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How do you use the normal distribution and standardising to find probabilities for continuous data?

Use the normal distribution: recognise its bell shape and symmetry, standardise values with the z-score, and find probabilities using the standard normal table.

A CCEA GCSE Further Mathematics answer on the normal distribution, covering its symmetric bell shape, standardising with the z-score, using the standard normal table, and finding probabilities for continuous data in the Statistics unit.

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  1. What this dot point is asking
  2. The shape of the distribution
  3. Standardising with the z-score
  4. Reading the table and using symmetry
  5. The empirical rule
  6. Why this matters

What this dot point is asking

The normal distribution is the bell-shaped model for continuous data, and CCEA GCSE Further Mathematics uses it to find probabilities. You must recognise its symmetric shape, standardise a value into a z-score so it can be looked up, use the standard normal table to find probabilities, and combine these for "greater than", "less than" and "between" questions. It is the continuous distribution of the Statistics unit and a major examined topic.

The shape of the distribution

The normal distribution describes many natural measurements, such as heights, masses and test scores. Its graph is a symmetric bell curve, highest at the mean and tailing off equally on both sides.

Because it is continuous, probability corresponds to an area under the curve between two values, not the height at a single point.

Standardising with the z-score

Tables exist only for the standard normal distribution, which has mean 00 and standard deviation 11. To use them, you convert any normal value into its z-score, which measures how many standard deviations it lies above or below the mean.

Reading the table and using symmetry

The standard normal table gives P(Z<z)P(Z < z), the area to the left of a z-value. From this single quantity you can find any probability using two facts: the complement for the area to the right, and the symmetry of the curve for negative z-values.

The empirical rule

A helpful property is that fixed proportions of a normal distribution lie within whole numbers of standard deviations: about 68%68\% within one standard deviation of the mean, about 95%95\% within two, and about 99.7%99.7\% within three. This gives a quick sense-check on table answers and an intuitive feel for how unusual a value is, which is exactly the kind of interpretation the exam rewards.

Why this matters

The normal distribution is the central continuous model in statistics and the climax of the Statistics unit's distribution work. Standardising reuses the idea of measuring distance in standard deviations from the measures of spread topic, and the distribution approximates the binomial for large nn, tying the discrete and continuous strands together. Confident standardising and careful use of the complement and symmetry are what these questions reward.

Exam-style practice questions

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

CCEA Unit 3 (style)4 marksThe heights of plants are normally distributed with mean 2020 cm and standard deviation 44 cm. Find the probability that a plant is taller than 2626 cm.
Show worked answer →

Standardise the value 2626: z=x−μσ=26−204=1.5z = \dfrac{x - \mu}{\sigma} = \dfrac{26 - 20}{4} = 1.5.

We want P(X>26)=P(Z>1.5)P(X > 26) = P(Z > 1.5). From the standard normal table, P(Z<1.5)=0.9332P(Z < 1.5) = 0.9332.

So P(Z>1.5)=1−0.9332=0.0668P(Z > 1.5) = 1 - 0.9332 = 0.0668.

Marks are for standardising, reading the table, and using the complement for the upper tail. Forgetting to subtract from 11 for "greater than" is the usual error.

CCEA Unit 3 (style)5 marksMarks in a test are normally distributed with mean 6060 and standard deviation 1212. Find the probability that a mark is between 4848 and 7272.
Show worked answer →

Standardise both bounds: z1=48−6012=−1z_1 = \dfrac{48 - 60}{12} = -1 and z2=72−6012=1z_2 = \dfrac{72 - 60}{12} = 1.

We want P(−1<Z<1)=P(Z<1)−P(Z<−1)P(-1 < Z < 1) = P(Z < 1) - P(Z < -1). From the table P(Z<1)=0.8413P(Z < 1) = 0.8413, and by symmetry P(Z<−1)=1−0.8413=0.1587P(Z < -1) = 1 - 0.8413 = 0.1587.

So P(−1<Z<1)=0.8413−0.1587=0.6826P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826.

Marks are for both z-scores, using symmetry for the negative value, and the subtraction. About 68%68\% of data lies within one standard deviation, which matches.

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