Mean, variance and standard deviation for continuous models | Year 12

01 / LEARN

Understand the idea.

A probability-weighted balance point

For a continuous density, $E (X) = \int x f (x) d x$ averages values using probability weights across the support. The mean is a balance point, while the median splits area in half. Neither must be the midpoint of a nonuniform support.

Compute the second moment $E (X^{2}) = \int x^{2} f (x) d x$ separately. Then use $Var (X) = E (X^{2}) - {[E (X)]}^{2}$ . The second moment combines centre and spread, so it is generally larger than variance.

\begin{aligned} E (X) & = \int x f (x) d x \\ Var (X) & = \int x^{2} f (x) d x \\ - {[E (X)]}^{2} \end{aligned}

Spread and changes of units

Standard deviation is the nonnegative square root of variance. Its units match the observations; variance has squared units. Preserve full precision in both moments before subtracting, since early rounding can noticeably affect a small variance.

For $Y = a X + b$ , use $E (Y) = a E (X) + b$ , $Var (Y) = a^{2} Var (X)$ and $SD (Y) = ∣ a ∣ SD (X)$ . A shift moves the centre without changing spread. These transformation rules do not require the density to be normal.

\begin{aligned} E (a X + b) & = a E (X) + b \\ Var (a X + b) & = a^{2} Var (X) \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Moments of a sloping density

For $f (x) = 2 x$ on [0,1], find the mean and variance.

Worked solution

1
Integrate the weighted values.
$E (X) = \int_{0}^{1} x (2 x) d x = {[\frac{2 x^{3}}{3}]}_{0}^{1} = \frac{2}{3}$
The extra factor of the value is required when computing a mean.
2
Integrate the weighted squared values.
$E (X^{2}) = \int_{0}^{1} x^{2} (2 x) d x = {[\frac{x^{4}}{2}]}_{0}^{1} = \frac{1}{2}$
This integral finds the second moment, rather than the variance directly.
3
Subtract the squared mean.
$Var (X) = \frac{1}{2} - {(\frac{2}{3})}^{2} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18}$
Exact fractions preserve the small difference between the two moments.

Answer

Mean $2 / 3$ ; variance $1 / 18$ .

EXAMPLE 02

Uniform measurement

X is uniform on [2,8]. Find its mean, variance and standard deviation.

Worked solution

1
Find the mean from uniform symmetry.
$E (X) = \frac{2 + 8}{2} = 5$
A constant density on the interval balances at its midpoint.
2
Calculate the second moment using the density.
$\begin{aligned} E (X^{2}) & = \int_{2}^{8} \frac{x^{2}}{6} d x \\ = {[\frac{x^{3}}{18}]}_{2}^{8} = \frac{512 - 8}{18} = 28 \end{aligned}$
The support width is six, so its uniform density is one sixth.
3
Find the variance.
$Var (X) = 28 - 5^{2} = 3$
This also agrees with the uniform formula ${(8 - 2)}^{2} / 12$ .
4
Find the standard deviation.
$SD (X) = \sqrt{3}$
The positive square root reports spread on the original measurement scale.

Answer

Mean 5; variance 3; standard deviation $\sqrt{3}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

If E(X)=4 and Var(X)=9, find the mean and variance of $2 X - 3$ .

Show answer & reasoning

Mean 5; variance 36.

For $Y = 2 X - 3$ , the mean is $2 (4) - 3 = 5$ and the variance is $2^{2} (9) = 36$ . The shift affects only the mean.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

For density $3 x^{2}$ on $[0, 1]$ , find $E (X)$ .
2 marks
For density $3 x^{2}$ on $[0, 1]$ , find $E (X^{2})$ and variance.
3 marks
Find the mean and variance of a uniform variable on $[0, 6]$ .
2 marks
For $E (X) = \frac{2}{3}$ and $Var (X) = \frac{1}{18}$ , find the mean and variance of $5 - 2 X$ .
2 marks
A length has variance 0.04 square metres. Find its standard deviation.
2 marks
For density $\frac{1}{2}$ on $[- 1, 1]$ , find mean and variance.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For density $3 x^{2}$ on $[0, 1]$ , find $E (X)$ .
2 marks
$\int_{0}^{1} 3 x^{3} d x = \frac{3}{4}$ .
$\frac{3}{4}$ .
For density $3 x^{2}$ on $[0, 1]$ , find $E (X^{2})$ and variance.
3 marks
$\int_{0}^{1} 3 x^{4} d x = \frac{3}{5}$ .
$\frac{3}{5} - {(\frac{3}{4})}^{2} = \frac{3}{80}$ .
$\frac{3}{5}$ and $\frac{3}{80}$ .
Find the mean and variance of a uniform variable on $[0, 6]$ .
2 marks
Use the midpoint and $\frac{6^{2}}{12} = 3$ .
Mean 3; variance 3.
For $E (X) = \frac{2}{3}$ and $Var (X) = \frac{1}{18}$ , find the mean and variance of $5 - 2 X$ .
2 marks
$5 - 2 (\frac{2}{3}) = \frac{11}{3}$ ; $\frac{{(- 2)}^{2}}{18} = \frac{2}{9}$ .
$\frac{11}{3}$ and $\frac{2}{9}$ .
A length has variance 0.04 square metres. Find its standard deviation.
2 marks
Take the positive square root and use the original unit.
0.2 m.
For density $\frac{1}{2}$ on $[- 1, 1]$ , find mean and variance.
3 marks
Symmetry gives mean zero.
$E (X^{2}) = \frac{1}{2} \int_{- 1}^{1} x^{2} d x = \frac{1}{3}$ .
0 and $\frac{1}{3}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

A probability-weighted balance point

Spread and changes of units

See the method in action.

Moments of a sloping density

Integrate the weighted values.

Integrate the weighted squared values.

Subtract the squared mean.

Uniform measurement

Find the mean from uniform symmetry.

Calculate the second moment using the density.

Find the variance.

Find the standard deviation.

Try it yourself.

Check your understanding

Keep practising