For X N(50, 5^2), find the z-value of X = 60. [2 marks]

Given (1) = 0.8413, find P(Z > 1). [1 mark]

CCEA CCEA 2019-style practice: The lengths of bolts are normally distributed with mean 50\,mm and standard deviation 2\,mm. Find the length exceeded by only 5\% of bolts.

We need the value x with P(X > x) = 0.05, so P(X < x) = 0.95. The standard normal value with P(Z < z) = 0.95 is z = 1.645 (inverse normal). Unstandardise: x = μ + zσ = 50 + 1.645 × 2 = 50 + 3.29 = 53.29\,mm. Markers reward converting to P(X < x) = 0.95, the critical z = 1.645, and unstandardising to find x = 53.3\,mm.

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How does the normal distribution model continuous data, and how do we calculate probabilities from it?

The normal distribution and its parameters, standardising to the standard normal variable Z, finding probabilities and values from the distribution, and the normal approximation to the binomial distribution.

A CCEA A-Level Mathematics answer on the normal distribution and its mean and standard deviation, standardising to the variable Z, finding probabilities and inverse-normal values, and using the normal distribution as an approximation to the binomial distribution.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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What this dot point is asking

CCEA wants you to understand the normal distribution and its parameters, standardise a normal variable to the standard normal $Z$ , find probabilities and values (forwards and inverse), and use the normal distribution to approximate the binomial. This is the principal continuous distribution of A2 statistics and underlies hypothesis testing.

The answer

The normal distribution

Standardising to Z

Finding probabilities and inverse values

To find $P(X > a)$ , standardise $a$ and use $P(Z > z) = 1 - \Phi(z)$ , exploiting symmetry for negative $z$ . To find a value from a given probability (the inverse problem), find the $z$ with the required $\Phi(z)$ , then unstandardise: $x = \mu + z\sigma$ . A quick sketch of the bell curve, shaded for the probability you want, prevents the common error of taking the wrong tail or forgetting to subtract from one.

The normal approximation to the binomial

Worked example: a between-probability

Find a probability between two values

The heights of a group are $X \sim N(170, 10^2)$ cm. Find $P(160 < X < 185)$ .

step Standardise both ends

z_1 = \frac{160 - 170}{10} = -1.0, \qquad z_2 = \frac{185 - 170}{10} = 1.5.

step Express as a difference of cumulative probabilities

P(160 < X < 185) = P(-1.0 < Z < 1.5) = \Phi(1.5) - \Phi(-1.0).

step Evaluate

Using $\Phi(1.5) = 0.9332$ and $\Phi(-1.0) = 1 - \Phi(1.0) = 1 - 0.8413 = 0.1587$ :

0.9332 - 0.1587 = 0.7745.

So about

77\%

of the group lie between those heights. The difference of two cumulative values gives the area of the strip between them.

Examples in context

Example 1. Quality limits. A factory setting tolerance limits for a product uses the normal distribution to find what fraction fall outside specification, and to set the mean so that a target proportion passes. The inverse-normal calculation chooses a machine setting to meet a quality target.

Example 2. Standardised test scores. Exam results scaled to a normal distribution let a grade boundary be set so that a fixed top percentage achieves it, found by the inverse-normal $z = 1.645$ for the top $5\%$ . Standardising compares scores from differently scaled tests on one footing.

Try this

Q1. For $X \sim N(50, 5^2)$ , find the $z$ -value of $X = 60$ . [2 marks]

Cue. $z = \frac{60 - 50}{5} = 2$ .

Q2. Given $\Phi(1) = 0.8413$ , find $P(Z > 1)$ . [1 mark]

Cue. $1 - 0.8413 = 0.1587$ .

Q3. State the mean and variance of the normal approximation to $B(100, 0.4)$ . [2 marks]

Cue. Mean $np = 40$ , variance $np(1-p) = 24$ .

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 20216 marksThe masses of apples are normally distributed with mean

160\,\text{g}

and standard deviation

20\,\text{g}

. Find the probability that a randomly chosen apple has mass greater than

190\,\text{g}

Show worked answer →

Let $X \sim N(160, 20^2)$ . Standardise $X = 190$ :

$Z = \frac{X - \mu}{\sigma} = \frac{190 - 160}{20} = \frac{30}{20} = 1.5.$

So $P(X > 190) = P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668$ .

Markers reward stating the distribution, the standardisation to $Z = 1.5$ , using the tables or calculator for $P(Z < 1.5)$ , and the final probability.

CCEA 20196 marksThe lengths of bolts are normally distributed with mean

50\,\text{mm}

and standard deviation

2\,\text{mm}

. Find the length exceeded by only

5\%

of bolts.

Show worked answer →

We need the value $x$ with $P(X > x) = 0.05$ , so $P(X < x) = 0.95$ .

The standard normal value with $P(Z < z) = 0.95$ is $z = 1.645$ (inverse normal).

Unstandardise: $x = \mu + z\sigma = 50 + 1.645 \times 2 = 50 + 3.29 = 53.29\,\text{mm}$ .

Markers reward converting to $P(X < x) = 0.95$ , the critical $z = 1.645$ , and unstandardising to find $x = 53.3\,\text{mm}$ .

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)