Mean and Percentiles for a Continuous Random Variable Revision Notes for VCE SSCE Mathematical Methods

Mean and Percentiles for a Continuous Random Variable

Introduction to measures of centre

When working with probability distributions, understanding the centre of the distribution is crucial. This tells us where the "middle" or "typical" values lie.

Two probability distributions can have identical shapes and spreads but be located at different positions on the number line. This difference in location is captured by measures of centre.

The most commonly used measure of centre for continuous random variables is the mean (also called the expected value). We'll also explore percentiles, which help us identify values that divide the distribution into specific proportions.

Mean (expected value)

What is the mean?

The mean of a continuous random variable represents the long-run average value we would expect to observe. Just as we can calculate a mean for discrete random variables, we can do the same for continuous random variables, but using integration instead of summation.

Definition

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For a continuous random variable $X$ with probability density function $f$ , the mean or expected value of $X$ is:

E(X) = \int_{-\infty}^{\infty} x f(x) \, dx

provided the integral exists.

The mean is denoted by the Greek letter $\mu$ (mu).

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If $f(x) = 0$ for all $x$ outside the interval $\[c, d]$ , then we can simplify this to:

E(X) = \int_{c}^{d} x f(x) \, dx

Interpretation

The expected value represents the average value of the random variable over a very long period of time. For instance, if we're measuring the daily demand for petrol at a service station, the mean tells us the average daily demand we'd observe over many years.

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Worked Example: Finding the Expected Value

Find the expected value of the random variable $X$ which has probability density function:

f(x) = \begin{cases} 0.5x & 0 \leq x \leq 2 \\ 0 & x < 0 \text{ or } x > 2 \end{cases}

Solution:

By definition:

E(X) = \int_{-\infty}^{\infty} x f(x) \, dx = \int_{0}^{2} x \times 0.5x \, dx

since $f(x) = 0$ elsewhere.

Using technology

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TI-Nspire:

Access the piecewise function template using the template key
Define the function $f$ with the appropriate pieces
Leave the domain for the last function piece blank (it will autofill as 'Else')
To get an exact answer, enter $\frac{1}{2}x$ instead of $0.5x$
Evaluate the definite integral $\int_{0}^{2} x \cdot f(x) \, dx$

Casio ClassPad:

Tap the piecewise template twice
Define the function $f$ as shown in the diagram
Find $E(X)$ by evaluating the definite integral
Note: Using the defined function gives the decimal answer only

Expected value of a function of X

Definition

Sometimes we need to find the expected value not of $X$ itself, but of some function applied to $X$ . If $g$ is a function and we want to find the expected value of $g(X)$ , we use:

E[g(X)] = \int_{-\infty}^{\infty} g(x)f(x) \, dx

provided the integral exists.

Important property

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Generally, the expected value of a function of $X$ is not equal to that function of the expected value of $X$ . That is:

E[g(X)] \neq g[E(X)]

However, there is one important exception: when $g$ is a linear function:

E(aX + b) = aE(X) + b

where $a$ and $b$ are constants.

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Worked Example: Expected Value of Functions of X

Let $X$ be a random variable with probability density function $f$ given by:

f(x) = \begin{cases} 0.5x & 0 \leq x \leq 2 \\ 0 & x < 0 \text{ or } x > 2 \end{cases}

Find: a) the expected value of $X^2$ and b) the expected value of $e^X$ .

Solution:

Part a:

E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx

Part b:

E(e^X) = \int_{-\infty}^{\infty} e^x f(x) \, dx

correct to three decimal places.

Percentiles

What are percentiles?

A percentile is a value that divides the distribution so that a specific percentage of values fall below it. For example, if you score at the 75th percentile on an exam, this means 75% of all students scored below you.

Definition

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The value $p$ of $X$ which is the solution to the equation:

\int_{-\infty}^{p} f(x) \, dx = q

is called a percentile of the distribution.

For example:

The 75th percentile is found by taking $q = 0.75$
The 90th percentile is found by taking $q = 0.90$
The 50th percentile is called the median

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Worked Example: Finding the 90th Percentile

The duration of telephone calls to the order department of a large company is a random variable, $X$ minutes, with probability density function:

f(x) = \begin{cases} \frac{1}{3}e^{-\frac{x}{3}} & x > 0 \\ 0 & x \leq 0 \end{cases}

Find the value of $a$ such that 90% of phone calls last less than $a$ minutes.

Solution:

We need to find $a$ such that 90% of values are below it. This means solving:

\int_{0}^{a} \frac{1}{3}e^{-\frac{x}{3}} \, dx = 0.9

Evaluating the integral:

\left[ -e^{-\frac{x}{3}} \right]_{0}^{a} = 0.9

-e^{-\frac{a}{3}} - (-e^0) = 0.9

1 - e^{-\frac{a}{3}} = 0.9

e^{-\frac{a}{3}} = 0.1

-\frac{a}{3} = \log_e 0.1

a = -3 \log_e 0.1

a = 3 \log_e 10

a = 6.908

correct to three decimal places.

Therefore, 90% of phone calls to this company last less than 6.908 minutes.

The median

What is the median?

The median is the 50th percentile - it's the middle value of the distribution. Half of all values fall below the median, and half fall above it.

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While the median is a useful concept, it is not required by the VCE Mathematical Methods Study Design. However, understanding it can deepen your comprehension of probability distributions.

Definition

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The median, $m$ , of a continuous random variable $X$ is the value such that:

\int_{-\infty}^{m} f(x) \, dx = 0.5

This means:

\text{Pr}(X \leq m) = \text{Pr}(X > m) = 0.5

Graphically, the median divides the area under the probability density function exactly in half.

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Worked Example: Finding the Median

Suppose the probability density function of weekly sales of topsoil, $X$ (in tonnes), is given by:

f(x) = \begin{cases} 2(1-x) & 0 \leq x \leq 1 \\ 0 & x < 0 \text{ or } x > 1 \end{cases}

Find the value of $m$ such that $\text{Pr}(X \leq m) = 0.5$ , and interpret.

Solution:

We need to solve:

\int_{0}^{m} 2(1-x) \, dx = 0.5

Evaluating the integral:

2 \left[ x - \frac{x^2}{2} \right]_{0}^{m} = 0.5

2 \left( m - \frac{m^2}{2} \right) = 0.5

2m - m^2 = 0.5

m^2 - 2m + 0.5 = 0

Using the quadratic formula:

m = 0.293 \text{ or } m = 1.707

Since $0 \leq x \leq 1$ , we must have $m = 0.293$ tonnes.

Interpretation: In the long run, 50% of weekly sales will be less than 0.293 tonnes, and 50% will be more than 0.293 tonnes.

Using technology for median calculations

infoNote

TI-Nspire:

Use the solve function with the definite integral equal to 0.5
Since $0 \leq m \leq 1$ , you can add the domain constraint $|0 \leq m \leq 1$

Casio ClassPad:

Define the function $f$ first
Use the solve function with the definite integral equal to 0.5 to find the median value

Remember!

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Key Points to Remember:

Mean (Expected Value): For a continuous random variable $X$ with pdf $f$ , the mean is $E(X) = \int_{-\infty}^{\infty} x f(x) \, dx$ . It represents the long-run average value.
Expected Value of Functions: $E[g(X)] = \int_{-\infty}^{\infty} g(x)f(x) \, dx$ . Generally, $E[g(X)] \neq g[E(X)]$ , except for linear functions where $E(aX + b) = aE(X) + b$ .
Percentiles: The value $p$ where $\int_{-\infty}^{p} f(x) \, dx = q$ gives the $(100q)$ th percentile. This tells you the value below which a certain percentage of the distribution falls.
Median: The 50th percentile, found by solving $\int_{-\infty}^{m} f(x) \, dx = 0.5$ . It divides the distribution exactly in half (not required by the study design but useful for understanding distributions).
Calculator skills: You can use your CAS calculator to evaluate the integrals needed for finding means and percentiles. Remember to define piecewise functions correctly and use the solve function for finding percentile values.

Mean and Percentiles for a Continuous Random Variable (VCE SSCE Mathematical Methods): Revision Notes