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Lesson 8 – 1R

Lesson 8 – 1R. Binomial Review. Knowledge Objectives. Describe the conditions that need to be present to have a binomial setting . Define a binomial distribution . Explain when it might be all right to assume a binomial setting even though the independence condition is not satisfied.

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Lesson 8 – 1R

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  1. Lesson 8 – 1R Binomial Review

  2. Knowledge Objectives • Describe the conditions that need to be present to have a binomial setting. • Define a binomial distribution. • Explain when it might be all right to assume a binomial setting even though the independence condition is not satisfied. • Explain what is meant by the sampling distribution of a count. • State the mathematical expression that gives the value of a binomial coefficient. Explain how to find the value of that expression. • State the mathematical expression used to calculate the value of binomial probability.

  3. Construction Objectives • Evaluate a binomial probability by using the mathematical formula for P(X = k). • Explain the difference between binompdf(n, p, X) and binomcdf(n, p, X). • Use your calculator to help evaluate a binomial probability. • If X is B(n, p), find µx and x (that is, calculate the mean and variance of a binomial distribution). • Use a Normal approximation for a binomial distribution to solve questions involving binomial probability

  4. Criteria for a Binomial Setting A random variable is said to be a binomial 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 Most important skill for using binomial distributions is the ability to recognize situations to which they do and don’t apply

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

  6. TI-83 Binomial Support • For P(X = k) using the calculator: 2nd VARS binompdf(n,p,k) • For P(k ≤ X) using the calculator: 2nd VARS binomcdf(n,p,k) • For P(X ≥ k) use 1 – P(k < X) = 1 – P(k-1 ≤ X)

  7. Binomial Mean and Std Dev A binomial experiment with n independent trials and probability of success p has Mean μx = np Standard Deviation σx = √np(1-p)

  8. Using Normal Apx to Binomials As binominal’s number of trials increases the formula for a binomial becomes unworkable (a situation alleviated with statistical software). So statisticians developed a procedure to use a continuous distribution, the normal, to estimate a discrete distribution. This procedure is used later with proportions.

  9. Example 1a Suppose that in its lifetime, a female kangaroo gives birth to exactly 10 young. Suppose further that each kangaroo baby, independently of all the others, has a 20% chance of surviving to maturity. • Find the probability that exactly four of the kangaroo’s young will survive to maturity p = 0.2 n = 10 x = 4 With x = 4, then we must use binompdf, binompdf(10, 0.2, 4) = 0.0881

  10. Example 1a Suppose that in its lifetime, a female kangaroo gives birth to exactly 10 young. Suppose further that each kangaroo baby, independently of all the others, has a 20% chance of surviving to maturity. (b) Find the probability that at least four of the kangaroo’s young will survive to maturity. p = 0.2 n = 10 x ≥ 4 With x ≥ 4, then we must use binomcdf and the complement rule, binomcdf(10, 0.2, 3) = 0.8791 P(x ≥ 4) = 1 – P(x ≤ 3) = 1 – 0.8791 = 0.1209

  11. Example 1c Suppose that in its lifetime, a female kangaroo gives birth to exactly 10 young. Suppose further that each kangaroo baby, independently of all the others, has a 20% chance of surviving to maturity. • What is the expected (mean) number of kangaroo young that will survive to maturity? p = 0.2 n = 10 x ≥ 4 E(x) = np = 10 (0.2) = 2

  12. Example 1d Suppose that in its lifetime, a female kangaroo gives birth to exactly 10 young. Suppose further that each kangaroo baby, independently of all the others, has a 20% chance of surviving to maturity. (d) What is the standard deviation of the number of kangaroo young that will survive to maturity? p = 0.2 n = 10 x ≥ 4 V(x) = n p (1 – p) = 10 (0.2) (0.8) = 1.6 σ(x) = 1.265

  13. Example 2 State the conditions that must be satisfied for a random variable X to have a binomial distribution. • Outcomes are mutually exclusive (success or failure) • Probability of success is the same for each event • Each event is independent • Fixed number of trials

  14. Example 3 What conditions must be satisfied for a normal distribution to provide a reasonable approximation for a binomial distribution? • The number of events must be large enough and the probability of success or failure not too small • or • np ≥ 10 • n(1 – p) ≥ 10

  15. Example 4a In a test for ESP, a subject is told that cards the experimenter can see (but the subject cannot see) contain either a star, circle, triangle, or square. As the experimenter looks at a card the subject names the shape on the card. The answers of a subject who does not have ESP should be independent observations, each with probability 1/4 of success. (a) Find the exact probability that a subject without ESP will be successful at most 30 times if he makes 100 attempts. Provide support for your work. p = 0.25, n = 100, and x ≤ 30 Meets binomial criteria: since x ≤ 30 we use binomcdf binomcdf (100, 0.25, 30) = 0.8962

  16. Example 4b In a test for ESP, a subject is told that cards the experimenter can see (but the subject cannot see) contain either a star, circle, triangle, or square. As the experimenter looks at a card the subject names the shape on the card. The answers of a subject who does not have ESP should be independent observations, each with probability 1/4 of success. (b) Use the normal approximation to estimate the probability that a subject without ESP will be successful at most 30 times if he makes 100 attempts. Show work! p = 0.25, n = 100, and x ≤ 30 μx = 25 σx = 4.33 Meets normal criteria for estimation : np = 25 > 10 n(1 – p) = 75 > 10 normcdf (-e99, 30, 25, 4.33) = 0.8759

  17. Example 4c In a test for ESP, a subject is told that cards the experimenter can see (but the subject cannot see) contain either a star, circle, triangle, or square. As the experimenter looks at a card the subject names the shape on the card. The answers of a subject who does not have ESP should be independent observations, each with probability 1/4 of success. (c) What is the probability that the subject’s first correct identification will occur on the 5th card? Geometric p = 0.25, n = 100, and x = 5 P(x=5) = (1 – p)  (1 – p)  (1 – p)  (1 – p)  (p) = (0.75)4(0.25) = 0.0791

  18. Example 4d In a test for ESP, a subject is told that cards the experimenter can see (but the subject cannot see) contain either a star, circle, triangle, or square. As the experimenter looks at a card the subject names the shape on the card. The answers of a subject who does not have ESP should be independent observations, each with probability 1/4 of success. (d) What is the probability that the subject’s first correct identification will occur after the 5th card? Geometric p = 0.25, n = 100, and x > 5 geometcdf(0.25, 5) = 0.7627 P(x > 5) = 1 – P(x ≤ 5) = = 1 – 0.7627 = 0.2373

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