8.1 The Binomial Distribution AP Statistics

Two of the most important and useful random variable distributions are the Binomial distribution and the Geometric distribution. In this chapter, we will examine both. The binomial distributions are an important class of discrete probability distributions.

The Binomial Distribution Four conditions must be present for a random variable distribution to be considered “binomial” The Binomial Distribution Each observation can be categorized as “success” or “failure.” The probability of success is the same for each observation. The observations are independent. There are a fixed number of observations.

A binomial setting will be noted as B(n, p) where B indicates the conditions for a binomial distribution are met with “n” as the fixed number of observations and “p” as the probability for success. Because the distribution is a random variable (X), the values of X will be whole numbers from 0 to n.

Defective or not defective 15% are defective Independent Example 1: Suppose you receive a shipment of five monkey-scooters (nothing is funnier than a monkey on a scooter). Each scooter has a 15% chance of not working. What is the probability 3 or more scooters in your shipment will be defective? Why does this situation satisfy the binomial setting? Defective or not defective 15% are defective Independent 5 observations X = # of defective scooters out of a shipment of 5 X is B(5, 0.15) 

P(3 or more scooters are defective) = P(X > 3) = 0.02661 Use methods from chapter 7 to complete the probability distribution for X = # defective scooters out of 5. Find: P(3 or more scooters are defective) = P(X > 3) = 0.02661 P(no more than 1 scooter is defective) = P(X < 1) = 0.83521 P(odd number of defective scooters) = X 1 2 3 4 5 P(X) 0.4437 0.3915 0.13818 0.024 0.00215 7.6e-5 0.415576

P(X = k) = binompdf(n, p, k) TI NOTE: (Woo Hoo!) Many binomial calculations can be done on your calculator… look under your DISTR menu:   P(X = k) = binompdf(n, p, k) Binompdf (# trials, probability, point of interest) P(X < k) = binomcdf(n, p, k) Binomcdf (#trials, probability, point of interest) NOTE: cdf calculates to the left

Practice Examples

X = # successful penalty shots X ~ B(9, 0.75) P(X < 5) = 0.1657 1. Suppose I am successful on 75% of my penalty shot attempts. What is the probability I will make 5 or fewer of my next 9 attempts? Define a binomial random variable and calculate the probability. X = # successful penalty shots X ~ B(9, 0.75) P(X < 5) = 0.1657 binomcdf(9, 0.75, 5)

X = # of passengers who prefer to sleep X ~ B(25, 0.80) 2. The Los Angeles Times reported that 80% of airline passengers prefer to sleep on long flights rather than watch movies, read, etc. Consider randomly selecting 25 passengers from a particular long flight. Define a random variable X and answer the following questions: X = # of passengers who prefer to sleep X ~ B(25, 0.80) Calculate and interpret P(X = 12) binompdf(25, 0.80, 12) Calculate and interpret P(X = 25) binompdf(25, 0.80, 25) Calculate and interpret P(X ≥ 20) 1 – binomcdf(25, 0.80, 19) 0.000293 0.003778 0.616689

Mean and Standard Deviation of a Binomial Distribution If a Random Variable X is B(n, p), what is the expected number of successes? How much variability will there be from trial to trial? Mean and Standard Deviation of a Binomial Random Variable If a Random Variable X is B(n, p), the mean and standard deviation of X are:

Normal Approximation to Binomial Distribution As the number of trials n gets larger, the binomial distribution of X gets closer to a normal distribution. B(n, p) =

When n is large, we can use normal probability calculations to approximate binomial probabilities. Rule of C: Use the normal approximation when np > 10 and n(1-p) > 10. NOTE: The normal approximation is most accurate when p is close to ½ and least accurate when p is close to 0 or 1.

Example 2: Sample surveys show that fewer people enjoy shopping than in the past. A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming”. The population that the poll wants to draw conclusions about is all U. S. Residents ages 18 and over. Suppose that in fact 60% of all adult U. S. residents would say “agree” if asked the same question. What is the probability that 1520 or more of the sample agree?

The difference in the two calculations is only .006 Justify using Binomial Distribution: Agree or Disagree 60% success Observations are independent Fixed number (2500) of observations Use the Normal distribution: np = 2500 (.6) = 1500 > 10 2500(.4) = 1000 > 10 N( np, ) = N ( 1500, 24.49) Normalcdf( 1520, 1E99, 1500, 24.49) = .2071 B( 2500, .60) 1 – Binomcdf (2500, .6, 1519) = .2131 The difference in the two calculations is only .006

Example 3: Suppose 55% of adults have credit card debt Example 3: Suppose 55% of adults have credit card debt. If we survey 3200 adults, what is the probability more than 1800 would have credit card debt? X = # of adults who have credit card debt out of 3200 X is B(3200, 0.55)   We want P(X > 1800). Can we use a normal approximation? What would the mean and standard deviation of the normal approximation be? Use this to calculate the probability.

np = (3200)(.55) = 1760 > 10 nq = (3200)(.45) = 1440 > 10 P(X > 1800) = .0776 Normalcdf(1800, 1E99, 1760, 28.14)

Youtube video from educator. com on binomial example: http://www