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

01 / LEARN

The idea, step by step

A probability-weighted balance point

The mean weights each possible value by its density and integrates across the support. A density that places more area near one end can have a mean away from the midpoint. The median instead splits the probability in half; it need not equal the mean.

Calculate the second moment separately. It describes the average of the squared values. To obtain variance, subtract the square of the mean. This subtraction is essential; a large mean alone should not make a distribution variable.

E (X) = \int x f (x) d x, Var (X) = \int x^{2} f (x) d x - {[E (X)]}^{2}

Spread and changes of units

Standard deviation is the positive square root of variance. If X is measured in seconds, its variance is in squared seconds, while its standard deviation is in seconds. Use unrounded moments when calculating the difference for variance.

Adding a constant shifts every value equally and does not change spread. Multiplying by a scales deviations by its absolute value, so variance scales by a squared. These transformation rules do not require normality.

E (a X + b) = a E (X) + b, Var (a X + b) = a^{2} Var (X)

A common trap

Treating the second moment as the variance.

Instead: Subtract the square of the mean before taking a square root for standard deviation.

02 / WORKED EXAMPLES

See the thinking

Example 1

Moments of a sloping density

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

$E (X) = \int_{0}^{1} 2 x^{2} d x = 2 / 3$ .
$E (X^{2}) = \int_{0}^{1} 2 x^{3} d x = 1 / 2$ .
$Var (X) = 1 / 2 - {(2 / 3)}^{2} = 1 / 18$ .

Mean

2 / 3

; variance

1 / 18

Example 2

Uniform measurement

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

The midpoint is $(2 + 8) / 2 = 5$ .
Uniform variance is ${(8 - 2)}^{2} / 12 = 3$ .
Standard deviation is $\sqrt{3}$ .

Mean 5; variance 3; standard deviation

\sqrt{3}

Check your understanding

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

Show answer

Mean 5; variance 36.

$2 (4) - 3 = 5$ and $2^{2} (9) = 36$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 squared) and variance.
3 marks
Find the mean and variance of a uniform variable on [0,6].
2 marks
For E(X)=2/3 and Var(X)=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 $1 / 2$ 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
$\int_{0}^{1} 3 x^{3} d x = 3 / 4$ .
$3 / 4$ .
For density $3 x^{2}$ on [0,1], find E(X squared) and variance.
3 marks
$\int_{0}^{1} 3 x^{4} d x = 3 / 5$ .
$3 / 5 - {(3 / 4)}^{2} = 3 / 80$ .
$3 / 5$ and $3 / 80$ .
Find the mean and variance of a uniform variable on [0,6].
2 marks
Use the midpoint and $6^{2} / 12 = 3$ .
Mean 3; variance 3.
For E(X)=2/3 and Var(X)=1/18, find the mean and variance of $5 - 2 X$ .
2 marks
$5 - 2 (2 / 3) = 11 / 3$ ; ${(- 2)}^{2} / 18 = 2 / 9$ .
$11 / 3$ and $2 / 9$ .
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 $1 / 2$ on [-1,1], find mean and variance.
3 marks
Symmetry gives mean zero.
$E (X^{2}) = \frac{1}{2} \int_{- 1}^{1} x^{2} d x = 1 / 3$ .
0 and $1 / 3$ .

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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 ↗

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