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Unit 2 Probability Distributions. Lesson 6 The Binomial Probability Distribution. To calculate a factorial on the TI-83, press , scroll over to the PRB menu, and choose 4: !. The Factorial Function.

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## Unit 2 Probability Distributions

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**Unit 2Probability Distributions**Lesson 6 The Binomial Probability Distribution**To calculate a factorial on the TI-83, press ,**scroll over to the PRB menu, and choose 4: ! The Factorial Function If n is a positive integer, n factorial, denoted n!, is the product of the integers from 1 to n, that is, The quantity 0! is defined to be 1.**(d)**Example 1: Computing Factorials Calculate: (a) 5! = 5 · 4 · 3· 2· 1 = 120 (b) 6! = 6 ·5 · 4 · 3· 2· 1 = 6 · 5! = 6 · 120 = 720 (c) 12! = 479,001,600**Combinations Rule**Suppose you have a set of n distinctly different elements, and you wish to select a subset of x elements without regard to order. This unordered arrangement is called a combination of n elements taken x at a time.**Pascal’s Triangle**To calculate a combination on the TI-83, press , scroll over to the PRB menu, and choose 3: nCr. The Combinations Rule The number of possible combinations of x elements chosen from a set of size n is denoted nCx. The combinations rule states that**Example 2: Computing with the Combinations Rule**Find the numerical values of (a) 6C3 (b) 100C98 (c) 50C50 (d) 50C0**Criteria for a Binomial Probability Experiment**An experiment is said to be a binomial experiment provided that 1. The experiment is performed a fixed number of times. Each repetition of the experiment is called a trial. 2. The trials are independent. This means the outcome of one trial will not affect the outcome of the other trials. 3. For each trial, there are two mutually exclusive outcomes, success (S) or failure (F). 4. The probability of success is fixed for each trial of the experiment.**Notation Used in the Binomial Probability Distribution**• There are n independent trials of the experiment. • The letter p represents the probability of success, so that 1 – p is the probability of failure. • There are 2n possible outcomes. • The binomial random variable x is the number of successes observed in the n trials of the experiment, so x = 0, 1, 2, …, n • For each value of x, there are nCx outcomes (combinations) of x successes in n trials of the experiment.**Illustration: Constructing a Binomial Distribution**A public interest research group reports that only 20% of voting-age Floridians favor proposed legislation allowing casino-style gambling. Let S represent an adult who supports the legislation and F represent one who opposes it. Thus, p = .20 and q = .80. Four adults are randomly and independently selected, and each is asked whether they support the casino gambling bill. Let x = the number of the four adults in the sample who support the bill. Find the probability distribution for the binomial random variable x.**x = 0**x = 1 x = 2 x = 3 x = 4 FFFF SFFF FSFF FFSF FFFS FFSS FSFS FSSF SFFS SFSF SSFF FSSS SFSS SSFS SSSF SSSS (.8)4 = .4096 4(.2)(.8)3= .4096 6(.2)2(.8)2= .1536 4(.2)3(.8) = .0256 (.2)4 = .0016 The probability distribution for a binomial experiment with n = 4 and p = .2 Illustration: Constructing a Binomial Distribution In table form, the probability distribution is given by:**P(x)**.4096 .4096 .4 .3 .2 .1536 .1 .0256 .0016 x 0 1 2 3 4 Number of adults in the sample of 4 who favor casino-style gambling Illustration: Constructing a Binomial Distribution In the form of a probability histogram, the probability distribution is given by:**The Binomial Probability Distribution Function**The probability of obtaining x successes in n independent trials of a binomial experiment where the probability of success is p is given by where x = 0, 1, 2, . . ., n, and**Press**DISTRibution Menu Binomial Distribution Function on the TI-83 The binomial distribution function on the TI-83 is binompdf(n,p,x) where n = the total number of trialsp = the probability of success on a single trialx = the number of successes (may be a list)**Example 3: Computing Binomial Probabilities**If x is a binomial random variable, compute P(x) for each of the following cases: (a) n = 5, x = 1, p = .2 P(1) = binompdf(5,.2,1) = .410 (b) n = 4, x = 2, q = .4 P(2) = binompdf(4,.6,2) = .346 (c) n = 3, x = 0, p = .7 P(0) = binompdf(3,.7,0) = .027 (d) n = 5, x = 3, p = .1 P(3) = binompdf(5,.1,3) = .008**P(x)**x 0 1 2 k Press to access the DISTR menu. Cumulative Binomial Probabilities Many problems relating to binomial distributions require the repeated use to the probability formula to sum up the probabilities for consecutive values of x. The cumulative binomial probability distribution function, binomcdf(n,p,k) calculates sums of binomial probabilities.**Summary of Cumulative Probabilities**For the binomial random variable x = 0, 1, 2, 3, ... , n, the cumulative probabilities are summarized as follows:**scientific notation**= 4.5×10–4 = .00045 Example 4: Computing Cumulative Binomial Probabilities If x is a binomial random variable, use the TI-83 to find the following probabilities: (a) P(x ≤ 5) for n = 15, p = .6 P(x ≤ 5) = binomcdf(15,.6,5) = .034 (b) P(x > 1) for n = 5, p = .1 P(x > 1) = 1 – P(x ≤ 1) = 1 – binomcdf(5,.1,1) = .081 (c) P(x < 10) for n = 25, p = .7 P(x < 10) = P(x ≤ 9) =binomcdf(25,.7,9) (d) P(x ≥ 10) for n = 15, p = .9 P(x ≥ 10) = 1 – P(x ≤ 9) = 1 – binomcdf(15,.9,9)= .998**Example 5:Using the Binomial Probability Distribution**Function According to the United States Census Bureau, 18.3% of all households have 3 or more cars. (a) In a random sample of 20 households, what is the probability that exactly 5 have 3 or more cars? The sample is a binomial experiment with n = 20 and p = .183 P(5) = 20C5 (.183)5 (.817)15 = (15504) (.0002052) (.04823) = .153 The probability of the sample containing exactly 5 homes with three or more cars is .153.**Example 5:Using the Binomial Probability Distribution**Function According to the United States Census Bureau, 18.3% of all households have 3 or more cars. (b) In a random sample of 20 households, what is the probability that less than 4 have 3 or more cars? P(x < 4) = P(0) + P(1) +P(2) +P(3) = binomcdf(20,.183,3) = .488 The probability of the sample containing fewer than 4 homes with three or more cars is .488.**Example 5:Using the Binomial Probability Distribution**Function According to the United States Census Bureau, 18.3% of all households have 3 or more cars. (c) In a random sample of 20 households, what is the probability that at least 4 have 3 or more cars? P(x≥ 4) = 1 – P(x < 4) = 1 – .488 = .512 The probability of the sample containing at least 4 homes with three or more cars is .512.**The Mean and Variance of a Binomial Random Variable**A binomial experiment with n independent trials and probability of success p will have a mean and standard deviation given by the formulas mean variance standard deviation**Example 6: Finding the Mean and Standard Deviation**According to the United States Census Bureau, 18.3% of all households have 3 or more cars. In a simple random sample of 400 households, determine the mean and standard deviation number of households that will have 3 or more cars. The sample is a binomial experiment with n = 400 and p = .183 μx = np = (400)(.183) = 73.2 households with 3 or more cars. The expected number of households in the samplewith 3+ cars is 73. = 7.7 households with 3+ cars.**p varies, n fixed**p = .05, .10, .15, … , .95; n = 10 Binomial Probability Distributions The following animations illustrate how the shape of a binomial probability distribution changes for varying values of n and p. Note that: • Values of p near 0 or 1 may result in skewed distributions. • Values of p near .5 result in symmetric, mound-shaped distributions.**n varies, p fixed**n = 2, 4, 6 … , 32; p = ¼ Binomial Probability Distributions The following animations illustrate how the shape of a binomial probability distribution changes for varying values of n and p. Note that: • Small values of n may result in skewed distributions. • Large values of n result in symmetric, mound-shaped distributions.**n**200 160 Large sample size 120 80 40 0 p 0 0.2 0.4 0.6 0.8 1 1 Bell-Shaped Binomial Distributions As the number of trials n in a binomial experiment increases, the probability distribution of the random variable x becomes bell-shaped. As a general rule of thumb, if np(1 – p) > 10, then the probability distribution will be approximately bell-shaped.**Chebyshev’s Rule**• np(1 – p)< 10 • Empirical Rule • np(1 – p)≥ 10 • Probability that x is within one standard deviation of the mean • Approximately 68% • Probability that x is within two standard deviations of the mean • At least 75% • Approximately 95% • Probability that x is within three standard deviations of the mean • At least 89% • Approximately 100% Interpreting Standard Deviation Chebyshev’s Rule and the Empirical Rule can be applied to binomial distributions to make statements about the likelihood of the random variable falling within one, two, or three standard deviations of the mean.**Example 7: Applying the Empirical Rule**According to the United States Census Bureau, in 2000, 18.3% of all households have 3 or more cars. Suppose a researcher selects a simple random sample of 400 households and found that 91 households had 3 or more cars. (a) Is the this binomial distribution bell-shaped? np(1 – p) = (400)(.183)(.817) = 59.8 Since np(1 – p) > 10, the binomial distribution is bell-shaped.**Example 7: Applying the Empirical Rule**According to the United States Census Bureau, in 2000, 18.3% of all households have 3 or more cars. Suppose a researcher selects a simple random sample of 400 households and found that 91 households had 3 or more cars. (b) Use the mean and standard deviation to compute the z-score for the observed value of x = 91. Interpret the result. The mean and standard deviation are μx = 73.2 and σx = 7.7. A sample of 91 households with 3+ cars is more than two standard deviations above the mean.**Example 7: Applying the Empirical Rule**According to the United States Census Bureau, in 2000, 18.3% of all households have 3 or more cars. Suppose a researcher selects a simple random sample of 400 households and found that 91 households had 3 or more cars. (c) Is this result unusual if the percentage of households with 3 or more cars is still 18.3%? By the Empirical Rule, only 2.5% of all observations will fall two standard deviations above the mean, so the result is unusual.

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