Standardisation and z-scores | Year 12

01 / LEARN

The idea, step by step

Remove location and scale

Subtracting the mean makes zero represent the centre. Dividing by standard deviation measures the distance from that centre in standard-deviation units. A positive score is above the mean and a negative score is below it.

For a normal X this transformed variable has the standard normal distribution, with mean zero and variance one. Standardisation itself is a change of scale; it does not make a non-normal distribution become normal.

Z = \frac{X - μ}{σ}, X \sim N (μ, σ^{2}) \Rightarrow Z \sim N (0, 1)

Translate the entire event

Convert both bounds of an interval, keeping the inequality directions because the standard deviation is positive. A probability calculation must still use the normal area function after the z-score has been found.

To reverse the change of scale, use x=mu+z sigma. The z-score has no measurement unit. When comparing performances on different scales, interpret the reference population: a larger z-score means more standard deviations above that population mean, not necessarily greater absolute achievement.

A common trap

Dividing by the variance when standardising.

Instead: Divide by standard deviation, the square root of variance.

02 / WORKED EXAMPLES

See the thinking

Example 1

A score two SDs above the mean

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

$σ = 3$ , so $z = (86 - 80) / 3 = 2$ .
$P (X < 86) = P (Z < 2)$ .
Normal cumulative calculation gives 0.9772.

z=2; probability 0.9772.

Example 2

Reverse the scale

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

$x = μ + z σ$ .
$x = 52 - 1.5 (4) = 46$ .
The measurement is 1.5 SD below the mean.

46.

Check your understanding

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

Show answer

-1.5.

$σ = 8$ and $(88 - 100) / 8 = - 1.5$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF 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
$(47.5 - 40) / 5 = 1.5$ .
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
$z = (44 - 50) / 6 = - 1$ .
$P (Z < - 1) = 0.1587$ .
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 ↗

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