Mean and Variance of a Distribution Revision Notes for HSC SSCE Mathematics Advanced

Mean and Variance of a Distribution

Introduction: from discrete to continuous distributions

When working with continuous probability distributions, we use similar concepts to discrete distributions, but with an important difference in how we calculate them. Instead of adding up values using summation ( $\sum$ ), we integrate over the continuous interval using $\int$ .

This makes sense because continuous distributions have infinitely many possible values, so we can't simply add them up one by one. Integration allows us to find the total effect across all possible values in a smooth, continuous way.

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Interestingly, both the Greek letter sigma ( $\sum$ ) for summation and the integral symbol ( $\int$ , which is an early form of the letter S) represent the same basic concept: adding things together. In continuous distributions, we're essentially adding up infinitely many infinitely small contributions.

Understanding the mean of a continuous distribution

The mean (also called the expected value) tells us the average value we would expect from a continuous probability distribution. It's the balancing point of the distribution.

For a discrete distribution, we calculate the mean by multiplying each value by its probability and adding them all up:

$E(X) = \sum xp(x)$

For a continuous distribution with probability density function $f(x)$ on interval $[a, b]$ , we replace the sum with an integral:

$E(X) = \int_a^b xf(x) \, dx$

chatImportant

This formula works by:

Taking each value $x$ in the interval
Weighting it by its probability density $f(x)$
Integrating (summing continuously) across the entire interval

The result gives us the mean value $\mu$ , which represents the centre of the distribution.

Understanding variance and standard deviation

The variance measures how spread out the distribution is. It quantifies how much the values typically differ from the mean. A larger variance means the distribution is more spread out.

Two equivalent formulas for variance

Just as with discrete distributions, we have two ways to calculate variance for continuous distributions:

Method 1: Using squared deviations from the mean

This method directly measures the average squared distance from the mean:

$\text{Var}(X) = E((X - \mu)^2) = \int_a^b (x - \mu)^2 f(x)\,dx$

This formula:

Takes the deviation of each value from the mean: $(x - \mu)$
Squares it to make all deviations positive: $(x - \mu)^2$
Weights it by the probability density: $f(x)$
Integrates across the entire interval

Method 2: Using the expected value of the square

This alternative formula is usually easier for calculations:

$\text{Var}(X) = E(X^2) - \mu^2 = \int_a^b x^2 f(x)\,dx - \mu^2$

This formula:

First finds the expected value of $X^2$ by integrating $x^2 f(x)$
Then subtracts the square of the mean

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Both methods give exactly the same answer, but Method 2 typically involves simpler algebra because you don't need to expand $(x - \mu)^2$ and deal with the resulting terms. This is the recommended approach for most calculations!

Standard deviation

The standard deviation $\sigma$ is simply the square root of the variance:

$\sigma = \sqrt{\text{Var}(X)}$

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The standard deviation is particularly useful because it has the same units as the original variable. For example, if $X$ measures distance in metres, then $\sigma$ is also in metres, while variance would be in square metres.

Key formulas summary

Let $f(x)$ be a probability density function on interval $[a, b]$ .

chatImportant

Mean or expected value:

$E(X) = \int_a^b x f(x) \, dx$

Variance (Method 1 - using deviations):

$\text{Var}(X) = E((X - \mu)^2) = \int_a^b (x - \mu)^2 f(x) \, dx$

Variance (Method 2 - usually easier):

$\text{Var}(X) = E(X^2) - \mu^2 = \int_a^b x^2 f(x) \, dx - \mu^2$

Standard deviation:

$\sigma = \sqrt{\text{Var}(X)}$

Worked example 1: calculating mean and variance

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Worked Example: Calculating Mean and Variance for a Linear PDF

Let's find the mean and variance for the probability density function $f(x) = \frac{1}{18}x$ on the interval $[0, 6]$ .

Finding the mean:

We calculate the expected value by integrating $x \times f(x)$ :

$\mu = \int_0^6 x \times \frac{1}{18}x \, dx$

$= \frac{1}{18} \int_0^6 x^2 \, dx$

$= \frac{1}{18} \left[\frac{x^3}{3}\right]_0^6$

$= \frac{1}{54} [x^3]_0^6$

$= \frac{1}{54}(216 - 0)$

$= \frac{216}{54}$

$= 4$

So the mean is 4 metres.

Finding the variance - Method 1 (using squared deviations):

Using the formula with $(x - \mu)^2$ :

$\sigma^2 = \int_0^6 (x - 4)^2 \times \frac{1}{18}x \, dx$

$= \frac{1}{18} \int_0^6 (x^3 - 8x^2 + 16x) \, dx$

$= \frac{1}{18}\left[\frac{1}{4}x^4 - \frac{8}{3}x^3 + 8x^2\right]_0^6$

$= \frac{1}{18}(324 - 576 + 288)$

$= \frac{1}{18}(36)$

$= 2$

Finding the variance - Method 2 (using $E(X^2) - \mu^2$ ):

Using the simpler formula:

$\sigma^2 = \int_0^6 x^2 \times \frac{1}{18}x \, dx - 4^2$

$= \frac{1}{18} \int_0^6 x^3 \, dx - 16$

$= \frac{1}{18} \left[\frac{x^4}{4}\right]_0^6 - 16$

$= \frac{1}{72}[x^4]_0^6 - 16$

$= \frac{1}{72}(1296 - 0) - 16$

$= 18 - 16$

$= 2$

Notice how Method 2 involved much simpler calculations! Both methods give us a variance of 2.

Finding the standard deviation:

$\sigma = \sqrt{2} \text{ metres}$

The standard deviation has the same units as the original variable (metres).

Worked example 2: uniform distribution

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Worked Example: Uniform Distribution

Find the mean and standard deviation for the probability density function $y = \frac{1}{8}$ on the interval $[0, 8]$ .

This represents a uniform distribution where all values are equally likely.

Finding the mean:

$\mu = \int_0^8 \frac{1}{8}x \, dx$

$= \frac{1}{8} \int_0^8 x \, dx$

$= \frac{1}{8}\left[\frac{x^2}{2}\right]_0^8$

$= \frac{1}{16}[x^2]_0^8$

$= \frac{1}{16}(64 - 0)$

$= 4$

The mean is 4, which makes sense as it's exactly halfway through the interval $[0, 8]$ .

Finding the variance:

Using the easier formula $\text{Var}(X) = E(X^2) - \mu^2$ :

$\sigma^2 = \int_0^8 \frac{1}{8}x^2 \, dx - 4^2$

$= \frac{1}{8}\left[\frac{x^3}{3}\right]_0^8 - 16$

$= \frac{1}{24}[x^3]_0^8 - 16$

$= \frac{1}{24}(512 - 0) - 16$

$= \frac{512}{24} - 16$

$= \frac{64}{3} - 16$

$= \frac{64 - 48}{3}$

$= \frac{16}{3}$

Finding the standard deviation:

$\sigma = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$

Worked example 3: linear probability density function

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Worked Example: Linear Probability Density Function

Find the mean and standard deviation for $y = \frac{1}{50}x$ on the interval $[0, 10]$ .

Finding the mean:

$\mu = \int_0^{10} \frac{1}{50}x^2 \, dx$

$= \frac{1}{50}\left[\frac{x^3}{3}\right]_0^{10}$

$= \frac{1}{150}[x^3]_0^{10}$

$= \frac{1}{150}(1000 - 0)$

$= \frac{1000}{150}$

$= \frac{20}{3}$

$= 6\frac{2}{3}$

Finding the variance:

Using $\text{Var}(X) = E(X^2) - \mu^2$ :

$\sigma^2 = \int_0^{10} \frac{1}{50}x^3 \, dx - \left(\frac{20}{3}\right)^2$

$= \frac{1}{50}\left[\frac{x^4}{4}\right]_0^{10} - \frac{400}{9}$

$= \frac{1}{200}[x^4]_0^{10} - \frac{400}{9}$

$= \frac{1}{200}(10000 - 0) - \frac{400}{9}$

$= 50 - \frac{400}{9}$

$= \frac{450 - 400}{9}$

$= \frac{50}{9}$

Finding the standard deviation:

$\sigma = \sqrt{\frac{50}{9}} = \frac{\sqrt{50}}{3} = \frac{5\sqrt{2}}{3}$

bookmarkSummary

Key Points to Remember:

From sums to integrals: When moving from discrete to continuous distributions, replace summation ( $\sum$ ) with integration ( $\int$ ).
Two variance formulas: You can calculate variance using either $\int_a^b (x - \mu)^2 f(x)\,dx$ or $\int_a^b x^2 f(x)\,dx - \mu^2$ . The second formula is usually simpler to work with.
Mean formula: $E(X) = \int_a^b xf(x) \, dx$ gives you the expected value by weighting each value by its probability density.
Units matter: Standard deviation has the same units as the original variable, while variance is in squared units. This makes standard deviation more interpretable.
Check your work: Both variance methods should give the same answer, so if you have time in an exam, you can verify your answer by calculating it both ways.

Mean and Variance of a Distribution (HSC SSCE Mathematics Advanced): Revision Notes