Standardisation and z-scores | Year 12

01 / LEARN

Understand the idea.

Remove location and scale

The z-score $z = (x - μ) / σ$ measures a value's signed distance from the mean in standard deviations. Subtract the mean to centre the scale, then divide by standard deviation. Positive scores are above the mean; negative scores are below it.

When $X$ is normal, $Z = (X - μ) / σ$ follows $N (0, 1)$ . Standardisation preserves shape apart from scale and location; it does not turn a non-normal population into a normal one. A z-score has no measurement units.

\begin{array}{c} Z = \frac{X - μ}{σ} \\ X \sim N (μ, σ^{2}) ⟹ Z \sim N (0, 1) \end{array}

Translate the entire event

Translate every bound in a probability event using the same mean and standard deviation. Inequality directions stay unchanged because $σ > 0$ . A z-score is a boundary, not a probability; use the normal area function after standardising.

Reverse the transformation with $x = μ + z σ$ . To compare values from different populations, interpret each relative to its own mean and spread. A larger z-score shows a higher relative position, which need not mean a larger original measurement.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A score two SDs above the mean

$X \sim N (80, 9)$ . Standardise X=86 and find P(X<86).

Worked solution

1
Identify the standard deviation.
$σ = \sqrt{9} = 3$
The model supplies variance as its second parameter.
2
Standardise the value and the event.
$\begin{aligned} z & = \frac{86 - 80}{3} = 2 \\ P (X < 86) & = P (Z < 2) \end{aligned}$
Subtracting the centre and dividing by a positive spread gives an equivalent event.
3
Evaluate the lower-tail normal area.
$P (Z < 2) = Φ (2) \approx 0.9772$
The function $Φ$ gives cumulative probability for a standard normal variable.

Answer

z=2; probability 0.9772.

EXAMPLE 02

Reverse the scale

A normal measurement has mean 52 and SD 4. Which measurement has z=-1.5?

Worked solution

1
Start with the standardisation equation.
$- 1.5 = \frac{x - 52}{4}$
The target lies one and a half standard deviations below the mean.
2
Undo the division and the shift.
$\begin{aligned} x - 52 & = (- 1.5) (4) = - 6 \\ x & = 52 - 6 = 46 \end{aligned}$
Multiplying by the standard deviation restores the measurement distance.
3
Check the recovered value.
$\frac{46 - 52}{4} = - 1.5$
Substitution verifies the original z-score and its negative sign.

Answer

46.

03 / YOUR TURN

Try it yourself.

Check your understanding

$X \sim N (100, 64)$ . Find the z-score for 88.

Show answer & reasoning

-1.5.

Here $σ = \sqrt{64} = 8$ , so $z = (88 - 100) / 8 = - 12 / 8 = - 1.5$ . The negative sign places $88$ below the mean.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

$X \sim N (40, 25)$ . Standardise $X = 47.5$ .
2 marks
$X \sim N (20, 4)$ . Write $P (18 < X < 23)$ using $Z$ .
2 marks
A normal variable has mean 70 and SD 12. Find the value with $z = 0.75$ .
2 marks
Two scores have z-values -0.4 and 1.2. Which is further above its own mean?
2 marks
Does standardising a strongly skewed variable guarantee a normal shape?
2 marks
$X \sim N (50, 36)$ . Find $P (X < 44)$ to four decimal places.
3 marks

Read the worked solutions

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

$X \sim N (40, 25)$ . Standardise $X = 47.5$ .
2 marks
$\frac{47.5 - 40}{5} = 1.5$ .
$z = 1.5$ .
$X \sim N (20, 4)$ . Write $P (18 < X < 23)$ using $Z$ .
2 marks
Standardise each bound using SD 2.
$P (- 1 < Z < 1.5)$ .
A normal variable has mean 70 and SD 12. Find the value with $z = 0.75$ .
2 marks
$70 + 0.75 (12) = 79$ .
79.
Two scores have z-values -0.4 and 1.2. Which is further above its own mean?
2 marks
It is 1.2 SD above its population mean.
The score with $z = 1.2$ .
Does standardising a strongly skewed variable guarantee a normal shape?
2 marks
Subtracting and scaling do not remove skewness.
No.
$X \sim N (50, 36)$ . Find $P (X < 44)$ to four decimal places.
3 marks
$z = \frac{44 - 50}{6} = - 1$ .
$P (Z < - 1) = 0.1587$ .
0.1587.

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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.

Remove location and scale

Translate the entire event

See the method in action.

A score two SDs above the mean

Identify the standard deviation.

Standardise the value and the event.

Evaluate the lower-tail normal area.

Reverse the scale

Start with the standardisation equation.

Undo the division and the shift.

Check the recovered value.

Try it yourself.

Check your understanding

Keep practising