Binomial setting and distributions Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of n independent trials. The observations must meet these requirements:  the total number of observations n is fixed in advance  each observation falls into just one of two categories: success and failure  the outcomes of all n observations are statistically independent  all n observations have the same probability p of “success”

Applications for binomial distributions Binomial distributions describe the possible number of times that a particular event will occur in a sequence of observations.  In a clinical trial, a patient’s condition may improve or not. The binomial distribution describes the number of patients who improved (not how much better they feel) among the study participants.  Is a child obese or not (based on their body mass index)? The binomial distribution describes the number of obese children in a random sample of school-age children.  In a quality control study, we assess the number of defective items in a lot of goods, irrespective of the type of defect.

We express a binomial distribution for the count X of successes among n observations as a function of the parameters n and p: X ~ B(n,p).  The parameter n is the total number of observations.  The parameter p is the probability of success on each observation.  The count of successes X can be any whole number between 0 and n. The CDC estimates that a third of adult men are obese. In a random sample of 10 adult men, each man is either obese or not. The variable X is the number of obese men among those 10 men sampled, our count of “successes.” For each man, the probability of success, “obese,” is 1/3. The number X of obese men among 10 men has the binomial distribution B(n = 10, p = 1/3). Binomial parameters

Binomial probabilities The number of ways of arranging k successes in a series of n observations (with constant probability p of success) is the number of possible combinations (unordered sequences). This can be calculated with the binomial coefficient: R: choose(n,k) The binomial coefficient “n_choose_k” uses the factorial notation “!”. The factorial n! for any strictly positive whole number n is: n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1 where k = 0, 1, 2,..., or n

The binomial coefficient counts the number of ways in which k successes can be arranged among n observations. The binomial probability P(X = k) is this count multiplied by the probability of any specific arrangement of the k successes: XP(X)P(X) 012…k…n012…k…n Total1 The probability that a binomial random variable takes any range of values is the sum of each probability for getting exactly that many successes in n observations. P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. In a group of 25 Caucasian American males, what is the probability that exactly five are color blind?  P(x = 5) = [n! / k!(n – k)!] p k (1 – p) n-k = (25! / 5!(20)!) = [21*22*23*24*24*25 / 1*2*3*4*5] = 53,130 * * =  Use technology > dbinom(5,25,.08) [1]

The probability that exactly 2 adults in the sample have depression is ???? The incidence of major depression in adults is about 10%. A random sample of 50 adults will be tested for depression. The variable X is the number of individuals diagnosed with depression among all 50 and has the binomial distribution Bin(n = 50, p = 0.1). A) B) C) D) E) 0.112

Binomial mean and variance The center and spread of the binomial distribution for a count X are defined by the mean  and standard deviation  The incidence of major depression in adults is about 10%. A random sample of 50 adults will be tested for depression. The variable X is the number of individuals diagnosed with depression among all 50 and has the binomial distribution Bin(n = 50, p = 0.1). Thus,

Effect of changing p when n is fixed Binomial distributions are skewed when p is close to 0 or close to 1 (especially if the sample is small).

Effect of changing n for a fixed value of p

Normal approximation to binomial Binomial distribution can be approximated by a Normal distribution, when both np ≥10 and n(1 − p) ≥10. The approximation can be improved by using a continuity correction to take into account the fact that the Normal distribution is continuous. Hint: P(X=x) = P(x-.5 ≤ X ≤ x+.5)

Count of adults diagnosed with depression in a sample of 20 adults, Bin(n = 20, p = 0.1). No Normal approximation Why?? The incidence of major depression in adults is about 10%. Count of adults diagnosed with depression in a sample of 100 adults, Bin(n = 100, p = 0.1). Normal approximation OK Why? Binomial, n=20, p=0.1 Binomial, n=100, p=0.1

The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is about 8%. We take a random sample of size 125 from this population. What is the probability that 6 individuals or fewer in the sample are color blind?  Distribution of the count X: B (n = 125, p = 0.08)  np = 10 P(X ≤ 6) = pbinom(6,size=125,prob=.08) in R [1] or about 12%  Normal approximation: N (np = 10, √np(1 − p) = 3.033) P(X ≤ 6) = pnorm(6, mean=10, sd=3.033) = or about 9% Or z = (x - µ)/σ = (6 − 10)/3.033 =  P(X ≤ 6) = from Table B The Normal approximation is reasonable, but not quite close to 12%. Here p =.08 is not close to 0.5, but np = 10 just meets the criterion. Using a continuity correction greatly improves the approximation:  P(X ≤ 6) = P(X≤6.5) = pnorm(6.5, mean=10, se=3.033) =

Distributions for the color blindness example. n = 50 n = 125 n = 1000 The larger the sample size the better the Normal approximation fits the binomial distribution.

The Poisson distributions A Poisson distribution describes the count X of occurrences of an event in fixed, finite intervals of time or space when  occurrences are all independent,  and the probability of an occurrence is the same over all possible intervals. Think of the Poisson distribution as describing the number of items in containers. Items Containers  Radioactive decays  Weeds  Fleas  Cardiovascular deaths  Second  Acre of farm land  Dog  County / year

If we divide a natural lawn into 1 ft 2 quadrants, we can count how many dandelions are in each quadrant. Dandelions seeds are wind-spread. The probabilities of a quadrant containing 0,1,2,3… dandelions are given by a Poisson distribution: (i) independence of dandelions: the presence of one dandelion in a quadrant does not make the presence of another more or less likely. (ii) homogeneity of quadrants: each quadrant is equally susceptible to contain dandelions.

Poisson probabilities If μ is the population mean number of occurrences for a specified interval of time or space, then the Poisson probability distribution of observing k occurrences (k = 0, 1, 2, …) at constant μ  (> 0) is: The Poisson distribution has mean μ and standard deviation σ:

Effect of changing μ: The Poisson distribution is skewed when μ  5.

The number of deer crossing a road at night during mating season in a particular rural area can be modeled with a Poisson distribution. A local survey conducted over 4 nights found a total of 20 deer crossings. Based on this information, what is the probability that fewer than three deer would cross on a given night during mating season in this area? To compute this probability using the Poisson distribution, we need to know μ. In this case μ = 20 / 4 = 5 deer crossings per night. > ppois(2,lambda=5) [1] x

Historical records over 20 years in a particular town indicate an average of 4 severe rainstorms per year. Modeling the occurrences of severe rainstorms with the Poisson distribution, the probability that there would be no severe rainstorm next year is P(X = 0) = (4) 0 e –4 / 0! = Probability of 5 severe rainstorms next year P(X = 5) = (4) 5 e –4 / 5! = Probability of 1 or more severe rainstorms next year P(X > 1) = 1 – P(X = 0) = 1 – = Probability of more than 5 severe rainstorms next year P(X > 5) = 1 – P(X ≤ 5) = 1 – = x P(X=x)P(X≤x) % 1.832% % 9.158% % % % % % % % % % % % % % % % % % % % % % % % % % %