Announcements U nit 2: P robability and distributions L ecture 2: B inomial and N ormal distribution S tatistics 101 Lab 2 Due today at 6 PM Nicole Dalzell May 20, 2015 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 2 / 1 Binary outcomes Conditional Probabilities Milgram experiment Stanley Milgram, a Yale University psychologist, conducted a series of experiments on obedience to Swaizland as the highest HIV prevalence in the world. 25.9% of this authority starting in 1963. country’s population is infected with HIV. The ELISA test is one of the Experimenter (E) orders the teacher finest and most accurate for detecting HIV. For those who carry HIV, (T), the subject of the experiment, to the test is 99.7% accurate. For those who do not carry HIV, the test is give severe electric shocks to a 96.2% accurate. If an individual from Swaizland tested positive on the learner (L) each time the learner ELISA test, what is the probability that he carries HIV? answers a question incorrectly. The learner is actually an actor, and the electric shocks are not real, but a prerecorded sound is played each time the teacher administers an electric shock. Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 3 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 4 / 1
Binary outcomes Binary outcomes Milgram experiment (cont.) Binary outcomes These experiments measured the willingness of study Each person in Milgram’s experiment can be thought of as a trial . participants to obey an authority figure who instructed them to A person is labeled a success if she refuses to administer a perform acts that conflicted with their personal conscience. severe shock, and failure if she administers such shock. Milgram found that about 65% of people would obey authority Since only 35% of people refused to administer a shock, and give such shocks, and only 35% refused. probability of success is p = 0 . 35. Over the years, additional research suggested this number is When an individual trial has only two possible outcomes, it is approximately consistent across communities and time. also called a Bernoulli random variable. Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 5 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 6 / 1 Binomial distribution Considering many scenarios Binomial distribution The binomial distribution Suppose we randomly select four individuals to participate in this ex- Binomial distribution periment. What is the probability that exactly 1 of them will refuse to administer the shock? The question from the prior slide asked for the probability of given Let’s call these people Allen (A), Brittany (B), Caroline (C), and Damian (D). Each number of successes, k , in a given number of trials, n , ( k = 1 one of the four scenarios below will satisfy the condition of “exactly 1 of them refuses success in n = 4 trials), and we calculated this probability as to administer the shock”: # of scenarios × P ( single scenario ) (A) refuse (B) shock (C) shock (D) shock Scenario 1 .35 × .65 × .65 × .65 = 0.0961 Scenario 2 (A) shock (B) refuse (C) shock (D) shock # of scenarios : there is a less tedious way to figure this out, .65 × .35 × .65 × .65 = 0.0961 we’ll get to that shortly... Scenario 3 (A) shock (B) shock (C) refuse (D) shock P ( single scenario ) = p k ( 1 − p ) ( n − k ) .65 × .35 × .35 × .65 = 0.0961 Scenario 4 (A) shock (B) shock (C) shock (D) refuse probability of success to the power of number of successes, probability of failure to the power of number of failures .65 .35 .65 .35 = 0.0961 × × × The Binomial distribution describes the probability of having exactly k The probability of exactly one 1 of 4 people refusing to administer the shock is the sum of all of these probabilities. successes in n independent Bernouilli trials with probability of success p . 0 . 0961 + 0 . 0961 + 0 . 0961 + 0 . 0961 = 4 × 0 . 0961 = 0 . 3844 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 7 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 8 / 1
Binomial distribution The binomial distribution Binomial distribution The binomial distribution Counting the # of scenarios Calculating the # of scenarios Choose function Earlier we wrote out all possible scenarios that fit the condition of The choose function is useful for calculating the number of ways to exactly one person refusing to administer the shock. If n was larger choose k successes in n trials. and/or k was different than 1, for example, n = 9 and k = 2: � n � n ! = RR SSSSSSS k !( n − k )! k S RR SSSSSS SS RR SSSSS · · · � 4 � 4 ! 4 × 3 × 2 × 1 k = 1, n = 4: = 1 !( 4 − 1 )! = 1 × ( 3 × 2 × 1 ) = 4 1 SS R SS R SSS � 9 � 2 !( 9 − 2 )! = 9 × 8 × 7 ! 9 ! 2 × 1 × 7 ! = 72 k = 2, n = 9: = 2 = 36 · · · 2 SSSSSSS RR Note: You can also use R for these calculations: writing out all possible scenarios would be incredibly tedious and > choose(9,2) prone to errors. [1] 36 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 9 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 10 / 1 Binomial distribution The binomial distribution Binomial distribution The binomial distribution Example Example At Duke University, 82% of students live in university owned or affiliated housing. A group of 12 students was chosen to speak to incoming students at orientation. What is the probability that zero, At Duke University, 82% of students live in university owned or one, four or more than four of these students live in student housing? affiliated housing. A group of 12 students was chosen to speak at Step 1: What is our response variable in this situation? orientation. What is the probability that (1) zero, (2) one, (3) two of Step 2: What is the probability that a student lives in university these students live in student housing? housing? We ask one student, and then a second, to come to the podium to speak. What is the probability that both of these students live in university housing? Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 11 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 12 / 1
Binomial distribution The binomial distribution Binomial distribution The binomial distribution Conditions for the Binomial Distribution Binomial distribution (cont.) Binomial probabilities If p represents probability of success, ( 1 − p ) represents probability of failure, n represents number of independent trials, and k represents The trials are independent. 1 number of successes The number of trials, n, is fixed. 2 � n � p k ( 1 − p ) ( n − k ) Each trial outcome can be classified as a success or failure. 3 P ( k successes in n trials ) = k The probability of a success, p, is the same for each trial. 4 # of scenarios × P ( single scenario ) Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 13 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 14 / 1 Binomial distribution The binomial distribution Binomial distribution The binomial distribution Example Example At Duke University, 82% of students live in university owned or Of the 12 students, we are going to randomly select 5 to be on a panel affiliated housing. A group of 12 students was chosen to speak to to speak to the parents. What is the probability that at least four (ie incoming students at orientation. What is the probability that (1) zero, four or more) of these selected students will live in student housing? (2) one, (3) two of these students live in student housing? P ( At least four ) = P ( Zero UH ) = P ( At least one ) = P ( One UH ) = P ( Two UH ) = Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 15 / 1 Statistics 101 (Nicole Dalzell) U2 - L2: Normal distribution May 20, 2015 16 / 1
Recommend
More recommend
