Parameters for a Binomial Distribution - AP Statistics Study Guide

Quick Summary

This guide will enable you to master the parameters of a binomial distribution. You will learn to calculate the mean (expected value) and standard deviation for a binomial random variable using simple formulas. More importantly, you will be able to interpret these values in context, explaining the long-run average number of successes and the typical variability around that average, a key skill for the AP exam.

Key Concepts

Before we can describe a binomial distribution with its parameters, we must first confirm that we are in a binomial setting. Remember the four conditions, easily recalled with the acronym BINS:

• Binary: Each trial has only two possible outcomes, typically labeled "success" or "failure."

• Independent: The outcome of one trial does not affect the outcome of any other trial.

• Number: There is a fixed number of trials, denoted by n.

• Success: The probability of success, denoted by p, is the same for each trial.

When these conditions are met, the count of successes, X, is a binomial random variable. Every binomial distribution is defined by two parameters: n (the number of trials) and p (the probability of success). From these two parameters, we can calculate the distribution's center and spread.

The Mean (Expected Value) of a Binomial Distribution

The mean of a random variable is its long-run average value. For a binomial random variable, it represents the average number of successes we would expect if we were to repeat the entire set of n trials many, many times. It is also called the expected value.

The formula is remarkably simple and intuitive.

• Formula for the Mean:

  • ${\mu }_X=E(X)=np$

  • Where:

    • ${\mu }_X$ is the mean of the random variable X.

    • $E(X)$ is the expected value of X.

    • $n$ is the number of trials.

    • $p$ is the probability of success on a single trial.

Example: An archer has a 70% chance of hitting the bullseye on any given shot. If she takes 20 shots, what is the expected number of bullseyes?

• Here, $n=20$ and $p=0.70$.

• ${\mu }_X=np=20\ast 0.70=14$.

• Interpretation: If this archer were to take 20 shots many times, we would expect her to get an average of 14 bullseyes per set of 20 shots.

The Standard Deviation of a Binomial Distribution

The standard deviation measures the typical or average distance of the outcomes (number of successes) from the mean. A larger standard deviation implies more variability in the number of successes from one set of n trials to the next.

• Formula for the Standard Deviation:

  • ${\sigma }_X=sqrt(np(1-p))$

  • Where:

    • ${\sigma }_X$ is the standard deviation of the random variable X.

    • $n$ is the number of trials.

    • $p$ is the probability of success.

    • $(1-p)$ is the probability of failure, sometimes denoted as $q$.

Important Note: The value inside the square root, $np(1-p)$, is the variance, denoted ${\sigma }_X^2$. The standard deviation is the square root of the variance.

Example (continued): For the same archer ($n=20$, $p=0.70$), what is the standard deviation of the number of bullseyes?

• ${\sigma }_X=sqrt(np(1-p))=sqrt(20\ast 0.70\ast (1-0.70))$

• ${\sigma }_X=sqrt(20\ast 0.70\ast 0.30)$

• ${\sigma }_X=sqrt(4.2)\approx 2.049$

• Interpretation: In many sets of 20 shots, the number of bullseyes the archer makes will typically vary from the mean of 14 by about 2.049 bullseyes.

[Image: A histogram of a binomial distribution, like B(20, 0.7). A vertical line marks the mean at μ=14. Horizontal lines extending from the mean show the distance of one standard deviation, σ \approx 2.05, in each direction, covering the range from about 12 to 16.]

Key Vocabulary

• Parameter: A value that describes a characteristic of a model or population. For a binomial distribution, the parameters are $n$ and $p$.

• Binomial Random Variable (X): A variable that counts the number of successes in a fixed number ($n$) of independent trials, where the probability of success ($p$) is constant.

• Mean (Expected Value) (μ_X): The long-run average value of a random variable. For a binomial variable, it is the expected number of successes, calculated as $np$.

• Standard Deviation (σ_X): A measure of the typical distance of the outcomes of a random variable from its mean. It quantifies the variability or spread of the distribution.

• Variance (σ_X^2): The square of the standard deviation, $np(1-p)$. It measures the average squared deviation from the mean.

• Number of Trials (n): The fixed number of times a chance process is repeated in a binomial setting.

• Probability of Success (p): The constant probability that a single trial results in a "success" in a binomial setting.

Calculator Tech (TI-84)

No major, specific calculator functions are required for this topic. The calculations for the mean ($np$) and standard deviation ($sqrt(np(1-p))$) are performed using standard arithmetic on the main calculator screen.

Example Calculation: To find the standard deviation for $n=20$ and $p=0.70$:

1. Press 2nd -> $x^2$ [√] to get the square root symbol.

2. Type the expression inside: $20\ast 0.70\ast (1-0.70)$

3. Your screen should show: $\surd (20\ast .7\ast (1-.7))$

4. Press ENTER. The calculator will display $\mathrm{2.04939...}$

Always perform calculations with unrounded values whenever possible and round only your final answer to an appropriate number of decimal places (usually 3 or 4 unless otherwise specified).

How to Show Work on the FRQ

To earn full credit on Free Response Questions involving the mean and standard deviation of a binomial distribution, you must show your calculations and interpret the results in context. Follow this two-part template for each parameter.

Calculating and Interpreting the Mean (μ_X)

1. State and Calculate:

   • Identify $n$ and $p$ from the problem context.

   • Write the formula: ${\mu }_X=np$.

   • Substitute the values: ${\mu }_X=(valueofn)\ast (valueofp)$.

   • State the final answer: ${\mu }_X=[result]$.

2. Interpret in Context:

   • Use the following sentence structure:

     "If we were to repeat the process of [describe the n trials in context] many times, we would expect the average number of [describe successes in context] to be about [value of μ_X]."

Calculating and Interpreting the Standard Deviation (σ_X)

1. State and Calculate:

   • Identify $n$ and $p$.

   • Write the formula: ${\sigma }_X=sqrt(np(1-p))$.

   • Substitute the values: ${\sigma }_X=sqrt((valueofn)\ast (valueofp)\ast (1-valueofp))$.

   • State the final answer: ${\sigma }_X=[result]$.

2. Interpret in Context:

   • Use the following sentence structure:

     "In many repetitions of [describe the n trials in context], the number of [describe successes in context] would typically vary from the mean of [value of μ_X] by about [value of σ_X]."

Practice Problems

Problem 1:

A recent survey found that 85% of all U.S. households have a streaming service subscription. A random sample of 50 households is selected. Let the random variable Y be the number of households in the sample that have a streaming service subscription.

(a) Calculate the mean of Y.

(b) Interpret the mean of Y in the context of the problem.

(c) Calculate the standard deviation of Y.

(d) Interpret the standard deviation of Y in the context of the problem.

Solution:

First, we confirm this is a binomial setting:

• Binary: A household either has a subscription ("success") or does not.

• Independent: The subscription status of one household is independent of another.

• Number: There is a fixed number of trials, $n=50$ households.

• Success: The probability of success is constant, $p=0.85$.

The random variable Y is binomial with $n=50$ and $p=0.85$.

(a) Calculate the mean of Y.

• Formula:${\mu }_Y=np$

• Substitution:${\mu }_Y=50\ast 0.85$

• Answer:${\mu }_Y=42.5$

(b) Interpret the mean of Y in context.

If we were to take many random samples of 50 households, we would expect the average number of households with a streaming service subscription to be about 42.5.

(c) Calculate the standard deviation of Y.

• Formula:${\sigma }_Y=sqrt(np(1-p))$

• Substitution:${\sigma }_Y=sqrt(50\ast 0.85\ast (1-0.85))$

• ${\sigma }_Y=sqrt(50\ast 0.85\ast 0.15)=sqrt(6.375)$

• Answer:${\sigma }_Y\approx 2.525$

(d) Interpret the standard deviation of Y in context.

In many random samples of 50 households, the number of households with a streaming service subscription would typically vary from the mean of 42.5 by about 2.525 households.

Problem 2:

A student is taking a 25-question multiple-choice test. Each question has four options (A, B, C, D), and the student decides to guess randomly on every question. Let X be the number of questions the student answers correctly.

(a) Calculate and interpret the expected number of correct answers for this student.

(b) Calculate and interpret the standard deviation of the number of correct answers.

Solution:

This is a binomial setting:

• Binary: An answer is either correct ("success") or incorrect.

• Independent: Guessing on one question doesn't affect the others.

• Number: Fixed $n=25$ questions.

• Success: The probability of guessing correctly is $p=1/4=0.25$.

The random variable X is binomial with $n=25$ and $p=0.25$.

(a) Calculate and interpret the expected number of correct answers.

• Calculation:

  • ${\mu }_X=np$

  • ${\mu }_X=25\ast 0.25=6.25$

• Interpretation:

  If a student were to guess on many 25-question multiple-choice tests, we would expect the average number of correct answers to be about 6.25.

(b) Calculate and interpret the standard deviation.

• Calculation:

  • ${\sigma }_X=sqrt(np(1-p))$

  • ${\sigma }_X=sqrt(25\ast 0.25\ast (1-0.25))$

  • ${\sigma }_X=sqrt(25\ast 0.25\ast 0.75)=sqrt(4.6875)$

  • ${\sigma }_X\approx 2.165$

• Interpretation:

  In many instances of a student guessing on a 25-question test, the number of correct answers would typically vary from the mean of 6.25 by about 2.165 questions.

Common Mistakes to Avoid

• Forgetting the Square Root: A very frequent error is to calculate the variance, $np(1-p)$, and report it as the standard deviation. Always remember to take the square root as the final step for ${\sigma }_X$.

• Misinterpreting the Mean as a Guarantee: The expected value is a long-run average, not what will happen in a single instance. Avoid saying, "The student will get 6.25 questions right." Instead, frame it as an average over many repetitions.

• Context is Not Optional: On the AP exam, failing to relate your interpretation back to the specifics of the problem (e.g., "households with subscriptions," "correct answers on the test") will result in a loss of credit. A generic interpretation like "the mean is 42.5" is incomplete.

• Using $p$ Incorrectly: Double-check that you are using the probability of success ($p$) in the formulas, not the probability of failure ($1-p$). The mean formula $np$ is particularly sensitive to this error.

• Reporting Variance Instead of Standard Deviation: Be sure you answer the question that is asked. If the question asks for standard deviation, do not provide the variance (${\sigma }_X^2$). They are different measures of spread with different units.