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Lesson 6 - 2

Lesson 6 - 2. Binomial Probability Distribution. Objectives. Determine whether a probability experiment is a binomial experiment Compute probabilities of binomial experiments Compute the mean and standard deviation of a binomial random variable Construct binomial probability histograms.

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Lesson 6 - 2

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  1. Lesson 6 - 2 Binomial Probability Distribution

  2. Objectives • Determine whether a probability experiment is a binomial experiment • Compute probabilities of binomial experiments • Compute the mean and standard deviation of a binomial random variable • Construct binomial probability histograms

  3. Vocabulary • Trial – each repetition of an experiment • Success – one assigned result of a binomial experiment • Failure – the other result of a binomial experiment • PDF – probability distribution function • CDF – cumulative (probability) distribution function, computes probabilities less than or equal to a specified value

  4. Criteria for a Binomial Probability Experiment An experiment is said to be a binomial experiment provided: • The experiment is performed a fixed number of times. Each repetition is called a trial. • The trials are independent • For each trial there are two mutually exclusive (disjoint) outcomes: success or failure • The probability of success is the same for each trial of the experiment

  5. Binomial Notation There are n independent trials of the experiment Let p denote the probability of success and then 1 – p is the probability of failure Let x denote the number of successes in n independent trials of the experiment. So 0 ≤ x ≤ n Determining probabilities: With your calculator: 2nd VARS 0 yields binompdf(n,p,x) Book: Table II in Appendix A Table III

  6. Binomial PDF The probability of obtaining x successes in n independent trials of a binomial experiment, where the probability of success is p, is given by: P(x) = nCx px (1 – p)n-x, x = 0, 1, 2, 3, …, n A binomial experiment with n independent trials and probability of success p has Mean μx = np Standard Deviation σx = √np(1-p)

  7. English Phrases ∑P(x) = 1 P(X) Cumulative probability or cdf P(x ≤ A) cdf(x > A) = 1 – P(x ≤ A) Values of Discrete Variable, X X=A

  8. Example 1 In the “Pepsi Challenge” a random sample of 20 subjects are asked to try two unmarked cups of pop (Pepsi and Coke) and choose which one they prefer. If preference is based solely on chance what is the probability that: a) 6 will prefer Pepsi? b) 12 will prefer Coke? P(d=P) = 0.5 P(x) = nCx px(1-p)n-x P(x=6 [p=0.5, n=20]) = 20C6 (0.5)6(1- 0.5)20-6 = 20C6 (0.5)6(0.5)14 = 0.037 P(x=12 [p=0.5, n=20]) = 20C12 (0.5)12(1- 0.5)20-12 = 20C12 (0.5)12(0.5)8 = 0.1201

  9. Example 1 cont P(d=P) = 0.5 P(x) = nCx px(1-p)n-x c) at least 15 will prefer Pepsi? d) at most 8 will prefer Coke? P(at least 15) = P(15) + P(16) + P(17) + P(18) + P(19) + P(20) Use cumulative PDF either Table III in book or calculator P(X ≥ 15) = 1 – P(X ≤ 14) = 1 – 0.9793 = 0.0207 P(at most 8) = P(0) + P(1) + P(2) + … + P(6) + P(7) + P(8) Use cumulative PDF either Table III in book or calculator P(X ≤ 8) = 0.2517

  10. Example 2 A certain medical test is known to detect 90% of the people who are afflicted with disease Y. If 15 people with the disease are administered the test what is the probability that the test will show that: a) all 15 have the disease? b) at least 13 people have the disease? P(x) = nCx px(1-p)n-x P(Y) = 0.9 P(x=15 [p=0.9, n=15]) = 15C15 (0.9)15(1- 0.9)15-15 = 15C15 (0.9)15(0.1)0 = 0.20589 P(at least 13) = P(13) + P(14) + P(15) Use cumulative PDF either Table III in book or calculator P(X ≥ 13) = 1 – P(X ≤ 12) = 1 – 0.1841 = 0.8159

  11. Example 2 cont P(Y) = 0.9 P(x) = nCx px(1-p)n-x c) 8 have the disease? P(x=8 [p=0.9, n=15]) = 15C8 (0.9)8(1- 0.9)15-8 = 15C8 (0.9)8(0.1)7 = 0.000277

  12. Example 3 A university claims that 80% of its basketball players get their degree. An investigation examines the fates of a random sample of 20 players who entered the program over a period of several years. Of these players, 10 graduated and 10 are no longer in school. If the university's claim is true, what is the probability that exactly 10 out of 20 graduate? Can you conclude anything about the university's claim?

  13. Example 3 cont If the university's claim is true, what is the probability that exactly 10 out of 20 graduate? Can you conclude anything about the university's claim? P(Y) = 0.8 P(x) = nCx px(1-p)n-x P(x=10 [p=0.8, n=20]) = 20C10 (0.8)10(1- 0.8)20-10 = 20C10 (0.8)10(0.2)10 = 0.00203 It is either a false claim or we got a very unusual sample! 10 – np 10 – 18 - 8 -8 Z = ----------- = ----------------- = -------- = --------------- = -4.47 np(1-p) 20(.8)(.2) 3.2 1.7889

  14. Summary and Homework • Summary • Binomial experiments have 4 specific criteria that must be met • E(X) = np and V(X) = np(1-p) • Calculator has pdf and cdf functions • Tables in book for pdf and cdf • Homework • pg 340-343: 11, 12, 17 – 20, 30 – 32, 41, 46

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