The Graph, Expectation, and Variance of a Binomial Distribution Revision Notes for VCE SSCE Mathematical Methods

The Graph, Expectation, and Variance of a Binomial Distribution

Introduction

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Building on our understanding of discrete probability distributions, we now explore specific properties of the binomial distribution. This section focuses on two key aspects:

Visualizing binomial distributions through graphs
Calculating expectation and variance using efficient formulas

The graph of a binomial probability distribution

A probability distribution can be represented in several ways: as a rule, a table, or a graph. Let's investigate how the shape of a binomial probability distribution graph changes with different values of the parameters $n$ (number of trials) and $p$ (probability of success).

How the value of p affects the graph shape

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The probability of success, $p$ , has a significant impact on the shape of the distribution:

When $p < 0.5$ : The graph is positively skewed (skewed to the right). Most of the probability is concentrated on the lower values, with a tail extending towards higher values.
When $p = 0.5$ : The graph is symmetrical. The probability of success equals the probability of failure, creating a balanced distribution.
When $p > 0.5$ : The graph is negatively skewed (skewed to the left). Most of the probability is concentrated on the higher values, with a tail extending towards lower values.

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Worked Example: Comparing distributions for different p values

Construct and compare the graph of the binomial probability distribution for 20 trials ( $n = 20$ ) with probability of success:

a) $p = 0.2$

b) $p = 0.5$

c) $p = 0.8$

Solution:

a) For $p = 0.2$ , the graph is positively skewed.

The distribution shows that mostly between 1 and 8 successes will be observed in 20 trials. The highest probabilities occur for lower numbers of successes, which makes sense because with only a 20% chance of success on each trial, we expect fewer total successes.

b) For $p = 0.5$ , the graph is symmetrical.

The probability of success is the same as the probability of failure, creating a balanced distribution. Mostly between 6 and 14 successes will be observed in 20 trials. The distribution peaks at 10 successes, which is exactly half of the 20 trials.

c) For $p = 0.8$ , the graph is negatively skewed.

With an 80% chance of success on each trial, mostly between 12 and 19 successes will be observed in 20 trials. The distribution is concentrated towards higher numbers of successes.

Key observation: Notice that the graphs for $p = 0.2$ and $p = 0.8$ are mirror images of each other. This symmetry makes sense because succeeding 80% of the time is the opposite of succeeding only 20% of the time.

Expectation and variance

Understanding expectation intuitively

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Consider this question: How many heads would you expect to obtain, on average, if a fair coin was tossed 10 times?

While the exact number of heads would vary from trial to trial (and could theoretically be anywhere from 0 to 10), intuition suggests that the long-run average would be 5 heads. This intuition is correct! For a binomial random variable $X$ with $n = 10$ and $p = 0.5$ :

E(X) = \sum_{x} x \cdot \Pr(X = x) = 5

Formulas for expectation and variance

For any binomial distribution, we can calculate the expected value and variance directly from the parameters $n$ and $p$ , without needing to work through the full probability distribution.

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If $X$ is the number of successes in $n$ trials, each with probability of success $p$ , then:

E(X) = np

\text{Var}(X) = np(1 - p)

These formulas provide a quick and efficient way to find the centre and spread of a binomial distribution.

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The term $(1 - p)$ represents the probability of failure, often denoted as $q$ . So the variance formula can also be written as $\text{Var}(X) = npq$ .

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Worked Example: Multiple-choice test

An examination consists of 30 multiple-choice questions, each question having three possible answers. A student guesses the answer to every question. Let $X$ be the number of correct answers.

a) How many will she expect to get correct? That is, find $E(X) = \mu$ .

b) Find $\text{Var}(X)$ .

Solution:

The number of correct answers, $X$ , is a binomial random variable with parameters $n = 30$ and $p = \frac{1}{3}$ (since there are three possible answers and only one is correct).

a) Using the formula for expectation:

\mu = E(X) = np

The student has an expected result of $\mu = 10$ correct answers.

Note: This is not enough to pass if the pass mark is 50% (which would require 15 correct answers).

b) Using the formula for variance:

\text{Var}(X) = np(1 - p)

\approx 6.67

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Worked Example: Influenza spread with standard deviation

The probability of contracting influenza this winter is known to be 0.2. Of the 100 employees at a certain business, how many would the owner expect to get influenza? Find the standard deviation of the number who will get influenza and calculate $\mu \pm 2\sigma$ . Interpret the interval $[\mu - 2\sigma, \mu + 2\sigma]$ for this example.

Solution:

The number of employees who get influenza is a binomial random variable, $X$ , with parameters $n = 100$ and $p = 0.2$ .

Finding the expectation:

\mu = E(X) = np

The owner will expect 20 of the employees to contract influenza.

Finding the variance and standard deviation:

\sigma^2 = \text{Var}(X) = np(1 - p)

Therefore, the standard deviation is:

\sigma = \sqrt{16} = 4

Calculating the interval:

\mu \pm 2\sigma = 20 \pm (2 \times 4)

This gives us the interval [12, 28].

Interpretation:

The owner of the business knows there is a probability of approximately 0.95 (or 95%) that between 12 and 28 of the employees will contract influenza this winter.

This interpretation uses the empirical rule (also called the 68-95-99.7 rule), which states that for many distributions, approximately 95% of values fall within two standard deviations of the mean.

Exam tips

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Essential Tips for Success:

Always identify the parameters $n$ and $p$ clearly before calculating expectation or variance.
Remember that $(1 - p)$ represents the probability of failure.
The expectation $E(X) = np$ gives the long-run average, not a guarantee of what will happen in any single set of trials.
The interval $\mu \pm 2\sigma$ captures approximately 95% of possible outcomes, making it useful for predicting reasonable ranges.
When the question asks for standard deviation, don't forget to take the square root of the variance!

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Key Points to Remember:

Graph shapes depend on $p$ : Positively skewed when $p < 0.5$ , symmetrical when $p = 0.5$ , and negatively skewed when $p > 0.5$ .
Expectation formula: $E(X) = np$ gives the expected number of successes in $n$ trials.
Variance formula: $\text{Var}(X) = np(1-p)$ measures how spread out the distribution is.
Standard deviation: Remember to take the square root of variance to find $\sigma = \sqrt{np(1-p)}$ .
The $\mu \pm 2\sigma$ interval: Contains approximately 95% of all possible outcomes, providing a useful range for predictions.

The Graph, Expectation, and Variance of a Binomial Distribution (VCE SSCE Mathematical Methods): Revision Notes