Chapter 12 Binomial Distributions BPS - 3rd Ed. Chapter 12 1 Binomial Setting Fixed number n of observations The n observations are independent Each observation falls into one of just two categories may be labeled success and failure The probability of success, p, is the same for each observation BPS - 3rd Ed. Chapter 12 2 Binomial Setting Examples In a shipment of 100 televisions, how many are defective? counting the number of successes (defective televisions) out of 100 A new procedure for treating breast cancer is tried on 25 patients; how many patients are cured? counting the number of successes (cured patients) out of 25 BPS - 3rd Ed. Chapter 12 3 Chapter 12 1
Binomial Distribution Let X = the count of successes in a binomial setting. The distribution of X is the binomial distribution with parameters n and p. n is the number of observations p is the probability of a success on any one observation X takes on whole values between 0 and n BPS - 3rd Ed. Chapter 12 4 Binomial Distribution not all counts have binomial distributions trials (observations) must be independent the probability of success, p, must be the same for each observation if the population size is MUCH larger than the sample size n, then even when the observations are not independent and p changes from one observation to the next, the change in p may be so small that the count of successes (X) has approximately the binomial distribution BPS - 3rd Ed. Chapter 12 5 An engineer selects a random sample of 10 switches from a shipment of 10,000 switches. Unknown to the engineer, 10% of the switches in the full shipment are bad. The engineer counts the number X of bad switches in the sample. BPS - 3rd Ed. Chapter 12 6 Chapter 12 2
X (the number of bad switches) is not quite binomial Removing one switch changes the proportion of bad switches remaining in the shipment (selections are not independent) However, removing one switch from a shipment of 10,000 changes the makeup of the remaining 9,999 very little the distribution of X is very close to the binomial distribution with n=10 and p=0.1 BPS - 3rd Ed. Chapter 12 7 Find the probability that a binomial random variable takes any particular value P(x successes out of n observations) =? need to add the probabilities for the different ways of getting exactly x successes in n observations BPS - 3rd Ed. Chapter 12 8 Example Each offspring hatched from a particular type of reptile has probability 0.2 of surviving for at least one week. If 6 offspring of these reptiles are hatched, find the probability that exactly 2 of the 6 will survive for at least one week. Label an offspring that survives with S for success and one that dies with F for failure. P(S) = 0.2 and P(F) = 0.8. BPS - 3rd Ed. Chapter 12 9 Chapter 12 3
Example (1) First, find probability that the two survivors are the first two offspring: Using the Multiplication Rule: P(SSFFFF) = (0.2)(0.2)(0.8)(0.8)(0.8)(0.8) = (0.2) 2 (0.8) 4 = 0.0164 BPS - 3rd Ed. Chapter 12 10 Example (2) Second, find the number of possible arrangements for getting two successes and four failures: SSFFFF SFSFFF SFFSFF SFFFSF SFFFFS FSSFFF FSFSFF FSFFSF FSFFFS FFSSFF FFSFSF FFSFFS FFFSSF FFFSFS FFFFSS There are 15 of these, and each has the same probability of occurring: (0.2) 2 (0.8) 4. Thus, the probability of observing exactly 2 successes out of 6 is: P(X=2) = 15(0.2) 2 (0.8) 4 = 0.246. BPS - 3rd Ed. Chapter 12 11 Binomial Coefficient The number of ways of arranging k successes among n observations is given by the binomial coefficient n n! = k k!( n k)! where n! is n factorial (see next slide). the binomial coefficient is read n choose k. BPS - 3rd Ed. Chapter 12 12 Chapter 12 4
Factorial Notation For any positive whole number n, its factorial n! is n! = n (n 1) (n 2) 3 2 1 Also, 0! = 1 by definition. Example: 6! = 6 5 4 3 2 1 = 720, and from the previous example: 6 = 2 6! 2!(6 = 2)! 6! 2!4! 6 5 4 3 2 1 6 5 30 = = = = 15 (2 1) ( 4 3 2 1) 2 1 2 BPS - 3rd Ed. Chapter 12 13 If X has the binomial distribution with n observations and probability p of success on each observation, the possible values of X are 0, 1, 2,, n. If k is any one of these values, then n k n k P( X = k) = p (1 k BPS - 3rd Ed. Chapter 12 14 The number X of bad switches has approximately the binomial distribution with n=10 and p=0.1. Find the probability of getting 1 or 2 bad switches in a sample of 10. P( X = 1or 2) = P( X = 1) + P( X = 2) 10 1 10-1 10 2 = (0.1) (1-0.1) + (0.1) (1-0.1) 1 2 10! 1 9 10! 2 8 = (0.1) (0.9) + (0.1) (0.9) 1!9! 2!8! = (10)(0.1)(0.3874) + (45)(0.01)(0.4305) = 0.3874 + 0.1937 = 0.5811 10-2 BPS - 3rd Ed. Chapter 12 15 Chapter 12 5
Mean and Standard Deviation If X has the binomial distribution with n observations and probability p of success on each observation, then the mean and standard deviation of X are µ = np σ = np(1 BPS - 3rd Ed. Chapter 12 16 The number X of bad switches has approximately the binomial distribution with n=10 and p=0.1. Find the mean and standard deviation of this distribution. µ = np = (10)(0.1) = 1 the probability of each being bad is one tenth; so we expect (on average) to get 1 bad one out of the 10 sampled σ = np(1 = (10)(0.1)(1 0.1) = 0.9 = 0.9487 BPS - 3rd Ed. Chapter 12 17 Probability Histogram n=10, p=0.1 BPS - 3rd Ed. Chapter 12 18 Chapter 12 6
Normal Approximation to the Binomial The formula for binomial probabilities becomes cumbersome as the number of trials n increases As the number of trials n increases, the binomial distribution gets close to a Normal distribution when n is large, Normal probability calculations can be used to approximate binomial probabilities BPS - 3rd Ed. Chapter 12 19 Normal Approximation to the Binomial The Normal distribution that is used to approximate the binomial distribution uses the same mean and standard deviation: µ = np and σ = np(1 When n is large, a binomial random variable X (with n trials and success probability is approximately Normal: X is approx. N ( np, np(1 ) BPS - 3rd Ed. Chapter 12 20 Normal Approximation to the Binomial (Sample Size) As a rule of thumb, we will use the Normal approximation to the Binomial when n is large enough to satisfy the following: np 10 and n(1 10 Note that these conditions also depend on the value of p (and not just on n) BPS - 3rd Ed. Chapter 12 21 Chapter 12 7
Hall, Trish. Shop? Many say Only if I must, New York Times, November 28, 1990. Nationwide random sample of 2500 adults were asked if they agreed or disagreed with the statement I like buying clothes, but shopping is often frustrating and time-consuming. Suppose that in fact 60% of the population of all adult U.S. residents would say Agree if asked this question. What is the probability that 1520 or more of the sample agree? BPS - 3rd Ed. Chapter 12 22 The responses of the 2500 randomly chosen adults (from over 210 million adults) can be taken to be independent. The number X in the sample who agree has a binomial distribution with n=2500 and p=0.60. To find the probability that at least 1520 people in the sample agree, we would need to add the binomial probabilities of all outcomes from X=1520 to X=2500 this is not practical. BPS - 3rd Ed. Chapter 12 23 Histogram of 1000 simulated values of the binomial variable X, and the density curve of the Normal distribution with the same mean and standard deviation: µ = np = 2500(0.6) = 1500 σ = np(1 = (2500)(0.6)(0.4) = 600 = 24.49 BPS - 3rd Ed. Chapter 12 24 Chapter 12 8
Assuming X has the N(1500, 24.49) distribution [np and n(1 are both 10], we have X µ 1520 1500 P( X 1520) = P σ 24.49 = P(Z 0.82) = 1 0.7939 = 0.2061 (from Standard Normal Table) BPS - 3rd Ed. Chapter 12 25 The probability of observing 1520 or more adults in the sample who agree with the statement has been calculated as 20.61% using the Normal approximation to the Binomial. Using a computer program to calculate the actual Binomial probabilities for all values from 1520 to 2500, the true probability of observing 1520 or more who agree is 21.31% This is a very good approximation! BPS - 3rd Ed. Chapter 12 26 Chapter 12 9