Variables Discrete Random Variable has 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 has infinitely many values, and those values can be associated with measurements on a continuous scale in such way that there are no gaps or interruptions.
7 Example Page 192, #2 Identify the given variable as being discrete or continuous. A. The cost of making a randomly selected movie. Discrete B. The number of movies currently being shown in U.S. theaters. Discrete C. The exact running time of a randomly selected movie. Continuous
8 Example Page 192, #2 D. The number of actors appearing in a randomly selected movie. Discrete E. The weight of the lead actor in a randomly selected movie. Continuous
9 Graphs The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities.
10 Requirements for a Discrete Probability Distribution 1. Px ( ) 1 where x assumes all possible values 2. 0 Px ( ) 1 for every individual value of x
11 Describing a Distributions Center Value that indicates where the middle of the data set is located. Variation Measures the amount that the values vary among themselves Distribution Shape of the distribution of data Outliers Sample values that lie very far away from the vast majority of the other sample values
12 Mean, Variance and Standard Deviation of a Probability Distribution [ x P( x)] Mean 2 x 2 P( x) Variance x P( x) x P( x) Variance (shortcut) Standard Deviation
13 Roundoff Rule for μ, σ, and σ² 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, σ, and σ² to one decimal place.
14 Example Page 192, #4 Determine whether a probability distribution is given. If probability distribution is described, find its mean and standard deviation. When manufacturing DVDs for Sony, batches of DVD s are randomly selected and the number of defects x is found for each batch. Not a probability distribution Px ( ) x P( x)
15 Example Page 192, #6 Determine whether a probability distribution is given. If probability distribution is described, find its mean and standard deviation. The Telektronic Company provides life insurance policies for its top four executives, and the random variable x is the number of those employees who live through next year. A probability distribution Px ( ) 1 x P( x)
16 Example Page 192, #6 x Px ( ) x P( x) x 2 P( x)
17 Example Page 192, #6 μ x P( x) σ 2 x 2 P( x) μ σ 2 σ
18 Example Page 192, #6 Use the TI to find the mean, variance, and standard deviations. x P(x) StatCALC
19 Identifying Unusual Results According to the range rule of thumb, most values should lie within 2 standard deviation of the mean. We can therefore identify unusual values by determining if they lie outside these limits: Maximum usual value = + 2 Minimum usual value = 2
20 Identifying Unusual Results Probabilities Rare Event Rule Given the assumption that boys and girls are equally likely the probability of a particular observed event (such as 13 girls in 14 births) 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) is very small (such as 0.05 or less). Unusually low x successes among n trials is an unusually low number of successes if P(x or fewer) is very small (such as 0.05 or less).
21 Example Page 194, #18 Assume that in a test of gender-selections technique, in clinical trial results in 12 girls and 14 births. Refer to Table 4-1 and find the indicated probabilities. A. Find the probability of exactly 12 girls in 14 births. Px ( 12) x (girls) P(x)
22 Example Page 194, #18 B. Find the probability of 12 or more girls in 14 births. Px ( 12) P( x 12) P( x 13) P( x 14) x (girls) P(x)
23 Example Page 194, #18 C. Which probability is relevant for determining whether 12 girls in 14 births is unusually high: the results from part (A) or part (B) Part (B), the occurrence of 12 girls among 14 would be unusually high if the probability of 12 or more girls is very small
24 Example Page 194, #18 D. Does 12 girls in 14 births suggest that the genderselection technique is effective? Why or why not? Yes, because the probability of 12 or more girls is very small (0.007). The result of 12 or more girls could not easily occur by chance.
25 Definition The expected value of a discrete random variable is denoted by E, and it represents the average value of the outcomes. It is obtained by find the value of E E x P ( x )
26 Example Page 192, #12 When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the probability of losing $5 is 20/38.
27 Example Page 192, #12 When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the probability of losing $5 is 20/38. A. If you bet $5 on the number 7, what is your expected value? x P( x) x P( x) 10 E The expected loss is $ for a $5 bet
28 Example Page 192, #12 When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the probability of losing $5 is 20/38. B. If you bet $5 that the outcome is an odd number, what is your expected value? x P( x) x P( x) 10 E The expected loss is $ for a $5 bet
29 Lesson 4-3 Binomial Probability Distribution
30 Binomial Probability Distribution Specific type of discrete probability distribution The outcomes belong to two categories pass or fail acceptable or defective success or failure
31 Example of a Binomial Distribution Suppose a cereal manufacturer puts pictures of famous athletes on cards in boxes of cereal, in the hope of increasing sales. The manufacture announces that 20% of the boxes contain a picture of Tiger Woods, 30% a picture of Lance Armstrong, and the rest a picture of Serena Williams. You buy 5 boxes of cereal. What s the probability you get exactly 2 pictures of Tigers Woods?
32 Requirements for a Binomial Probability Distribution 1) The procedure has a fixed number of trials 2) Trials are 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 4) Probabilities must remain constant for each trial.
33 Example Page 203, #4 Determine whether the given procedure results in a binomial distribution. Rolling a loaded dice 50 times and find the number of times that 5 occurs. Binomial
34 Notation for a Binomial Distribution There are n number of fixed trials Let p denote the probability of success Let q denote the probability of failure q = 1 p Let x denote the number of success in n independent trials: 0 x n Let P(x) denotes the probability of getting exactly x successes among the n trials
35 Important Hints Be sure that x and p both refer to the same category being called a success. When sampling without replacement, the events can be treated as if they were independent if the sample size is no more than 5% of the population size. 0.05N Where n
36 Mathematical Symbols Phrase at least or no less than Math Symbols more than or greater than > fewer than or less than < no more than or at most exactly =
37 Methods for Finding Probabilities of a Binomial Distribution Using the Binomial Probability Formula Using Table A-1 in Appendix A Using the TI
38 Binomial Probability Formula Number of outcomes with exactly x successes among n trials Probability of x successes among n trials for any particular order Px n! ( ) x n x! x! p q nx P( x) C p q n x x nx
39 Binomial Probability Formula P( x) C p q n x x nx for x = 0, 1, 2,,n where n = number of trials x = number of success among n trials p = probability of success in any one trial q = probability of failure in any one trial (q = 1 p)
40 Example 1 Cereal Suppose you buy 5 boxes of cereal. Where n = 5 and p = 0.2. What s the probability you get exactly 2 pictures of Tiger Woods? First find the total number of outcomes. 5 5! C 2! 5 2! There are 10 ways to get 2 Tiger pictures in 5 boxes.
41 Example 1 Cereal Suppose you buy 5 boxes of cereal. Where n = 5 and p = 0.2. What s the probability you get exactly 2 pictures of Tiger Woods? There are 10 ways to get 2 Tiger pictures in 5 boxes. 2 3 PX ( 2) 10(0.20) (0.80) nd Vars
42 Example Cereal The following table show the probability distribution function (p.d.f) for the binomial random variable, X. X = Tiger P(X) Sum 1 Binompdf Binompdf Binompdf Binompdf Binompdf Binompdf (5,0.20,0) (5,0.20,1) (5,0.20, 2) (5,0.20,3) (5,0.20, 4) (5,0.20,5)
43 Example Cereal The following table show the cumulative distribution function (c.d.f) for the binomial random variable, X. X PX ( ) pdf PX ( ) cdf P( X 0) P( X 1) P( X 2) P( X 3) P( X 4) P( X 5) Binomcdf Binomcdf Binomcdf (5,0.20,0) (5,0.20,1) (5,0.20, 2) Binomcdf Binomcdf (5,0.20, 4) (5,0.20,5) 1 Binomcdf (5,0.20,3)
44 Example Cereal Construct a histogram of the pdf and cdf using X[0, 6] 1 and Y[0, 1] 0.01 X PX ( ) pdf PX ( ) cdf P( X 0) P( X 1) P( X 2) P( X 3) P( X 4) P( X 5) pdf cdf
45 TI Binomial Probability Computing exact probabilities 2 nd /Vars/binompdf binompdf(n, p, x) pdf: probability distribution function Computing less than or equal to probabilities 2 nd /Vars/binomcdf binomcdf(n, p, x) cdf: cumulative distribution function
46 Example Page 204, #18 Use the Binomial Probability Formula. Assume that a procedure yields a binomial distribution. n 6, x 2, p 0.45 P( x) C p q 6 2 x nx Px ( 2) 15(0.45)
47 Binomial Probability Table
48 Example Page 204, #12 Using Table A-1. Assume that a procedure yields a binomial distribution. n 7, x 2, p 0.01 Px ( 2) 0.002
49 Example Page 204, #14 Using Table A-1. Assume that a procedure yields a binomial distribution. n 6, x 5, p 0.99 Px ( 5) 0.057
50 Example - Nielson Media According to Nielson Media research, 75% of US households have cable TV. A). In a random sample of 15 households, what is the probability that 10 have cable? x = households, p = 0.75, n = 15 Px ( 10) binompdf (15,0.75,10)
51 Example - Nielson Media The probability of getting exactly 10 households out 15 with cable is In 100 trials of this experiment we would expect (100)(0.1651) = 17 trials to result in 10 households with cable.
52 Example - Nielson Media According to Nielson Media research, 75% of US households have cable TV. B). In a random sample of 15 households, what is the probability that fewer than 13 have cable? P( x 13) P( x 0) P( x 1)... P( x 12) Px ( 12) binomcdf(15, 0.75, 12)
53 Example - Nielson Media There is probability that in a random sample of 15 households, less than 13 will have cable. In 100 trials of this experiment, we would expect (0.7639)(100) = 76 trials to result in fewer than 13 households that have cable.
54 Example Nielson Media According to Nielson Media research, 75% of US households have cable TV. C). In a random sample of 15 households, what is the probability that at least 13 have cable? P( x 13) P( x 13) P( x 14) P( x 15) 1 Px ( 12) binomcdf (15,0.75,12)
55 Example Nielson Media There is probability that in a random sample of 15 households, at least 13 will have cable. In 100 trials at of this experiment we would expect about (100)(0.2361) = 24 trials to result in at least 13 households have cable
56 Example Page 205, #28 An article in USA Today stated that Internal surveys paid by the directory assistance providers show that even the most accurate companies give out wrong numbers 15% of the time. Assume that you are testing such a provider by making 10 requests and also assume that the provider gives the wrong telephone number 15% of the time. x = the number of wrong answers, n = 10, p = 0.15 A). Find the probability of getting one wrong number. Px ( 1) binompdf (10,0.15,1)
57 Example Page 205, #28 B). Find the probability of getting at most one wrong number. P( x 1) P( x 0) P( x 1) binomcdf (10,0.15,1) C). If you do get at most one wrong number, does it appear that the rate of wrong numbers is not 15%, as claimed? No, since > 0.05 at most one wrong number is not an unusual occurrence when the error rate is 15%
58 Example Page 206, #32 A study was conducted to determine whether there were significant difference between medical students admitted through special programs (such as affirmative action) and medical students admitted through the regular admission criteria. It was found that the graduation rate was 94% for the medical students admitted through special programs. x = the number of special program student who graduate n = 10, p = 0.94 A). If 10 students from the special programs are randomly selected, find the probability that at least nine of them graduated. P( x 9) P( x 9) P( x 10) 1 Px ( 8) 1 binomcdf (10,0.94,8)
59 Example Page 206, #32 B). Would it be unusual to randomly select 10 students from the special programs and get only seven that graduate? Why or why not? Px ( 7) binompdf (10,0.94,7) Yes, since < 0.05 getting only 7 graduates would be unusual result.
60 Lesson 4-4 Binomial Distribution: Mean, Variance and Standard Devation
61 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
62 Binomial Distribution Mean Variance µ = n p 2 = n p q Std. Dev. = 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
63 Interpretation of Results The range rule of thumb suggests that values are unusual if they lie outside of these limits: Maximum usual values = µ + 2 Minimum usual values = µ 2
64 Example Page 210, #2 Finding,, and Unusual Values. Assume that a procedure yields a binomial distributions with n trials and the probability of success for one trial is p. n = 250, p = 0.45 np 250(0.45) min (7.86) 96.8 max (7.86) npq (250)(0.45)(1 0.45) 7.86
65 Example Page 210, #6 Several students are unprepared for a multiple-choice quiz with 10 questions, and all of their answers are guesses. Each question has five possible answers, and only one of them is correct. A). Find the mean and standard deviation for the number of correct answers for such students. n 10 p q μ np 10(0.20) 2 σ npq (10)(0.20)(0.80)
66 Example Page 210, #6 B). Would it be unusual for a student to pass by guessing and getting at least 7 correct answers? Why or why not? max 2 2 2(1.3) 4.6 Yes, since 7 is greater than or equal to the maximum usual value, it would be unusual for a student to pass by getting at least 7 correct answers.
67 Example Page 210, #10 The Central Intelligence Agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. In standard English text, the letter r is used at a rate of 7.7% A). Find the mean and standard deviation for the number of times the letter r will be found on a typical page of 2600 characters. n 2600 p q μ np 2600(0.077) σ npq (2600)(0.077)(0.923) 13.6
68 Example Page 210, #10 B). In the intercepted message sent to Iraq, a page of 2600 characters is found to a have the letter r occurring 175 times. Is this unusual? μ 2σ unusual values are outside of these boundaries (13.6) 173 and No, since 175 is within the above limits it would not be considered an unusual result.
69 Example Page 212, #16 Car Crashes: For drivers in the age bracket, there is a 34% rate of car accidents in one year (based on the data from the National Safety Council). An insurance investigator finds that in a group of 500 randomly selected drivers age living in New York City, 42% had accidents in the last year. A) How many drivers in the New York City group of 500 had accidents in the last year? (500) (0.42) 210
70 Example Page 212, #16 B) Assuming that the same 34% rate applies to New York City, find the mean and standard deviation for the number of people in groups of 500 that can expected to have accidents np 500(0.34) npq 500(0.34)(1 0.34) 10.6
71 Example Page 212, #16 C) Based on the preceding results, does 42% result for the New York City drivers appear to be unusually high when compared to the 34% rate for the general population? Does it appear that higher insurance rates for New York City drivers justified? max (10.6) Yes, the result of 210 accidents for NYC drivers are unusually high if the 34% rate for the general population applies. Yes, it appears that the higher rates for NYC drivers are justified.
72 Lesson 4-5 The Poisson Distribution
73 Examples of Poisson Distribution Planes arriving at an airport Arrivals of people in a line Cars pulling into a gas station
74 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. Formula P(x) = µ x e -µ where e x!
75 Requirements for a Poisson Distribution Random variable (x) is the number of occurrences of an event over some interval. Occurrences must be random Occurrences must be independent of each other Occurrences must be uniformly distributed over the interval being used. mean is μ Standard deviation is σ μ
76 Difference from a Binomial Distribution The binomial distribution is affected by the sample size n and the probability p, whereas the Poisson distribution is affected only by the mean. 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.
77 Example Page 216, #2 Use a Poisson Distribution to find probability. If μ = 0.5, find P(2). Px ( ) μ x e x! μ 0.5 e P(2) 2!
78 Example Page 216, #6 Currently, 11 babies are born in the village of Westport (population 760) each year. A). Find the mean number of births per day. 11 μ
79 Example Page 216, #6 B). Find the probability that on a given day, there is no births. 11 x = births 365 P( x 0) poissonpdf (11/ 365,0) Px ( 0) nd vars/c:poissonpdf(μ,x)
80 Example Page 216, #6 C). Find the probability that on a given day, there is at least one birth. P( x 1) P( x 1) P( x 2)... P( x 11) 1 Px ( 0) 1 poissonpdf (11/ 365,0)
81 Example Page 216, #6 D). Based on the preceding results, should medical birthing personnel be on permanent standby, or should they be called in as needed? Does this mean that Westport mothers might not get the immediate medical attention they would be likely to get in a more populated area? The personnel should be called as needed. Yes; this does mean that women giving birth might not get the immediate attention available in more populated areas.
82 Example Page 216, #8 Homicide Deaths: In one year, there were 116 homicide deaths in Richmond, Virginia (Based on A Classroom Note on the Poisson Distribution: A Model for Homicidal Deaths in Richmond, VA for 1991, in Mathematics and Computer Education, by Winston A. Richards). For a randomly selected day, find the probability that the number of homicide deaths is a. 0 b. 1 c. 2 d. 3 e. 4 Compare the calculated probabilities to these actual results: 268 days (no homicides); 79 days (1 homicides); 17 days (2 homicides); 1 day (3 homicides); no days with more than 3 homicides.
83 Example Page 216, #8 Let x = the number of homicides per day a. P(x = 0) = poissonpdf(0.3178, 0) = b. P(x = 1) = poissonpdf(0.3178, 1) = c. P(x = 2) = poissonpdf(0.3178, 2) = d. P(x = 3) = poissonpdf(0.3178, 3) = e. P(x = 4) = poissonpdf(0.3178, 4) =
84 Example Page 216, #8 x frequency Relative frequency P(x) = 268/ or more (by subtraction) The agreement between the observed relative frequencies and the probabilities predicted by the Poisson formula is very good.
85 Poisson as Approximation to Binomial The Poisson distribution is sometimes used to approximate the binomial distribution when n is large and p is small. Rule of thumb n 100 np 10 If the two conditions are satisfied we can use the Poisson distribution as an approximation to the binomial distribution where = np
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