What is the normal distribution in the context of Edexcel International A Levels Statistics 1?

The normal distribution is a continuous probability distribution that is symmetrical and bell-shaped, representing the distribution of many types of data. In the Edexcel International A Levels Statistics 1 curriculum, it is used to model real-world phenomena where data tends to cluster around a mean. It is characterized by its mean (μ) and standard deviation (σ).

How do you calculate probabilities using the normal distribution?

To calculate probabilities using the normal distribution, you typically use the standard normal distribution, which has a mean of 0 and a standard deviation of 1. You convert a normal distribution to a standard normal distribution using the z-score formula: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. You then use z-tables or statistical software to find the probability associated with the z-score.

What is the significance of the mean and standard deviation in a normal distribution?

In a normal distribution, the mean (μ) determines the center of the distribution, while the standard deviation (σ) measures the spread or dispersion of the data around the mean. A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation suggests more spread out data. These parameters are crucial for understanding the distribution's shape and for calculating probabilities.

How is the normal distribution used in hypothesis testing in Statistics 1?

In Statistics 1, the normal distribution is used in hypothesis testing to determine if there is enough evidence to reject a null hypothesis. By assuming that the test statistic follows a normal distribution under the null hypothesis, you can calculate the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from your sample data. This probability, known as the p-value, helps in deciding whether to reject the null hypothesis.

What are some real-world examples of phenomena that follow a normal distribution?

Many real-world phenomena follow a normal distribution, making it a fundamental concept in Statistics 1. Examples include heights of people, test scores, measurement errors, and IQ scores. These examples typically exhibit the characteristic bell-shaped curve of the normal distribution, where most observations cluster around the mean, with fewer observations appearing as you move away from the mean.

Why is "Using $\sigma^{2}$ where $\sigma$ is needed" a common mistake in The normal distribution?

Why it happens: Confusion between the two parameters of N(, ^2). How to avoid it: Always take the square root before dividing: X N(50, 64) has = 8, not = 64. Write 'so = 8' as the first line of working. Source: WST01/01 examiner reports — annual

Why is "Looking up $\Phi(-1.4)$ directly in the table" a common mistake in The normal distribution?

Why it happens: IAL tables only give z 0. How to avoid it: Use symmetry: (-z) = 1 - (z). So (-1.4) = 1 - 0.9192 = 0.0808. Sketching the curve confirms which tail you want.

Why is "Confusing $P(X > a)$ with $P(X < a)$" a common mistake in The normal distribution?

Why it happens: Tables give (z) = P(Z < z); students forget the direction. How to avoid it: Sketch the curve. Shade the region you want. If shading is right of z, probability is 1 - (z). If left, it's (z).

Why is "Forgetting the negative sign on a left-tail inverse" a common mistake in The normal distribution?

Why it happens: Inverse-normal tables typically give positive z for probabilities > 0.5. How to avoid it: If P(X < x) < 0.5, the corresponding z is NEGATIVE. Example: P(Z < z) = 0.05 z = -1.645, not +1.645. Source: WST01/01 examiner reports — recurring

Why is "Sign slips in the simultaneous-equation step for finding $\mu$ and $\sigma$" a common mistake in The normal distribution?

Why it happens: Two equations with two unknowns and at least one negative z. How to avoid it: Write each equation as x_i - = z_i . Subtract one from the other so cancels — solving for first is usually cleanest. Then substitute back for .

Why is "Giving the probability without an interpretive sentence" a common mistake in The normal distribution?

Why it happens: Focus on the number. How to avoid it: Cite the variable in the answer: 'About 12.1\% of women are taller than 172 cm'. The contextual sentence often carries the final A-mark.

Edexcel International A Levels Mathematics (XMA01-YMA01)

The normal distribution

Q: Why is "Confusing $P(X > a)$ with $P(X < a)$" a common mistake in The normal distribution?

Why it happens: Tables give (z) = P(Z < z); students forget the direction. How to avoid it: Sketch the curve. Shade the region you want. If shading is right of z, probability is 1 - (z). If left, it's (z).

Q: Why is "Forgetting the negative sign on a left-tail inverse" a common mistake in The normal distribution?

Why it happens: Inverse-normal tables typically give positive z for probabilities > 0.5. How to avoid it: If P(X < x) < 0.5, the corresponding z is NEGATIVE. Example: P(Z < z) = 0.05 z = -1.645, not +1.645. Source: WST01/01 examiner reports — recurring

Q: Why is "Sign slips in the simultaneous-equation step for finding $\mu$ and $\sigma$" a common mistake in The normal distribution?

Why it happens: Two equations with two unknowns and at least one negative z. How to avoid it: Write each equation as x_i - = z_i . Subtract one from the other so cancels — solving for first is usually cleanest. Then substitute back for .

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Short Notes - The normal distribution

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Detailed Study Notes

Detailed notes on Statistics 1 for Edexcel International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

The Normal Distribution Study Notes — Edexcel IAL Mathematics S1 (WST01/01, 2018 spec — 2026 onwards)

Standardisation $Z = (X - \mu)/\sigma$ , lookups of $\Phi(z)$ in the IAL formula booklet (or via calculator), symmetry for negative $z$ , inverse-normal for percentiles, and simultaneous equations when both $\mu$ and $\sigma$ are unknown. The capstone of S1 — and the foundation of S2/S3.

At a glance

Notation: $X \sim N(\mu, \sigma^{2})$ — note variance, not SD, in the brackets.
Standardisation: $Z = (X - \mu)/\sigma$ , $Z \sim N(0, 1)$ .
$\Phi(z)$ : standard normal CDF; tables in the booklet for $z \ge 0$ .
Symmetry: $\Phi(-z) = 1 - \Phi(z)$ .
Forward: $P(X > a) = 1 - \Phi((a-\mu)/\sigma)$ .
Inverse: given $p$ , find $z = \Phi^{-1}(p)$ , then $x = \mu + z\sigma$ .
Unknowns: two probabilities give two linear equations in $\mu$ and $\sigma$ — solve simultaneously.
Sanity check: $68$ / $95$ / $99.7$ rule for $\pm 1, \pm 2, \pm 3$ SDs.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

S1 7.1 — Understand the normal distribution and its parameters $\mu, \sigma^{2}$ .
S1 7.2 — Standardise to the standard normal $Z$ and use the tables for $\Phi(z)$ .
S1 7.3 — Calculate forward probabilities $P(X > a)$ , $P(a < X < b)$ and inverse probabilities.
S1 7.4 — Find unknown $\mu$ or $\sigma$ given probabilities; solve simultaneous equations when both are unknown.

Shape and notation

Bell-shaped, symmetric about $\mu$ , spread controlled by $\sigma$ .

The normal density. $X \sim N(\mu, \sigma^{2})$ has the bell-shaped curve

$f(x) = \dfrac{1}{\sigma \sqrt{2\pi}} \exp\left(-\dfrac{(x - \mu)^{2}}{2\sigma^{2}}\right).$

The S1 specification does NOT require you to integrate this PDF — use tables / calculator for probabilities instead.

Key properties.

Symmetric about $\mu$ .
Mean $=$ median $=$ mode $= \mu$ .
Inflexion points at $\mu \pm \sigma$ .
Total area $= 1$ .

Notation gotcha. The second argument is the variance $\sigma^{2}$ , not the standard deviation. So $N(50, 64)$ has $\sigma^{2} = 64$ and $\sigma = 8$ .

Sketching. Always sketch the curve, marking $\mu$ at the centre and the region of interest. The sketch helps you avoid tail-direction errors.

Normal N(μ, σ²) is symmetric about μ with 68/95/99.7% of area inside μ ± 1σ, 2σ, 3σ.

Bell-shaped, symmetric about $\mu$ .
Notation $N(\mu, \sigma^{2})$ — second slot is VARIANCE.
Mean = median = mode = $\mu$ .
Inflexion points at $\mu \pm \sigma$ .

Standardisation — converting to $Z$

$Z = (X - \mu)/\sigma$ converts any normal to $N(0, 1)$ for table lookup.

Why. Statistical tables only tabulate the standard normal $Z \sim N(0, 1)$ . Any normal $X \sim N(\mu, \sigma^{2})$ is converted by

$Z = \dfrac{X - \mu}{\sigma}.$

The new variable $Z$ measures how many SDs $X$ is from $\mu$ .

Probability translation.

$P(X \le a) = P\left(Z \le \dfrac{a - \mu}{\sigma}\right) = \Phi\left(\dfrac{a - \mu}{\sigma}\right).$

$\Phi$ from the booklet table. For $z \ge 0$ , look up $\Phi(z) = P(Z \le z)$ directly.

Negative $z$ via symmetry.

$\Phi(-z) = 1 - \Phi(z).$

Worked example. $X \sim N(50, 64)$ , find $P(X > 62)$ .

$\sigma = 8$ , $z = (62 - 50)/8 = 1.5$ .
$P(X > 62) = P(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$ .

Modern calculators. A Casio FX-991 or equivalent has direct normal CDF: NormCD $(\text{lower}, \text{upper}, \sigma, \mu)$ . Show the standardisation step anyway — mark schemes give M-marks for explicit working.

$Z = (X - \mu)/\sigma$ .
$\Phi(z)$ is the standard normal CDF — tabulated for $z \ge 0$ .
Symmetry: $\Phi(-z) = 1 - \Phi(z)$ .
Always show the standardisation step, even if using calculator.

Forward probabilities $P(X > a)$ , $P(a < X < b)$

Standardise, look up $\Phi$ , manipulate with symmetry and complements.

Right tail.

$P(X > a) = 1 - \Phi\left(\dfrac{a - \mu}{\sigma}\right).$

Left tail.

$P(X < a) = \Phi\left(\dfrac{a - \mu}{\sigma}\right).$

Interval.

$P(a < X < b) = \Phi\left(\dfrac{b - \mu}{\sigma}\right) - \Phi\left(\dfrac{a - \mu}{\sigma}\right).$

Strict vs non-strict. For a continuous distribution, $P(X < a) = P(X \le a)$ — the boundary has probability zero. So ' $<$ ' and ' $\le$ ' are interchangeable.

Worked example. $X \sim N(180, 25)$ , find $P(173 < X < 190)$ .

$\sigma = 5$ . $z_1 = (173 - 180)/5 = -1.4$ , $z_2 = (190 - 180)/5 = 2$ .
$\Phi(2) = 0.9772$ , $\Phi(-1.4) = 1 - \Phi(1.4) = 0.0808$ .
$P(173 < X < 190) = 0.9772 - 0.0808 = 0.8964$ .

Sketch. Draw the standard normal curve, mark $z_1$ and $z_2$ , shade the region between them. The picture confirms the subtraction direction.

$P(X > a) = 1 - \Phi(z)$ where $z = (a - \mu)/\sigma$ .
$P(X < a) = \Phi(z)$ .
$P(a < X < b) = \Phi(z_2) - \Phi(z_1)$ .
Continuous: ' $<$ ' and ' $\le$ ' give the same probability.

Inverse probabilities — finding $x$ from $p$

Inverse-normal $z = \Phi^{-1}(p)$ , then $x = \mu + z\sigma$ .

Setup. Given $P(X < x) = p$ , find $x$ .

Method.

Find $z$ such that $\Phi(z) = p$ (inverse normal tables or calculator).
Back-transform: $x = \mu + z\sigma$ .

Important sign. If $p < 0.5$ then $z < 0$ . The inverse normal table (in the IAL booklet) gives positive $z$ for $p > 0.5$ ; for $p < 0.5$ , look up $1 - p$ and negate the $z$ :

$\Phi^{-1}(0.05) = -\Phi^{-1}(0.95) = -1.6449.$

Worked example. $X \sim N(75, 100)$ , $P(X < x) = 0.9$ .

$z = \Phi^{-1}(0.9) = 1.2816$ (or $1.282$ , both accepted).
$x = 75 + 1.2816 \times 10 = 87.82$ .

Common percentiles.

Probability	$z$
$0.90$	$+1.2816$
$0.95$	$+1.6449$
$0.975$	$+1.96$
$0.99$	$+2.3263$
$0.05$	$-1.6449$
$0.025$	$-1.96$

These are worth memorising for confidence-interval work in S2.

Inverse normal: $z = \Phi^{-1}(p)$ .
Back-transform: $x = \mu + z\sigma$ .
$p < 0.5$ $\Rightarrow$ $z < 0$ .
Memorise common $z$ : $1.282$ ( $p = 0.9$ ), $1.645$ ( $p = 0.95$ ), $1.96$ ( $p = 0.975$ ).

Finding unknown $\mu$ and $\sigma$

Two probabilities $\to$ two equations $x_i = \mu + z_i \sigma$ $\to$ solve.

Setup. Given two probability statements, e.g.

$P(X < a_1) = p_1, \quad P(X > a_2) = p_2,$

find $\mu$ and $\sigma$ .

Method.

Convert each probability to a $z$ -value via the inverse normal.
Write each as a linear equation: $a_i = \mu + z_i \sigma$ .
Subtract to eliminate $\mu$ and find $\sigma$ .
Substitute back to find $\mu$ .

Worked example. $P(X < 50) = 0.2$ , $P(X > 80) = 0.1$ .

$\Phi^{-1}(0.2) = -0.8416$ , so $50 = \mu - 0.8416\sigma$ … (i)
$P(X > 80) = 0.1 \Rightarrow P(X < 80) = 0.9 \Rightarrow \Phi^{-1}(0.9) = 1.2816$ , so $80 = \mu + 1.2816\sigma$ … (ii)
Subtract (i) from (ii): $30 = 2.1232 \sigma$ , so $\sigma \approx 14.13$ .
From (i): $\mu = 50 + 0.8416 \times 14.13 \approx 61.9$ .

Single-unknown case. Same method, but only one equation needed. E.g. given $P(X < a) = p$ and known $\sigma$ (or $\mu$ ), one linear equation solves directly.

Two probabilities $\to$ two linear equations.
Subtract to eliminate $\mu$ , solve for $\sigma$ .
Substitute back to find $\mu$ .
Watch signs when one of the $z$ -values is negative.

The $68$ - $95$ - $99.7$ rule

Quick sanity check: $\mu \pm 1, 2, 3 \sigma$ holds $\approx 68$ , $95$ , $99.7\%$ .

Statement. For $X \sim N(\mu, \sigma^{2})$ :

$P(\mu - \sigma < X < \mu + \sigma) \approx 0.682$ .
$P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.954$ .
$P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 0.997$ .

Uses.

Sanity check. If you compute $P(\mu < X < \mu + 2\sigma) = 0.7$ , suspect an error — the correct value is about $0.477$ .
Range of typical values. Almost all data ( $99.7\%$ ) lies within $\mu \pm 3\sigma$ . Useful for plausibility comments.
Modelling refinement. If the normal model predicts impossible values (e.g. $\mu - 3\sigma < 0$ for a non-negative variable), the model is suspect.

Exact tail values.

Interval	Exact probability
$\mu \pm 1\sigma$	$0.6827$
$\mu \pm 2\sigma$	$0.9545$
$\mu \pm 3\sigma$	$0.9973$

$\mu \pm 1\sigma$ : $\approx 68\%$ .
$\mu \pm 2\sigma$ : $\approx 95\%$ .
$\mu \pm 3\sigma$ : $\approx 99.7\%$ .
Use as sanity check, not exam-grade precision.

Quick recap

$X \sim N(\mu, \sigma^{2})$ — second slot is variance.
Standardise: $Z = (X - \mu)/\sigma$ .
Forward: $P(X > a) = 1 - \Phi(z)$ .
Symmetry: $\Phi(-z) = 1 - \Phi(z)$ .
Interval: $\Phi(z_2) - \Phi(z_1)$ .
Inverse: $z = \Phi^{-1}(p)$ , then $x = \mu + z\sigma$ .
Two probabilities $\to$ solve for $\mu$ and $\sigma$ simultaneously.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Standardisation: $Z = (X - \mu)/\sigma$ .
$\Phi$ symmetry: $\Phi(-z) = 1 - \Phi(z)$ .
Right tail: $P(X > a) = 1 - \Phi((a - \mu)/\sigma)$ .
Inverse: $x = \mu + z\sigma$ .
Common $z$ : $1.282$ ( $0.9$ ), $1.645$ ( $0.95$ ), $1.96$ ( $0.975$ ), $2.326$ ( $0.99$ ).
$68$ - $95$ - $99.7$ rule.

How it’s examined

The normal distribution is the headline topic of WST01/01 — typically the longest question on the paper, worth $10$ - $14$ marks across $4$ - $6$ sub-parts. Standard sub-parts: forward probability $P(X > a)$ or $P(a < X < b)$ ( $4$ - $5$ marks); inverse percentile ( $4$ - $5$ marks); find unknown $\mu$ or $\sigma$ ( $3$ - $4$ marks); find both via simultaneous equations ( $6$ - $7$ marks); contextual application with a comment on the model's plausibility ( $3$ - $5$ marks). Mark schemes require explicit standardisation in working; relying entirely on calculator output without showing the $z$ -value loses M-marks. A clean sketch of the curve with shaded region routinely earns a B-mark.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — The normal distribution

Step-by-step solutions to past-paper-style questions on the normal distribution, written exactly the way a tutor would explain them at the board.

1Forward probability $P(X > a)$ (4 marks, S1)
Core• Adapted from WST01/01 January 2024 Q10• normal, standardisation, tail probability
Question
$X \sim N(50, 64)$ . Find $P(X > 62)$ .
Step-by-step solution
1. Step 1
  $\sigma^{2} = 64 \Rightarrow \sigma = 8$ . Standardise:
  $Z = \dfrac{X - \mu}{\sigma} = \dfrac{62 - 50}{8} = 1.5$
2. Step 2
  Express the probability in standard form:
  $P(X > 62) = P(Z > 1.5) = 1 - \Phi(1.5)$
3. Step 3
  From the standard normal tables (or calculator): $\Phi(1.5) = 0.9332$ .
4. Step 4
  Evaluate:
  $P(X > 62) = 1 - 0.9332 = 0.0668$
Answer
$P(X > 62) = 0.0668$ ( $4$ dp).
Examiner tip
M1 for standardising. A1 for $Z = 1.5$ . M1 for $1 - \Phi(1.5)$ . A1 for $0.0668$ . Mark schemes accept $0.0668$ or $0.0668$ to $3$ sf $= 0.0668$ . Sketch a standard normal curve and shade the right tail — examiners look for a diagram for full marks in some sittings.
2 $P(a < X < b)$ (5 marks, S1)
Extended• Adapted from WST01/01 June 2024 Q10• normal, interval probability
Question
$X \sim N(180, 25)$ . Find $P(173 < X < 190)$ .
Step-by-step solution
1. Step 1
  $\sigma = 5$ . Standardise both ends.
  $z_1 = \dfrac{173 - 180}{5} = -1.4, \quad z_2 = \dfrac{190 - 180}{5} = 2$
2. Step 2
  Express as a difference of CDFs:
  $P(173 < X < 190) = \Phi(2) - \Phi(-1.4)$
3. Step 3
  From tables: $\Phi(2) = 0.9772$ . By symmetry, $\Phi(-1.4) = 1 - \Phi(1.4) = 1 - 0.9192 = 0.0808$ .
4. Step 4
  Subtract:
  $P(173 < X < 190) = 0.9772 - 0.0808 = 0.8964$
Answer
$P(173 < X < 190) = 0.8964$ .
Examiner tip
M1 for both standardisations. A1 for $z_1$ and $z_2$ . M1 for $\Phi(2) - \Phi(-1.4)$ . M1 for symmetry use $\Phi(-1.4) = 1 - \Phi(1.4)$ . A1 for $0.8964$ . Forgetting symmetry for negative $z$ is the most common error.
3Inverse — find $x$ given $P(X < x) = p$ (5 marks, S1)
Extended• Adapted from WST01/01 January 2023 Q9• normal, inverse, percentile
Question
$X \sim N(75, 100)$ . Find the value of $x$ such that $P(X < x) = 0.9$ .
Step-by-step solution
1. Step 1
  $\sigma = 10$ . Use the inverse standard normal: find $z$ such that $\Phi(z) = 0.9$ .
2. Step 2
  From inverse normal tables (or calculator): $z = 1.2816$ (often quoted as $z = 1.282$ or $1.28$ ).
3. Step 3
  Back-transform:
  $x = \mu + z\sigma = 75 + 1.2816 \times 10 \approx 87.8$
Answer
$x \approx 87.8$ (or $87.82$ to $4$ sf).
Examiner tip
M1 for using inverse tables. A1 for $z = 1.2816$ (or $1.282$ / $1.28$ , all accepted). M1 for $x = \mu + z\sigma$ . A1 for $87.8$ . The inverse tables are in the IAL formula booklet — but most candidates use the calculator's inverse-normal function.
4Find $\mu$ and $\sigma$ from two probabilities (7 marks, S1)
Challenge• Adapted from WST01/01 June 2023 Q9• normal, find mu sigma, simultaneous equations
Question
$X \sim N(\mu, \sigma^{2})$ . Given $P(X < 50) = 0.2$ and $P(X > 80) = 0.1$ , find $\mu$ and $\sigma$ .
Step-by-step solution
1. Step 1
  First probability: $P(X < 50) = 0.2 \Rightarrow \Phi(z_1) = 0.2$ where $z_1 = (50 - \mu)/\sigma$ . Inverse normal: $z_1 \approx -0.8416$ .
2. Step 2
  So
  $\dfrac{50 - \mu}{\sigma} = -0.8416 \;\Rightarrow\; 50 - \mu = -0.8416 \sigma \;\;\text{(i)}$
3. Step 3
  Second probability: $P(X > 80) = 0.1 \Rightarrow P(X < 80) = 0.9 \Rightarrow \Phi(z_2) = 0.9$ , $z_2 \approx 1.2816$ .
4. Step 4
  So
  $\dfrac{80 - \mu}{\sigma} = 1.2816 \;\Rightarrow\; 80 - \mu = 1.2816\sigma \;\;\text{(ii)}$
5. Step 5
  Subtract (i) from (ii):
  $(80 - \mu) - (50 - \mu) = 1.2816\sigma - (-0.8416\sigma)$
6. Step 6
  Simplify:
  $30 = 2.1232 \sigma \;\Rightarrow\; \sigma \approx 14.13$
7. Step 7
  From (i): $\mu = 50 + 0.8416 \times 14.13 \approx 50 + 11.89 = 61.9$ .
Answer
$\mu \approx 61.9$ , $\sigma \approx 14.1$ ( $3$ sf).
Examiner tip
M1 for inverse-normal of $0.2$ and of $0.9$ . A1 for each $z$ value. M1 for setting up each linear equation (one each). M1 for solving simultaneously. A1 for $\sigma$ . A1 for $\mu$ . Mark schemes accept $z = 0.84$ and $1.28$ (less precise rounding) provided the candidate is consistent.
5Use the normal model in context (5 marks, S1)
Extended• Adapted from WST01/01 January 2024 Q11• normal, context, interpretation
Question
The heights of adult women in a population are modelled as $H \sim N(165, 36)$ (cm). (a) Find the proportion of women taller than $172$ cm. (b) The shortest $5\%$ of women have heights below what value? (c) Comment briefly on whether the model is plausible.
Step-by-step solution
1. Step 1
  $\sigma = 6$ . (a) $P(H > 172) = P(Z > (172 - 165)/6) = P(Z > 7/6) = P(Z > 1.167)$ .
2. Step 2
  From tables: $\Phi(1.17) \approx 0.879$ . So $P(H > 172) \approx 1 - 0.879 = 0.121$ .
3. Step 3
  (b) $5\%$ percentile: $\Phi(z) = 0.05 \Rightarrow z = -1.6449$ (or $-1.645$ ).
4. Step 4
  Back-transform:
  $h = 165 + (-1.6449)(6) \approx 165 - 9.87 = 155.1\,\text{cm}$
5. Step 5
  (c) Plausibility. Adult heights are commonly modelled as normal because the distribution is roughly symmetric, unimodal and bell-shaped. The model assigns positive probability to negative heights, but the probability is astronomically small ( $z \approx -27.5$ for $h = 0$ ), so this is practically irrelevant. The model is plausible.
Answer
(a) $\approx 0.121$ . (b) $\approx 155.1$ cm. (c) Plausible — heights are approximately symmetric and bell-shaped; the model's tiny probability of negative heights is negligible.
Examiner tip
M1 for standardising. A1 for $0.121$ . M1 for inverse $-1.645$ . A1 for $155.1$ . B1 for a plausibility comment that references symmetry / bell shape / negligible negative-height probability. Examiners want a substantive comment, not 'yes it's plausible'.

Key Formulae — The normal distribution

The formulae you need to memorise for the normal distribution on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Standardisation
$Z = \dfrac{X - \mu}{\sigma}, \quad X \sim N(\mu, \sigma^{2}) \Rightarrow Z \sim N(0, 1)$
$Z$
standard normal random variable
$\mu$
mean of $X$
$\sigma$
standard deviation of $X$
When to use
First step in every normal-distribution calculation. Converts an arbitrary $N(\mu, \sigma^{2})$ probability to a $\Phi$ -function lookup. IS in the IAL formula booklet.
Example
$X \sim N(50, 64)$ , $X = 62$ : $Z = (62 - 50)/8 = 1.5$ .
Standard normal CDF $\Phi(z)$
$\Phi(z) = P(Z \le z), \quad Z \sim N(0, 1)$
$\Phi(z)$
CDF of standard normal at $z$
When to use
Read off the standard normal tables in the IAL formula booklet, or from a calculator. Tail probabilities: $P(Z > z) = 1 - \Phi(z)$ .
Example
$\Phi(1.5) = 0.9332$ . $P(Z > 1.5) = 0.0668$ .
Symmetry of the standard normal
$\Phi(-z) = 1 - \Phi(z), \quad P(Z < -z) = P(Z > z)$
$z$
non-negative input
When to use
Whenever $z$ is negative (tables only give $z \ge 0$ ). NOT explicitly in the booklet — but standard property of any symmetric density.
Example
$\Phi(-1.4) = 1 - \Phi(1.4) = 1 - 0.9192 = 0.0808$ .
Inverse standard normal
$P(Z < z) = p \Rightarrow z = \Phi^{-1}(p)$
$p$
probability in $(0, 1)$
When to use
Whenever you are given a probability and need the corresponding $z$ (or $x$ ). Use the inverse-normal tables in the IAL booklet, or your calculator's invNorm.
Example
$p = 0.9$ : $z \approx 1.282$ . $p = 0.05$ : $z \approx -1.645$ .
$68$ - $95$ - $99.7$ rule (approximate)
$P(\mu - \sigma < X < \mu + \sigma) \approx 0.68$
$X$
$N(\mu, \sigma^{2})$ random variable
When to use
Quick sanity check on probabilities. $\mu \pm 2\sigma$ holds about $95\%$ , $\mu \pm 3\sigma$ about $99.7\%$ . NOT in the booklet — memorise as a rough guide.
Example
For $X \sim N(50, 64)$ , about $95\%$ of values lie between $50 - 16 = 34$ and $50 + 16 = 66$ .

Key Definitions and Keywords — The normal distribution

Definitions to memorise and the exact keywords mark schemes credit for the normal distribution answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Normal distribution
Examiner keyword
Continuous distribution $X \sim N(\mu, \sigma^{2})$ with bell-shaped density symmetric about $\mu$ . Two parameters: mean $\mu$ (location) and variance $\sigma^{2}$ (spread).
Standard normal $Z$
Examiner keyword
Normal distribution with $\mu = 0$ and $\sigma = 1$ : $Z \sim N(0, 1)$ . CDF $\Phi(z)$ is tabulated in the IAL formula booklet.
Standardisation
Examiner keyword
Transformation $Z = (X - \mu)/\sigma$ that converts any normal $X$ to the standard normal $Z$ . The $z$ -value measures how many standard deviations $X$ is from its mean.
$\Phi(z)$
Examiner keyword
CDF of the standard normal: $\Phi(z) = P(Z \le z)$ . Tabulated for $z \ge 0$ in the IAL booklet. For $z < 0$ : $\Phi(-z) = 1 - \Phi(z)$ .
Percentile
Value $x$ such that $P(X < x) = p$ . Inverse-normal calculation: find $z$ with $\Phi(z) = p$ , then $x = \mu + z\sigma$ .
$68$ - $95$ - $99.7$ rule
Approximate proportions of a normal distribution within $1$ , $2$ , $3$ standard deviations of the mean. Useful for sanity-checking results.

Common Mistakes and Misconceptions — The normal distribution

The traps other students keep falling into on the normal distribution questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Using $\sigma^{2}$ where $\sigma$ is needed
WST01/01 examiner reports — annual
Why it happens
Confusion between the two parameters of $N(\mu, \sigma^{2})$ .
How to avoid it
Always take the square root before dividing: $X \sim N(50, 64)$ has $\sigma = 8$ , not $\sigma = 64$ . Write 'so $\sigma = 8$ ' as the first line of working.
✕Looking up $\Phi(-1.4)$ directly in the table
Why it happens
IAL tables only give $z \ge 0$ .
How to avoid it
Use symmetry: $\Phi(-z) = 1 - \Phi(z)$ . So $\Phi(-1.4) = 1 - 0.9192 = 0.0808$ . Sketching the curve confirms which tail you want.
✕Confusing $P(X > a)$ with $P(X < a)$
Why it happens
Tables give $\Phi(z) = P(Z < z)$ ; students forget the direction.
How to avoid it
Sketch the curve. Shade the region you want. If shading is right of $z$ , probability is $1 - \Phi(z)$ . If left, it's $\Phi(z)$ .
✕Forgetting the negative sign on a left-tail inverse
WST01/01 examiner reports — recurring
Why it happens
Inverse-normal tables typically give positive $z$ for probabilities $> 0.5$ .
How to avoid it
If $P(X < x) < 0.5$ , the corresponding $z$ is NEGATIVE. Example: $P(Z < z) = 0.05 \Rightarrow z = -1.645$ , not $+1.645$ .
✕Sign slips in the simultaneous-equation step for finding $\mu$ and $\sigma$
Why it happens
Two equations with two unknowns and at least one negative $z$ .
How to avoid it
Write each equation as $x_i - \mu = z_i \sigma$ . Subtract one from the other so $\mu$ cancels — solving for $\sigma$ first is usually cleanest. Then substitute back for $\mu$ .
✕Giving the probability without an interpretive sentence
Why it happens
Focus on the number.
How to avoid it
Cite the variable in the answer: 'About $12.1\%$ of women are taller than $172$ cm'. The contextual sentence often carries the final A-mark.

The normal distribution — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

The normal distribution

Short Notes - The normal distribution

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Forward probability P(X>a)P(X > a)P(X>a) (4 marks, S1)

2P(a<X<b)P(a < X < b)P(a<X<b) (5 marks, S1)

3Inverse — find xxx given P(X<x)=pP(X < x) = pP(X<x)=p (5 marks, S1)

4Find μ\muμ and σ\sigmaσ from two probabilities (7 marks, S1)

5Use the normal model in context (5 marks, S1)

Standardisation

Standard normal CDF Φ(z)\Phi(z)Φ(z)

Symmetry of the standard normal

Inverse standard normal

686868-959595-99.799.799.7 rule (approximate)

Normal distribution

Standard normal ZZZ

Standardisation

Φ(z)\Phi(z)Φ(z)

Percentile

686868-959595-99.799.799.7 rule

✕Using σ2\sigma^{2}σ2 where σ\sigmaσ is needed

✕Looking up Φ(−1.4)\Phi(-1.4)Φ(−1.4) directly in the table

✕Confusing P(X>a)P(X > a)P(X>a) with P(X<a)P(X < a)P(X<a)

✕Forgetting the negative sign on a left-tail inverse

✕Sign slips in the simultaneous-equation step for finding μ\muμ and σ\sigmaσ

✕Giving the probability without an interpretive sentence

The normal distribution — frequently asked questions

1Forward probability $P(X > a)$ (4 marks, S1)

2 $P(a < X < b)$ (5 marks, S1)

3Inverse — find $x$ given $P(X < x) = p$ (5 marks, S1)

4Find $\mu$ and $\sigma$ from two probabilities (7 marks, S1)

Standard normal CDF $\Phi(z)$

$68$ - $95$ - $99.7$ rule (approximate)

Standard normal $Z$

$\Phi(z)$

$68$ - $95$ - $99.7$ rule

✕Using $\sigma^{2}$ where $\sigma$ is needed

✕Looking up $\Phi(-1.4)$ directly in the table

✕Confusing $P(X > a)$ with $P(X < a)$

✕Sign slips in the simultaneous-equation step for finding $\mu$ and $\sigma$