Skip to main content
SingaporeMathsSyllabus dot point

How do we compute probabilities for a normally distributed variable using standardisation?

Model continuous data with the normal distribution, standardise to the Z-distribution to find probabilities, and find values from given probabilities

A focused answer to the H2 Mathematics outcome on the normal distribution. The bell curve and its parameters, standardising to Z, finding probabilities and inverse problems, and combining normal variables.

Generated by Claude Opus 4.810 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 answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to model continuous data with the normal distribution, standardise to the standard normal ZZ-distribution to find probabilities, solve inverse problems (find a value given a probability), and combine independent normal variables.

The answer

The normal distribution

A continuous variable XN(μ,σ2)X \sim \mathrm{N}(\mu, \sigma^2) has a symmetric bell-shaped curve centred at the mean μ\mu with spread set by the standard deviation σ\sigma. Probabilities are areas under the curve, so P(X=a)=0\mathrm{P}(X = a) = 0 for any single point and only intervals carry probability.

Standardising to Z

Any normal variable converts to the standard normal ZN(0,1)Z \sim \mathrm{N}(0, 1) by

Z=Xμσ.Z = \frac{X - \mu}{\sigma}.

A zz-score measures how many standard deviations a value lies from the mean. Then P(X<a)=P(Z<aμσ)\mathrm{P}(X < a) = \mathrm{P}\left(Z < \dfrac{a - \mu}{\sigma}\right), read from the standard normal (or the graphing calculator).

Finding probabilities

Sketch the bell curve, shade the required region, and express it using the cumulative function. Use symmetry (P(Z<z)=P(Z>z)\mathrm{P}(Z < -z) = \mathrm{P}(Z > z)) and the complement for tails. The graphing calculator gives normal probabilities directly.

Inverse problems

To find a value aa such that P(X<a)=p\mathrm{P}(X < a) = p, find the zz-value with that cumulative probability, then unstandardise: a=μ+zσa = \mu + z\sigma.

Combining normal variables

If XN(μX,σX2)X \sim \mathrm{N}(\mu_X, \sigma_X^2) and YN(μY,σY2)Y \sim \mathrm{N}(\mu_Y, \sigma_Y^2) are independent, then aX+bYaX + bY is normal with mean aμX+bμYa\mu_X + b\mu_Y and variance a2σX2+b2σY2a^2\sigma_X^2 + b^2\sigma_Y^2 (variances add, with squared coefficients).

Examples in context

Example 1. Setting a pass mark. An examiner choosing a mark so that the top 15%15\% achieve a distinction solves an inverse normal problem, finding the zz for the 8585th percentile and unstandardising to the raw mark.

Example 2. Tolerance in manufacturing. A component dimension N(μ,σ2)\mathrm{N}(\mu, \sigma^2) is acceptable within a tolerance band; standardising the limits gives the proportion within specification, the everyday quality-control calculation.

Try this

Q1. For XN(100,152)X \sim \mathrm{N}(100, 15^2), find the zz-score of X=130X = 130. [1 mark]

  • Cue. z=13010015=2z = \dfrac{130 - 100}{15} = 2.

Q2. For ZN(0,1)Z \sim \mathrm{N}(0, 1), find P(Z>1)\mathrm{P}(Z > 1). [2 marks]

  • Cue. 10.8413=0.15871 - 0.8413 = 0.1587.

Q3. Independent variables XN(5,4)X \sim \mathrm{N}(5, 4) and YN(3,9)Y \sim \mathrm{N}(3, 9). State the distribution of X+YX + Y. [2 marks]

  • Cue. N(8,13)\mathrm{N}(8, 13) (means and variances add).

Exam-style practice questions

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

Original4 marksThe heights of a population are normally distributed with mean 170 cm170\ \text{cm} and standard deviation 8 cm8\ \text{cm}. Find the probability that a randomly chosen person is taller than 182 cm182\ \text{cm}.
Show worked answer →

Let XN(170,82)X \sim \mathrm{N}(170, 8^2). Standardise: Z=1821708=128=1.5Z = \dfrac{182 - 170}{8} = \dfrac{12}{8} = 1.5.

P(X>182)=P(Z>1.5)=1P(Z<1.5)=10.9332=0.0668\mathrm{P}(X > 182) = \mathrm{P}(Z > 1.5) = 1 - \mathrm{P}(Z < 1.5) = 1 - 0.9332 = 0.0668.

Markers reward standardising with the correct zz-score, reading or computing the tail probability, and the value 0.06680.0668.

Original5 marksA machine fills bottles with volume XN(500,42) mlX \sim \mathrm{N}(500, 4^2)\ \text{ml}. Find the volume vv such that 90%90\% of bottles contain more than vv.
Show worked answer →

We need P(X>v)=0.9\mathrm{P}(X > v) = 0.9, so P(X<v)=0.1\mathrm{P}(X < v) = 0.1.

The zz-value with P(Z<z)=0.1\mathrm{P}(Z < z) = 0.1 is z=1.2816z = -1.2816.

Unstandardise: v=μ+zσ=500+(1.2816)(4)=5005.13=494.9 mlv = \mu + z\sigma = 500 + (-1.2816)(4) = 500 - 5.13 = 494.9\ \text{ml}.

Markers reward translating the probability to the lower tail, finding the inverse zz-value, and unstandardising to the volume.

Related dot points