5.2 Binomial Distribution

How many of 400 randomly sampled adults are left-handed? How many of 20 randomly selected parts from an assembly line are defective? These “count of successes” questions arise constantly in statistics and are answered by the binomial distribution. In this chapter, we identify the four conditions that define a binomial setting, develop the formula for computing binomial probabilities, and introduce the mean and standard deviation.

Key Concepts

  • Identify a binomial random variable
  • Compute probabilities for a binomial random variable
  • Compute the mean and/or standard deviation for a binomial random variable

The binomial setting

A binomial random variable measures the number of successes in a fixed number of trials.

Random variables often follow simple, predictable patterns. In these cases, calculating probabilities and the mean or standard deviation are much simpler and do not require a table to be created. One type of random variable with a simple pattern is called a “Binomial Random Variable.” Suppose we call the occurrence of an event a success. “Success” does not necessarily indicate the outcome was positive (e.g., if we are investigating cancer in a sample of mice, success could be defined as the presence of cancer). A binomial random variable measures the number of successes in a fixed number of trials.

In order to be considered a binomial random variable, the following properties must hold:

  1. Fixed number of trials: The number of trials \(n\) is fixed in advanced
  2. Binary outcomes: Each trial results in one of two outcomes, which we label “success” and “failure.”
  3. Independence: The trials are independent, the outcome of one trial does not affect the outcome of any other trial.
  4. Constant probability: The probability of success \(p\) is the same for every trial.

Class Example 5.2.1: Is it binomial?

For each of the following, determine if the random variable is binomial. If it is, determine \(n\) and \(p\).

  1. On average, 25% of La Crosse residents recycle. 500 residents of La Crosse were asked if they recycle. \(A\) is the number of ‘yes’ answers in the sample.
  1. A box contains 50 marbles, 35 green and 15 white. Fifteen marbles are selected without replacement. \(B\) is the number of white marbles in the sample.
  1. An American roulette wheel consists of 18 red, 18 black, and 2 green numbers. Suppose you observed red on 10 consecutive spins. J is the number of additional spins needed until you observe black.
  1. \(D\) is the number of 5s in ten rolls of a fair die.
  1. Customers arrive at Alice’s with a rate of 5 per hour. \(E\) is the number of customers that enter Alice’s between 2 A.M. and 4 A.M.

Computing binomial probabilites

A health insurance company found that 70% of the people they insure stay below their deductible in any given year. Consider a random sample of four individuals. Let’s find the probability that exactly 1 of 4 randomly selected insured individuals exceeds the deductible (so 3 “successes” — staying below — and 1 “failure”).

Consider the specific scenario where person A exceeds and persons B, C, D do not: \[P(A = \text{exceed}, B = \text{not}, C = \text{not}, D = \text{not}) = (0.3)(0.7)(0.7)(0.7) = (0.7)^3(0.3)^1 = 0.103\]

But there are four such scenarios — any one of the four people could be the one who exceeds. Each scenario has the same probability. So: \[P(\text{exactly 3 successes in 4 trials}) = 4 \times (0.7)^3(0.3)^1 = 0.412\]

The general pattern is: \[P(X = k) = [\text{number of arrangements}] \times [\text{probability of one arrangement}]\]

The number of ways to arrange \(k\) successes in \(n\) trials is given by a combination: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

For our example: \(\binom{4}{3} = \frac{4!}{3! \cdot 1!} = \frac{24}{6 \times 1} = 4\), confirming the four arrangements we identified.

For a binomial random variable \(X\) with \(n\) trials and probability of success \(p\),

  • \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\)for k = 0, 1, 2, …, n.
  • Mean:\(\mu=E[X]=np\)
  • Standard Deviation:\(\sigma=sd(X)=\sqrt{np(1-p)}\)

Class Example 5.2.2: Testing ESP

To test for ESP, we have 4 cards, each with different symbols. These cards are shuffled and one is chosen at random (symbol down). Each time, you guess the symbol on the card. Suppose that you repeat this test 8 times and you do not have ESP.

  1. Let \(X\) be the number of times you guess correctly. Is \(X\) binomial? If so, what are \(n\) and \(p\)?
  1. What is the expected value of \(X\)?
  1. What is the standard deviation of \(X\)?
  1. To be certified as having ESP, you need to correctly identify at least 6 cards of the 8 drawn. What is the probability you get certified?

Class Example 5.2.3: Customer service

Joan takes 10 phone calls per day from customers, and, on average, 30% of these callers are angry. If 7 or more of these callers in a given day are angry, Joan has a bad day.

  1. Let \(C\) be the number of angry callers in a day. Is \(C\) binomial? If so, what are \(n\) and \(p\)?
  1. How many angry callers will Joan speak to on average each day? What is the standard deviation for the number of angry callers?
  1. What is the probability that Friday is a good day?
  1. What is the probability that 3 of 5 days in her workweek are good?

Class Example 5.2.4: Jordan Love

In the 2023 NFL season, Jordan Love threw 579 passes. A fair-weather fan happened to only see a few games that season. The fan only saw Jordan Love throw 33 times, and of those 18 were completed (caught).

  1. What would the fan estimate is Jordan Love’s probability to complete a pass?
  1. Based on the probability in part (a), how many passes would the fan expect Jordan Love to complete in the next 10 throws?

Summary

  • The binomial distribution counts the number of successes in \(n\) independent trials, each with the same probability of success \(p\).
  • Four conditions: fixed \(n\), binary outcomes, independence, constant \(p\).
  • Binomial probability formula: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
  • Mean and standard deviation: \(\mu = np\) and \(\sigma = \sqrt{np(1-p)}\).