Skip to main content
SAMath MethodsSyllabus dot point

How do we find exact normal probabilities for any value using standardisation?

A z-score standardises a normal value to the standard normal distribution, enabling exact probabilities to be found by technology or tables.

How to standardise a normal value with a z-score, interpret z as standard deviations from the mean, and find exact normal probabilities and inverse-normal values.

Generated by Claude Opus 4.78 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The z-score
  3. Finding probabilities
  4. The inverse problem
  5. Common errors
  6. Why it matters

What this dot point is asking

The empirical rule only handles whole-number multiples of σ\sigma. For any value, you standardise it to a z-score and read the probability from the standard normal distribution N(0,1)N(0,1).

The z-score

Standardising maps XN(μ,σ2)X\sim N(\mu,\sigma^2) onto ZN(0,1)Z\sim N(0,1), which has mean 00 and standard deviation 11. Equal z-scores represent equally extreme positions in different distributions, so z-scores let you compare values from different normal distributions.

Finding probabilities

In Stage 2 you use a calculator's normal CDF (or the standard normal table) to find the area. The standardisation tells you which area to look up.

The inverse problem

Sometimes you know the probability and want the value - the inverse normal. Find the z-score for that area, then unstandardise with x=μ+zσx=\mu+z\sigma.

Common errors

Why it matters

Z-scores are the exact-probability engine of Topic 5 and a guaranteed exam skill, both forward (value to probability) and inverse (probability to value). Standardisation is also the mechanism behind Topic 6's confidence intervals, where critical z-values such as 1.961.96 come straight from inverse-normal reasoning.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 SACE Stage 21 marksLet X be a normally distributed random variable with a mean of 54 and a standard deviation of 6. Given that Pr(X > k) = 0.10, determine the value of k.
Show worked answer →

This is an inverse-normal (find-the-value) problem. Pr(X > k) = 0.10 means k lies in the upper tail, so Pr(X <= k) = 0.90.

The z-score with 0.90 of the area below it is z = 1.2816 (the 90th percentile of the standard normal).

Convert back to an X value using x = mean + z times sd:

k = 54 + 1.2816 times 6 = 54 + 7.69 = 61.7 (to three significant figures).

So k = 61.7. The single mark is for the correct value. Sense check: only 10% of values exceed k, so k must sit well above the mean of 54, which 61.7 does. Using the calculator's inverse-normal with area 0.90 (or 0.10 from the right) gives the same answer.

2023 SACE Stage 21 marksThe daily milk production of a cow is normally distributed with a mean of 21.9 litres and a standard deviation of 3.26 litres. Given that 15% of randomly selected cows have a daily milk production of c litres or more, determine the value of c.
Show worked answer →

"15% have c litres or more" means Pr(X >= c) = 0.15, so c is the upper-tail cut-off with Pr(X < c) = 0.85.

The z-score for the 85th percentile is z = 1.0364.

Convert to a milk-production value with c = mean + z times sd:

c = 21.9 + 1.0364 times 3.26 = 21.9 + 3.38 = 25.3 litres (to three significant figures).

So c = 25.3 litres. The single mark is for the correct value. Sense check: only the top 15% exceed c, so c should be above the mean of 21.9, which it is. Equivalently, use the calculator's inverse-normal with a lower area of 0.85.