A Level Math Revision
Updated October 2026 · By Pete Bromfield
In S1, \(X \sim N(\mu, \sigma^2)\) models a continuous variable with a symmetric bell-shaped distribution. Standardise with \(Z = \frac{X - \mu}{\sigma}\) to find probabilities, work backwards from a probability to a value, and set up simultaneous equations to find an unknown \(\mu\) or \(\sigma\). The second number in \(N(\mu, \sigma^2)\) is the variance, so take its square root.
The curve is symmetric about the mean \(\mu\), so mean, median and mode are equal. About \(68\%\) of values lie within one standard deviation of the mean and about \(95\%\) within two. \(X \sim N(50, 16)\) has variance 16, so \(\sigma = 4\).
\(Z = \frac{X - \mu}{\sigma} \sim N(0, 1)\). For \(X \sim N(50, 4^2)\): \(P(X < 56) = P(Z < 1.5) = 0.9332\). For an upper tail use \(P(Z > z) = 1 - P(Z < z)\), and for a negative \(z\) use symmetry: \(P(Z < -z) = P(Z > z)\).
Find the \(z\)-value from the probability, then \(x = \mu + z\sigma\). For \(X \sim N(50, 4^2)\), the value exceeded by \(10\%\) of the distribution has \(z = 1.2816\), so \(x = 50 + 1.2816 \times 4 = 55.1\).
Each probability you are given becomes one equation \(\frac{x - \mu}{\sigma} = z\). One unknown needs one equation; both \(\mu\) and \(\sigma\) need two, solved simultaneously.
The diameters of bolts made by a machine are normally distributed with mean $12\text{ mm}$ and standard deviation $0.2\text{ mm}$. A bolt is rejected if its diameter is less than $11.7\text{ mm}$ or more than $12.3\text{ mm}$. Find the expected number of rejected bolts in a batch of $1000$.
$z = \pm\frac{0.3}{0.2} = \pm 1.5$ M1
$P(Z \gt 1.5) = 1 - 0.9332 = 0.0668$ A1
$1000 \times 2 \times 0.0668$ M1
$= 133.6$, so about $134$ bolts A1
The lengths of a species of fish are normally distributed with mean $\mu$ cm and standard deviation $\sigma$ cm. $10\%$ of the fish are longer than $80$ cm and $20\%$ are shorter than $60$ cm. Find the values of $\mu$ and $\sigma$.
$z = 1.2816$ for the upper $10\%$ B1
$z = -0.8416$ for the lower $20\%$ B1
$80 - \mu = 1.2816\sigma$ and $60 - \mu = -0.8416\sigma$ M1
subtracting: $20 = 2.1232\sigma$ M1
$\sigma = 9.42$ A1
$\mu = 67.9$ A1
The random variable $X \sim N(40, \sigma^2)$ and $P(X \gt 45) = 0.3$.
(a) Find the value of $\sigma$. (3)
(b) Hence find $P(X \lt 30)$. (2)
$P(Z \lt z) = 0.7 \Rightarrow z = 0.5244$ B1
$\frac{45 - 40}{\sigma} = 0.5244$ M1
$\sigma = 9.53$ A1
$P\left(Z \lt \frac{30 - 40}{9.53}\right) = P(Z \lt -1.049)$ M1
$= 0.147$ A1
The points lie on the curve of a normal distribution. Change m (the mean) and s (the standard deviation) until the curve passes through every point. What does changing m do to the curve? What does changing s do?
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Yes. A calculator is allowed in every IAL maths unit, S1 included, but write down the standardisation and the probability you are finding: method marks are given for that working.
Turn the probability into a z-value, then solve (x − μ) / σ = z for σ. If μ is unknown too, use two probabilities to get two equations and solve them simultaneously.
Yes. A calculator is allowed in every Edexcel IAL maths unit, S1 included. Method marks are for the working you write down, so show it.
Each IAL maths unit, S1 included, is one written paper of 1 hour 30 minutes worth 75 marks.
In our mapping of the 4 Pearson S1 papers from June 2024 to June 2025, the normal distribution was tested in 4 of them (6 questions in all). Any topic can be examined in any series.
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