Lesson 8 – R Taken from And modified slightlyhttp:// Review of Chapter 8 Discrete PDFs Binomial and Geometeric

Objectives Explain what is meant by a binomial setting and binomial distribution. Use technology to solve probability questions in a binomial setting. Calculate the mean and variance of a binomial random variable. Solve a binomial probability problem using a Normal approximation. Explain what is meant by a geometric setting. Solve probability questions in a geometric setting. Calculate the mean and variance of a geometric random variable.

Binomial PDF Objectives Identify a random variable as binomial by verifying four conditions: two outcomes (success and failure); fixed number of trials; independent trials; and the same probability of success for each trial. Use technology or the formula to determine binomial probabilities and to construct probability distribution tables and histograms. Calculate cumulative distribution functions for binomial random variables, and construct cumulative distribution tables and histograms. Calculate means (expected values) and standard deviations of binomial random variables. Use a Normal approximation to the binomial distribution to compute probabilities.

Geometric PDF Objectives Identify a random variable as geometric by verifying four conditions: two outcomes (success and failure); independent trials; the same probability of success for each trial; and the count of interest is the number of trials required to get the first success. Use formulas or technology to determine geometric probabilities and to construct probability distribution tables and histograms. Calculate cumulative distribution functions for geometric random variables, and construct cumulative distribution tables and histograms. Calculate expected values and standard deviations of geometric random variables.

English Phrases Math Symbol English Phrases ≥At leastNo less thanGreater than or equal to >More thanGreater than <Fewer thanLess than ≤No more thanAt mostLess than or equal to =ExactlyEqualsIs ≠Different from Cumulative probability or cdf P(x ≤ A) P(x > A) = 1 – P(x ≤ A) Values of Discrete Variable, X P(X) ∑P(x) = 1 X=A P(x ≤ A) = cdf (A)P(x = A) = pdf (A)

Binomial Probability Criteria A random variable is said to be a binomial provided: 1.For each trial there are two mutually exclusive (disjoint) outcomes: success or failure 2.The trials are independent 3.The probability of success is the same for each trial of the experiment 4.The experiment is performed a fixed number of times. Each repetition is called a trial Most important skill for using binomial distributions is the ability to recognize situations to which they do and don’t apply

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) = n C x p x (1 – p) n-x, x = 0, 1, 2, 3, …, n n C x is also called a binomial coefficient and is defined by combination of n items taken x at a time or where n! is n  (n-1)  (n-2)  …  2  1 n n! = k k! (n – k)!

Geometric Probability Criteria A random variable is said to be a geometric provided: 1.For each trial there are two mutually exclusive (disjoint) outcomes: success or failure 2.The trials are independent 3.The probability of success is the same for each trial of the experiment 4.We repeat the trials until we get a success

Geometric PDF When we studied the Binomial distribution, we were only interested in the probability for a success or a failure to happen. The geometric distribution addresses the number of trials necessary before the first success. If the trials are repeated k times until the first success, we will have had k – 1 failures. If p is the probability for a success and q (1 – p) the probability for a failure, the probability for the first success to occur at the kth trial will be (where x = k) P(x) = p(1 – p) x-1, x = 1, 2, 3, … The probability that more than n trials are needed before the first success will be P(k > n) = q n = (1 – p) n

Means and Normal Apx to Binomial Means and Standard Deviations –Binomial Mean: E(X) = μ = np Variance:  ² = np(1 – p) –Geometric Mean: E(X) = μ =1/p Variance:  ² = (1- p) / p² Normal distribution N(μ,σ) can approximate a Binomial curve, when conditions are met 1.n < 0.10N (sample small enough – independence) 2.np ≥ 10 and n(1-p) ≥ 10 (for normality)

TI-83 Reminders Binomial –N: number of trials –P: probability of success –X: number of successes Geometric –P: probability of success –X: number of trials until first success Remember to use catalog help PDF  X = # CDF  X ≤ # Complement Rule for X ≥ #

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

TI-83 Geometric Support For P(X = k) using the calculator: 2 nd VARS geometpdf(p,k) For P(k ≤ X) using the calculator: 2 nd VARS geometcdf(p,k) For P(X > k) use 1 – P(k ≤ X) or (1- p) k

Non AP Distributions - ID Hypergeometric –Small population sampling without replacement –Example: drawing names out of a hat Negative Binomial –Number of trials until the n th success –Example: number of foul shots until his 3 rd successful one –Geometric is a special case of this (n = 1) Poisson –Successes spread over spatial random variable (time or area) –Example: arrivals per minute at McD, potholes per mile on I-81

Example 1a/b: Which PDF? Determine which probability distribution (Binomial, Negative Binomial, Geometric, Hyper-geometric, and Poisson) best fits the following. Use only once. a. A stats class using a bucket filled with 20 red and 20 green balls, pulls a ball out of the bucket and records its color. They record the number of pulls required until they have 5 green balls pulled and repeat whole process 50 times. b. A stats class using a bucket filled with 20 red and 20 green balls, drops the bucket and scatters the balls across the room. They record the number of balls per floor tile and repeat this 50 times.

Which PDF? Determine which probability distribution (Binomial, Negative Binomial, Geometric, Hyper-geometric, and Poisson) best fits the following. Use only once. c. A stats class using a bucket filled with 20 red and 20 green balls, pulls a ball out of the bucket, records its color, replaces it and repeats until they pull a green ball. They record the number of pulls before the green ball is pulled out. d. A stats class using a bucket filled with 20 red and 20 green balls, pulls a ball out of the bucket and records its color and repeat it 50 times

Which PDF? Determine which probability distribution (Binomial, Negative Binomial, Geometric, Hyper-geometric, and Poisson) best fits the following. Use only once. e. A stats class using a bucket filled with 20 red and 20 green balls, pulls 5 balls out of the bucket and records the number of red balls and repeat it 50 times. a. Negative Binomial (pull until r th success) b. Poisson (successes over an area) c. Geometric (pulls till first success) d. Binomial (with n=1) e. Hyper-geometric (w/o replacement)

Summary and Homework Summary –Use pdf for an X = # –Use cdf for an X ≤ # –Use complement rule for X ≥ # P(X ≥ #) = 1 – P(X < #) P(X > #) = 1 – P( –Binomial – Bernoulli with fixed # of trials Mean: np Variance: np(1-p) –Geometric – Bernoulli until first success Mean: 1/p Variance: (1-p)/p² Homework: pg 556 – 59;

Problem 1 (a)The binomial setting and the geometric setting are similar in that they both involve 1) 2) 3) (b) How do the binomial and geometric settings differ? success or failure (mutually exclusive or binary outcomes) probability of success is constant independent outcomes (trials) Binomial Geometric fixed number of trials repeat trials until first success

Problem 2 According to the manufacturers, 13% of the M&M’s produced today are brown. (Did you know that at one time all M&M’s were brown?) Assume that all large bags of M&M’s contain 13% brown. Suppose you start taking individual candies out of a large bag, hoping for a brown one. Let X represent the number of the draw on which you get your first brown M&M. (a) On average, how many M&M’s would you expect to select in order to find a brown one? (b) Construct a table showing the probability distribution for X (up through X = 5). Show work for probabilities in the space below. Round probabilities to 3 decimal places. X = Probability = X~G(0.13) E(X) = 1/p = 1/(0.13) =

Problem 2 cont According to the manufacturers, 13% of the M&M’s produced today are brown. (Did you know that at one time all M&M’s were brown?) Assume that all large bags of M&M’s contain 13% brown. Suppose you start taking individual candies out of a large bag, hoping for a brown one. Let X represent the number of the draw on which you get your first brown M&M. (c) Construct a histogram that shows the cumulative probability distribution for X (up through X = 5). Label the height of each bar in addition to providing a scale on the vertical axis Nr of trials until first brown M&M Probability

Problem 3 When an oil company conducts exploratory oil drilling, each well is classified as a producer well or a dry well. Past experience shows that 15% of all wells drilled are producer wells. The company has plans to drill at 12 new locations. (a) What is the probability that exactly three wells will be producer wells? Be sure to provide support for your answer. (b) Calculate the probability that at least three wells will be producer wells. Be sure to provide support for your answer. X ~ B(0.15,12) P(X=3) = binompdf(12,.15,3) X ~ B(0.15,12) P(X≥3) = 1 – P(X<3) = 1 – P(X ≤ 2) = binomcdf(12,.15,2)

Problem 4 A seed producer claims that 95% of a certain type seed will germinate under ideal conditions. A testing agency attempts to germinate 3000 of these seeds. (a) Give the mean and standard deviation for the number of seeds that would germinate if the producer’s claim is correct. Mean = Standard deviation = (b) 2830 of the testing agency’s seeds eventually germinate. Use a normal approximation to estimate the probability that 2830 or fewer seeds would germinate if the producer’s claim is correct. Show work. X ~ B(0.95,3000) X ~ N(2850,11.94) P(X≤2830) = normalcdf(-E99,2830,2850,11.94) np =.95(3000) = 2850 √np(1-p) √3000(.95(.05) = Check conditions: assume > seeds produced np ≥ 10 n(1-p) ≥ ≥ ≥ 10

Problem 4 A seed producer claims that 95% of a certain type seed will germinate under ideal conditions. A testing agency attempts to germinate 3000 of these seeds. (a) Give the mean and standard deviation for the number of seeds that would germinate if the producer’s claim is correct. Mean = Standard deviation = (b) 2830 of the testing agency’s seeds eventually germinate. Use a normal approximation to estimate the probability that 2830 or fewer seeds would germinate if the producer’s claim is correct. Show work. X ~ B(0.95,3000) X ~ N(2850,11.94) P(X≤2830) = normalcdf(-E99,2830,2850,11.94) np =.95(3000) = 2850 √np(1-p) √3000(.95(.05) = Check conditions: np ≥ 10 n(1-p) ≥ ≥ ≥ 10