Chapter 6 Section 2

Chapter 6Section 2 The Binomial Probability Distribution

1 2 3 4 Chapter 6 – Section 2 • Learning 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

Chapter 6 – Section 2 • A binomial experiment has the following structure • A binomial experiment has the following structure • The first test is performed … the result is either a success or a failure • A binomial experiment has the following structure • The first test is performed … the result is either a success or a failure • The second test is performed … the result is either a success or a failure. This result is independent of the first and the chance of success is the same • A binomial experiment has the following structure • The first test is performed … the result is either a success or a failure • The second test is performed … the result is either a success or a failure. This result is independent of the first and the chance of success is the same • A third test is performed … the result is either a success or a failure. The result is independent of the first two and the chance of success is the same

Chapter 6 – Section 2 • Example • A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit • Example • A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit • The card is then put back into the deck • Example • A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit • The card is then put back into the deck • A second card is drawn from the deck with the same definition of success. • Example • A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit • The card is then put back into the deck • A second card is drawn from the deck with the same definition of success. • The second card is put back into the deck • Example • A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit • The card is then put back into the deck • A second card is drawn from the deck with the same definition of success. • The second card is put back into the deck • We continue for 10 cards

Chapter 6 – Section 2 • A binomialexperiment is an experiment with the following characteristics • A binomialexperiment is an experiment with the following characteristics • The experiment is performed a fixed number of times, each time called a trial • A binomialexperiment is an experiment with the following characteristics • The experiment is performed a fixed number of times, each time called a trial • The trials are independent • A binomialexperiment is an experiment with the following characteristics • The experiment is performed a fixed number of times, each time called a trial • The trials are independent • Each trial has two possible outcomes, usually called a success and a failure • A binomialexperiment is an experiment with the following characteristics • The experiment is performed a fixed number of times, each time called a trial • The trials are independent • Each trial has two possible outcomes, usually called a success and a failure • The probability of success is the same for every trial

Chapter 6 – Section 2 • Notation used for binomial distributions • The number of trials is represented by n • The probability of a success is represented by p • The total number of successes in n trials is represented by X • Because there cannot be a negative number of successes, and because there cannot be more than n successes (out of n attempts) 0 ≤ X ≤ n

Chapter 6 – Section 2 • In our card drawing example • Each trial is the experiment of drawing one card • The experiment is performed 10 times, so n = 10 • In our card drawing example • Each trial is the experiment of drawing one card • The experiment is performed 10 times, so n = 10 • The trials are independent because the drawn card is put back into the deck • In our card drawing example • Each trial is the experiment of drawing one card • The experiment is performed 10 times, so n = 10 • The trials are independent because the drawn card is put back into the deck • Each trial has two possible outcomes, a “success” of drawing a heart and a “failure” of drawing anything else • The probability of success is 0.25, the same for every trial, so p = 0.25 • In our card drawing example • Each trial is the experiment of drawing one card • The experiment is performed 10 times, so n = 10 • The trials are independent because the drawn card is put back into the deck • Each trial has two possible outcomes, a “success” of drawing a heart and a “failure” of drawing anything else • The probability of success is 0.25, the same for every trial, so p = 0.25 • X, the number of successes, is between 0 and 10

Chapter 6 – Section 2 • The word “success” does not mean that this is a good outcome or that we want this to be the outcome • A “success” in our card drawing experiment is to draw a heart • If we are counting hearts, then this is the outcome that we are measuring • There is no good or bad meaning to “success”

Chapter 6 – Section 2 • We would like to calculate the probabilities of X, i.e. P(0), P(1), P(2), …, P(n) • Do a simpler example first • For n = 3 trials • With p = .4 probability of success • Calculate P(2), the probability of 2 successes

Chapter 6 – Section 2 • For 3 trials, the possible ways of getting exactly 2 successes are • S S F • S F S • F S S • For 3 trials, the possible ways of getting exactly 2 successes are • S S F • S F S • F S S • The probabilities for each (using the multiplication rule) are • 0.4 • 0.4 • 0.6 = 0.096 • 0.4 • 0.6 • 0.4 = 0.096 • 0.6 • 0.4 • 0.4 = 0.096

Chapter 6 – Section 2 • The total probability is P(2) = 0.096 + 0.096 + 0.096 = 0.288 • The total probability is P(2) = 0.096 + 0.096 + 0.096 = 0.288 • But there is a pattern • Each way had the same probability … the probability of 2 success (0.4 times 0.4) times the probability of 1 failure (0.6 times 0.6) • The total probability is P(2) = 0.096 + 0.096 + 0.096 = 0.288 • But there is a pattern • Each way had the same probability … the probability of 2 success (0.4 times 0.4) times the probability of 1 failure (0.6 times 0.6) • The probability for each case is 0.42 • 0.61

Chapter 6 – Section 2 • There are 3 cases • S S F could represent choosing a combination of 2 out of 3 … choosing the first and the second • S F S could represent choosing a second combination of 2 out of 3 … choosing the first and the third • F S S could represent choosing a third combination of 2 out of 3 • There are 3 cases • S S F could represent choosing a combination of 2 out of 3 … choosing the first and the second • S F S could represent choosing a second combination of 2 out of 3 … choosing the first and the third • F S S could represent choosing a third combination of 2 out of 3 • These are the 3 = 3C2 ways to choose 2 out of 3

Chapter 6 – Section 2 • Thus the total probability P(2) = .096 + .096 + .096 = .288 can also be written as P(2) = 3C2• .42 • .61 • Thus the total probability P(2) = .096 + .096 + .096 = .288 can also be written as P(2) = 3C2• .42 • .61 • In other words, the probability is • The number of ways of choosing 2 out of 3, times • The probability of 2 successes, times • The probability of 1 failure

Chapter 6 – Section 2 • The general formula for the binomial probabilities is just this • The general formula for the binomial probabilities is just this • For P(x), the probability of x successes, the probability is • The number of ways of choosing x out of n, times • The probability of x successes, times • The probability of n-x failures • The general formula for the binomial probabilities is just this • For P(x), the probability of x successes, the probability is • The number of ways of choosing x out of n, times • The probability of x successes, times • The probability of n-x failures • This formula is P(x) = nCx px (1 – p)n-x

Chapter 6 – Section 2 • Example • A student guesses at random on a multiple choice quiz • There are n = 10 questions in total • There are 5 choices per question so that the probability of success p = 1/5 = .2 • What is the probability that the student gets 6 questions correct?

Chapter 6 – Section 2 • Example continued • This is a binomial experiment • There are a finite number n = 10 of trials • Each trial has two outcomes (a correct guess and an incorrect guess) • The probability of success is independent from trial to trial (every one is a random guess) • The probability of success p = .2 is the same for each trial

Chapter 6 – Section 2 • Example continued • The probability of 6 correct guesses is P(x) = nCx px (1 – p)n-x • Example continued • The probability of 6 correct guesses is P(x) = nCx px (1 – p)n-x = 6C10 .26 .84 = 210 • .000064 • .4096 = .005505 • Example continued • The probability of 6 correct guesses is P(x) = nCx px (1 – p)n-x = 6C10 .26 .84 = 210 • .000064 • .4096 = .005505 • This is less than a 1% chance • Example continued • The probability of 6 correct guesses is P(x) = nCx px (1 – p)n-x = 6C10 .26 .84 = 210 • .000064 • .4096 = .005505 • This is less than a 1% chance • In fact, the chance of getting 6 or more correct (i.e. a passing score) is also less than 1%

Chapter 6 – Section 2 • Binomial calculations can be difficult because of the large numbers (the nCx) times the small numbers (the px and (1-p)n-x) • It is possible to use tables to look up these probabilities • It is best to use a calculator routine or a software program to compute these probabilities

Chapter 6 – Section 2 • We would like to find the mean of a binomial distribution • Example • There are 10 questions • The probability of success is .20 on each one • Then the expected number of successes would be10 • .20 = 2 • The general formula μX = n p

Chapter 6 – Section 2 • We would like to find the standard deviation and variance of a binomial distribution • This calculation is more difficult • The standard deviation is σX = √ n p (1 – p) and the variance is σX2 = n p (1 – p)

Chapter 6 – Section 2 • For our random guessing on a quiz problem • n = 10 • p = .2 • x = 6 • For our random guessing on a quiz problem • n = 10 • p = .2 • x = 6 • Therefore • The mean is np = 10 • .2 = 2 • The variance is np(1-p) = 10 • .2 • .8 = .16 • The standard deviation is √.16 = .4 • For our random guessing on a quiz problem • n = 10 • p = .2 • x = 6 • Therefore • The mean is np = 10 • .2 = 2 • The variance is np(1-p) = 10 • .2 • .8 = .16 • The standard deviation is √.16 = .4 • Remember the empirical rule? A passing grade of 6 is 10 standard deviations from the mean …

Chapter 6 – Section 2 • With the formula for the binomial probabilities P(x), we can construct histograms for the binomial distribution • With the formula for the binomial probabilities P(x), we can construct histograms for the binomial distribution • There are three different shapes for these histograms • When p < .5, the histogram is skewed right • When p = .5, the histogram is symmetric • When p > .5, the histogram is skewed left

Chapter 6 – Section 2 • For n = 10 and p = .2 (skewed right) • Mean = 2 • Standard deviation = .4

Chapter 6 – Section 2 • For n = 10 and p = .5 (symmetric) • Mean = 5 • Standard deviation = .5

Chapter 6 – Section 2 • For n = 10 and p = .8 (skewed left) • Mean = 8 • Standard deviation = .4

Chapter 6 – Section 2 • Despite binomial distributions being skewed, the histograms appear more and more bell shaped as n gets larger • This will be important!

Summary: Chapter 6 – Section 2 • Binomial random variables model a series of independent trials, each of which can be a success or a failure, each of which has the same probability of success • The binomial random variable has mean equal to np and variance equal to np(1-p)

Examples • After World War II, under the rule of the Communist party, some opponents of the Romanian government were imprisoned and tortured. A random sample of 59 former political detainees in Romania (none of whom know each other) was examined and the number of former detainees who suffered from lifetime posttraumatic stress disorder (PTSD) was determined. Answer the following questions. (Source: Bichescu D, et al. (2005) Long-term consequences of traumatic experiences: an assessment of former political detainees in Romania. Clinical Practice and Epidemiology in Mental Health (1)17.) • a. Explain why this is a binomial experiment. • b. Suppose that half of the former political detainees in Romania suffer from PTSD, what is the mean of X, the number of people suffering from PTSD in a sample of 59 former detainees? • c. Interpret the mean. • d. Compute the standard deviation of X. • e. Would it be unusual to find 32 individuals suffering from PTSD in a group of 59 former political detainees?

Examples - Answer • a. (There are a fixed number of trials (presence of PTSD in the 59 former detainees), the trials are independent (the incidence of PTSD is independent for each subject), there are only two outcomes (presence or absence of PTSD), the probability of developing PTSD is the same for each former detainee, and we are interested in the number of subjects who suffer from PTSD.) • b. (29.5) • c. (Assuming that half of the former detainees suffer from PTSD, we expect that in a sample of size 59, we expect that 29.5 of the subjects will suffer from PTSD.) • d. (3.84) • e. (No, since this is less than one standard deviation above the mean.)

Example • To avoid unpleasant surprises when the statement comes, you try to record all your credit card transactions in a ledger. Unfortunately, you tend to neglect recording about 5% of your purchases. Suppose that last month, you had 25 purchases on you credit card account. When the statement arrives, you count the number of purchases you forgot to record. The random variable X represents the number of unrecorded purchases in a month with 25 transactions. • a. Explain why this is a binomial experiment. • b. Find and interpret the mean of X. • c. Compute the standard deviation of X. • d. Find the probability that you would record all 25 purchases. • e. Find the probability that exactly 4 purchases would have been unrecorded. • f. Find the probability that fewer than 4 purchases would have been unrecorded. • g. Find the probability that at least 4 purchases would have been unrecorded. • h. Would it be unusual to find 4 unrecorded purchases in a month with 25 purchases?

Example • a. (There is a fixed number of trials (25), the trials are independent (whether we record a transaction or not does not depend on recording another transaction), there are only two outcomes (recorded or not), the probability is the same for each trial (0.05), and we are interested in the number of transactions that are not recorded.) • b. (1.25; this is the expected number of transactions that would be unrecorded) • c. (1.09) • d. (0.2774) • e. (0.0269) • f. (0.9659) • g. (0.0341) • h. (Yes)

Chapter 6 Section 2