Basic Practice of Statistics - 3rd Edition Binomial Distributions Chapter 13 Binomial Distributions BPS - 5th Ed. Chapter 13 Chapter 12

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 - 5th Ed. Chapter 13

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 - 5th Ed. Chapter 13

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 - 5th Ed. Chapter 13

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 - 5th Ed. Chapter 13

Case Study Inspecting Switches 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 - 5th Ed. Chapter 13

Case Study Inspecting Switches 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 - 5th Ed. Chapter 13

Binomial Probabilities 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 - 5th Ed. Chapter 13

Binomial Probabilities 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 - 5th Ed. Chapter 13

Binomial Probabilities 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 - 5th Ed. Chapter 13

Binomial Probabilities 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 - 5th Ed. Chapter 13

Binomial Coefficient The number of ways of arranging k successes among n observations is given by the binomial coefficient where n! is “n factorial” (see next slide). the binomial coefficient is read “n choose k”. BPS - 5th Ed. Chapter 13

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: BPS - 5th Ed. Chapter 13

Binomial Probabilities 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 BPS - 5th Ed. Chapter 13

Case Study Inspecting Switches 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. BPS - 5th Ed. Chapter 13

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 BPS - 5th Ed. Chapter 13

Case Study Inspecting Switches µ = np = (10)(0.1) = 1 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 BPS - 5th Ed. Chapter 13

Probability Histogram Case Study Inspecting Switches Probability Histogram n=10, p=0.1 BPS - 5th Ed. Chapter 13

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 - 5th Ed. Chapter 13

Normal Approximation to the Binomial The Normal distribution that is used to approximate the binomial distribution uses the same mean and standard deviation: When n is large, a binomial random variable X (with n trials and success probability p) is approximately Normal: BPS - 5th Ed. Chapter 13

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p) ≥ 10 Note that these conditions also depend on the value of p (and not just on n) BPS - 5th Ed. Chapter 13

Case Study Shopping Attitudes 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 - 5th Ed. Chapter 13

Case Study Shopping Attitudes 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 - 5th Ed. Chapter 13

Case Study Shopping Attitudes Chapter 13 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: Find probability of getting at least 1520: µ = np = 2500(0.6) = 1500 BPS - 5th Ed. Chapter 13

Case Study Shopping Attitudes Assuming X has the N(1500, 24.49) distribution [np and n(1p) are both ≥ 10], we have BPS - 5th Ed. Chapter 13

Case Study Shopping Attitudes 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 - 5th Ed. Chapter 13