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Method · Probability Distribution

Using the normal distribution

For a continuous variable XN(μ,σ2)X\sim N(\mu,\sigma^{2}), convert to the standard normal by Z=XμσZ=\frac{X-\mu}{\sigma}, then read the probability from the standard normal table.

What the normal distribution is for

The normal distribution models a continuous quantity, something measured rather than counted, such as height, mass, time, or a test score, whose values cluster symmetrically around a mean, thinning out towards the extremes to give the familiar bell-shaped curve. We write XN(μ,σ2)X\sim N(\mu,\sigma^{2}), where μ\mu is the mean and σ\sigma is the standard deviation.

Because there is a different bell curve for every μ\mu and σ\sigma, we do not have a table for each one. Instead we convert any normal variable into the single standard normal variable ZZ, which has mean 00 and standard deviation 11, using the Z-score Z=XμσZ=\frac{X-\mu}{\sigma}.

One standard table then answers every question, letting us find the probability that XX lies above a value, below a value, or between two values.

When to reach for it

Reach for the normal distribution when the variable is a continuous measurement and the question states, or clearly implies, that it is normally distributed with a given mean and standard deviation. Signal words include 'normally distributed', 'mean', 'standard deviation', and a request for the probability that a measurement exceeds, falls below, or lies between certain values.

This is the opposite situation to the binomial distribution, which counts successes in a fixed number of trials. If you are counting whole 'successes', think binomial; if you are measuring a quantity that can take any value on a scale, think normal.

The tell-tale first move for a normal question is always to standardise, convert the given XX-value into a ZZ-score before touching the table.

The method, step by step

Standardising (Z-score)Must memorise
Z=XμσZ=\frac{X-\mu}{\sigma}
  1. 1

    Write down the parameters

    Note the mean μ\mu, the standard deviation σ\sigma, and the XX-value(s) in the question.

  2. 2

    Standardise

    Convert each XX-value to a Z-score with Z=XμσZ=\frac{X-\mu}{\sigma}.

  3. 3

    Sketch and shade

    Draw the standard normal curve and shade the region whose probability you want.

  4. 4

    Read the table

    Look up the table value for the relevant zz. The SPM table gives the upper-tail area P(Z>z)P(Z>z).

  5. 5

    Use symmetry or complement

    Adjust with the total area 11 and the curve's symmetry to get P(Z<z)P(Z<z) or a 'between' probability.

  6. 6

    State the probability

    Write the final probability, matching it back to the original XX description.

Worked example

Q1[4 marks]

A continuous variable XX is normally distributed with mean μ=50\mu=50 and standard deviation σ=8\sigma=8. (a) Find P(X>58)P(X>58).

(b) Find P(X<42)P(X<42). Use P(Z>1)=0.1587P(Z>1)=0.1587.

Show worked solution

(a) Standardise X=58X=58 using Z=XμσZ=\frac{X-\mu}{\sigma}.

Z=58508=88=1Z=\frac{58-50}{8}=\frac{8}{8}=1

So P(X>58)=P(Z>1)P(X>58)=P(Z>1). The table gives the upper-tail area directly:

P(X>58)=P(Z>1)=0.1587P(X>58)=P(Z>1)=0.1587

(b) Standardise X=42X=42.

Z=42508=88=1Z=\frac{42-50}{8}=\frac{-8}{8}=-1

So P(X<42)=P(Z<1)P(X<42)=P(Z<-1). By the symmetry of the curve, P(Z<1)=P(Z>1)P(Z<-1)=P(Z>1).

P(X<42)=P(Z<1)=P(Z>1)=0.1587P(X<42)=P(Z<-1)=P(Z>1)=0.1587

Both probabilities equal 0.15870.1587, a neat illustration that the curve is symmetric about the mean.

Common mistakes to avoid

  • Forgetting to standardise and trying to read the table with the raw XX-value.
  • Dividing by the variance σ2\sigma^{2} instead of the standard deviation σ\sigma when finding ZZ.
  • Sign slips when X<μX<\mu: the Z-score should be negative, and its size still matters.
  • Confusing 'area to the left' with 'area to the right', know exactly which tail your table gives.
  • For a 'between' probability, subtracting the wrong two areas instead of sketching first to see what to combine.

How one-to-one teaching helps

The step that costs the most marks is the shading: students standardise correctly, then take the wrong area from the table because they never drew the curve. In a one-to-one lesson our teachers insist on a quick sketch every time and talk you through which region matches P(X>k)P(X>k), P(X<k)P(X<k) or a 'between' probability, so symmetry and complements stop feeling like guesswork.

We also check that you divide by σ\sigma, not σ2\sigma^{2}. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach the normal distribution, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why do I need to standardise?

Every mean and standard deviation gives a different bell curve, so there cannot be a table for each one. Standardising with Z=XμσZ=\frac{X-\mu}{\sigma} turns any normal variable into the single standard normal variable ZZ, which the one standard table describes.

What does the standard normal table give?

In the SPM Add Math papers the table gives the upper-tail area P(Z>z)P(Z>z), the probability to the right of zz. To get P(Z<z)P(Z<z) use 1P(Z>z)1-P(Z>z), and use the symmetry P(Z<z)=P(Z>z)P(Z<-z)=P(Z>z) for negative values.

How do I find a 'between' probability like P(a<X<b)P(a<X<b)?

Standardise both ends to z1z_{1} and z2z_{2}, sketch the curve, and combine the tail areas. For example P(z1<Z<z2)=P(Z>z1)P(Z>z2)P(z_{1}<Z<z_{2})=P(Z>z_{1})-P(Z>z_{2}) when both are positive.

A sketch makes the right subtraction obvious.

How is the normal distribution different from the binomial?

The binomial counts successes in a fixed number of trials and is discrete; the normal describes a continuous measurement such as length or mass. If you are measuring on a scale rather than counting whole outcomes, use the normal distribution.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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