Review of Chapter 6 Discrete PDFs Binomial and Geometeric Lesson 6 - R Review of Chapter 6 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.

Vocabulary None new

Using your TI-83 calculator We can use 1-Var-Stats to calculate the mean and standard deviation of a discrete random variable given it’s outcomes and probability Type in outcomes in L1 Type in corresponding probabilities in L2 Use 1-Var-Stats L1, L2 to get statistics Notes: Discrete Random Variables have countable (finite) values Continuous Random Variables have an interval of values (infinite) Ranges of Random Variables are determined by minimum or maximum values that they can take on

Discrete Random Variable - Mean The mean, or expected value [E(x)], of a discrete random variable is given by the formula μx = ∑ [x ∙P(x)] where x is the value of the random variable and P(x) is the probability of observing x (multiply them together and add all of them up)   Mean of a Discrete Random Variable Interpretation: If we run an experiment over and over again, the law of large numbers helps us conclude that the difference between x and ux gets closer to 0 as n (number of repetitions) increases

Discrete Random Variable - Variance Variance and Standard Deviation of a Discrete RV: The variance of a discrete random variable is given by:   σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2x and standard deviation is √σ2 Note: round the mean, variance and standard deviation to one more decimal place than the values of the random variable

Means and Variances Rules for Means Rules for Variances Means follow the rules for linear combinations (from Algebra) When you linearly combine two or more (rules give only the 2 case example) random variables, you combine their means in the same manner E(a + X + bY) = a + E(X) + bE(Y) Rules for Variances Adding a number to a random variable does not change its variance Multiply a random variable by a number changes the variance by the square of that number V(a + X + bY) = V(X) + b²V(Y) When you combine random variables, you always add the variances V(X - Y) = V(X) + V(Y) = V(X + Y)

Greater than or equal to English Phrases Math Symbol English Phrases ≥ At least No less than Greater than or equal to > More than Greater than < Fewer than Less than ≤ No more than At most Less than or equal to = Exactly Equals Is ≠ Different from 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

Binomial Probability Criteria A random variable is said to be a binomial provided: For each trial there are two mutually exclusive (disjoint) outcomes: success or failure The trials are independent The probability of success is the same for each trial of the experiment 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) = nCx px (1 – p)n-x, x = 0, 1, 2, 3, …, n nCx 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: For each trial there are two mutually exclusive (disjoint) outcomes: success or failure The trials are independent The probability of success is the same for each trial of the experiment 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) = qn = (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 n < 0.10N (sample small enough – independence) np ≥ 10 and n(1-p) ≥ 10 (for normality)

TI-83 Reminders Binomial Geometric Remember to use catalog help N: number of trials P: probability of success X: number of successes Geometric 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: 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)

TI-83 Geometric Support For P(X = k) using the calculator: 2nd VARS geometpdf(p,k) For P(k ≤ X) using the calculator: 2nd 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 nth success Example: number of foul shots until his 3rd 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 rth 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 Homework: 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:

Problem 1a The random variable X represents the number of people that you have to wait behind in line when you go to the post office to buy stamps at lunch time. The probability distribution of X is provided below:   X = 0 1 2 3 Probability = .1 .5 .3 .1 Find the mean number of people that will be in front of you in the stamp line. Use the definition and show work. Mean: ∑ [x ∙P(x)] = (.1)(0) + (.5)(1) + (.3)(2) + (.1)(3) = 0 + .5 + .6 + .3 = 1.4

Problem 1b The random variable X represents the number of people that you have to wait behind in line when you go to the post office to buy stamps at lunch time. The probability distribution of X is provided below:   X = 0 1 2 3 Probability = .1 .5 .3 .1 (b) Find the standard deviation for the number of people in the line in front of you. Use the definition and show work. Var: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx2 = (0 + .5 + .3(4) + .1(9) ) – 1.96) = 2.6 – 1.96 = 0.64 St Dev = 0.8

Problem 2 From the previous problem, let f(X) = 2X + 0.5 represent the amount of time (in minutes) required for the clerks to process X people. Show your work and use the shortcut methods (not the definitions) to find:   The mean number of minutes that you will have to wait. The standard deviation of the number of minutes you will have to wait. μX = 1.4 so μf(X) = 0.5 + 2 μX = 0.5 + 2(1.4) = 3.3 minutes σX = 0.8 so σ²f(X) = 2² σ²X = 4 (0.8)² = 2.56 minutes σf(X) = 2.56 = 1.6 minutes

Problem 3 While you are at the post office you also need to pick up a package. The random variable Y represents the number of people you have to wait behind in the pickup line. The probability distribution of Y is provided below:   Y = 0 1 2 Probability = .2 .3 .5 Use your calculator to find the mean number of people that will be in front of you in this line Use your calculator to find the standard deviation of the number of people in this line Mean: ∑ [x ∙P(x)] = 1.3 Var: ∑[x2 ∙ P(x)] – μ2x = 0.781

Problem 4 Suppose that the numbers of people in the two lines are independent of each other. Let Z = X + Y represent the total number of people you will have to wait behind at the post office. Use the rules we discussed in class to find:   The mean or expected value of Z. The standard deviation of Z. E(Z) = E(X) + E(Y) = 1.4 + 1.3 = 2.7 V(Z) = V(X) + V(Y) = 0.8² + 0.781² = 1.25 σ(Z) = 1.25 = 1.118

Problem 5 The weight of eggs produced by a certain breed of hen is normally distributed with mean of μ = 65 grams and standard deviation of σ = 5 grams. What is the probability that the weight of a dozen (12) randomly selected eggs is between 750 grams and 800 grams? E(D) = μD = μE1 + μE2 + … + μE12 = 12  μE = 12  65 = 780 grams V(D) = σ²D = σ²E1 + σ²E2 + σ²E3 … + σ²E12 = 12  σ²E = 12  5² = 300 grams σD = 300 = 17.32 750 800 780 normalcdf(750, 800, 780, 17.32) = 83.43%

Problem 6 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 7 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) = 7.69 1 2 3 4 5 0.130 0.113 0.098 0.086 0.074

Problem 7 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. 1 2 3 4 5 0.130 0.113 0.098 0.086 0.074 1 2 3 4 5 Nr of trials until first brown M&M 0.5 0.4 Probability 0.3 0.2 0.1 0.13 0.24 0.35 0.44 0.51

Problem 8 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) = 0.1720 binompdf(12,.15,3) X ~ B(0.15,12) P(X≥3) = 1 – P(X<3) = 1 – P(X ≤ 2) = 0.2642 1 - binomcdf(12,.15,2)

Problem 9 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) np = .95(3000) = 2850 √np(1-p) √3000(.95(.05) = 11.94 Check conditions: assume > 30000 seeds produced np ≥ 10  n(1-p) ≥ 10  2850 ≥ 10 150 ≥ 10 X ~ N(2850,11.94) P(X≤2830) = 0.047 normalcdf(-E99,2830,2850,11.94)