# Normal distribution

> CIE A-Level Mathematics · 9709
> Source: https://www.owlsprep.com/study/cie-9709-u4-normal-distribution/

This sub-topic covers the properties of the normal distribution, the most widely used continuous probability distribution. You will learn to calculate probabilities and solve inverse quantile problems, a core exam requirement for CIE A-Level Stats 1.

**Prerequisites:** [Discrete random variables (mean, variance & the binomial)](https://www.owlsprep.com/study/cie-9709-u4-discrete-random-variables/); [Basic probability rules](https://www.owlsprep.com/study/cie-9709-u4-probability/)

## Learning objectives

- Identify key properties of the normal distribution
- Standardize any normal variable to the standard normal distribution
- Calculate probabilities for normally distributed random variables
- Solve inverse normal problems to find quantiles
- Apply the normal approximation to the binomial distribution (valid when np > 5 and nq > 5), using a continuity correction

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 S1 past papers (2016–2025): **Normal distribution** appears **124 times** in the last 10 years — **24.9%** of all S1 questions (124 of 438).

Most-tested forms: Computing normal probabilities (65), Inverse normal (find x given p) (63), Normal approximation to binomial (58), Z-score conversion (1), Standard normal distribution (1).

## Properties of the Normal Distribution

**Normal Distribution** — A continuous probability distribution with a symmetric, bell-shaped probability density function. $\mu$ is the population mean (equal to median and mode due to symmetry), and $\sigma^2$ is the population variance.

*Notation:* $X \sim N(\mu, \sigma^2)$

*Example:* Heights of adult humans, measurement errors, and test scores are commonly modelled as normally distributed.

The normal distribution has several key properties that are frequently tested: it is symmetric about its mean, the total area under the curve equals 1, and almost all (99.7%) of the distribution lies within 3 standard deviations of the mean. The spread of the curve increases as the variance increases.

**Worked example:** Identify the mean, variance, and standard deviation of the distribution $X \sim N(30, 36)$.

1. Recall that the notation $X \sim N(\mu, \sigma^2)$ lists the mean first, followed by the variance:
2. $$\mu = 30, \quad \sigma^2 = 36$$
3. Calculate the standard deviation as the square root of the variance:
4. $$\sigma = \sqrt{\sigma^2} = \sqrt{36} = 6$$
5. Final answer:
6. Mean = 30, Variance = 36, Standard deviation = 6

> **Exam tip:** Always check if the question asks for variance or standard deviation. Many students lose marks by mixing these up in answers.

## Standardization and Probability Calculation

**Z-Score** — A standardized value that converts any normally distributed variable $X$ to the standard normal variable $Z \sim N(0, 1)$, which can be looked up in standard normal tables.

*Notation:* $z = \frac{X - \mu}{\sigma}$

Any normal distribution can be converted to the standard normal distribution using the z-score formula. This allows us to use precomputed tables to find cumulative probabilities. The cumulative probability for a z-score $z$ is written $\Phi(z) = P(Z \leq z)$. For negative z-scores, we use the symmetry of the normal distribution: $\Phi(-z) = 1 - \Phi(z)$.

**Worked example:** Given $X \sim N(50, 100)$, calculate $P(X < 62)$.

1. Extract the parameters from the notation:
2. $$\mu = 50, \quad \sigma^2 = 100 \implies \sigma = 10$$
3. Calculate the z-score for $X = 62$:
4. $$z = \frac{62 - 50}{10} = 1.2$$
5. Look up $z=1.2$ in the standard normal table to get $\Phi(1.2) = 0.8849$, so $P(X < 62) = 0.8849$

**Worked example:** For the same distribution $X \sim N(50, 100)$, calculate $P(X > 45)$.

1. Calculate the z-score:
2. $$z = \frac{45 - 50}{10} = -0.5$$
3. Use the symmetry property for negative z:
4. $$P(X > 45) = P(Z > -0.5) = 1 - \Phi(-0.5) = \Phi(0.5)$$
5. Look up $\Phi(0.5) = 0.6915$, so $P(X > 45) = 0.6915$

> **Exam tip:** Draw a quick sketch of the normal curve to shade the region you need, this avoids mistakes with tail probabilities.

## Inverse Normal Distribution Problems

Inverse normal problems reverse the process: you are given a probability, and need to find the corresponding value of $X$. These questions are very common in CIE exams, often worth 5-6 marks.

**Inverse Normal** — For a cumulative probability $p = P(X < x)$, inverse normal finds the value of $x$ that corresponds to the $p$-th quantile of the normal distribution.

**Worked example:** Given $X \sim N(20, 25)$, find $x$ such that $P(X > x) = 0.10$.

1. Convert the upper tail probability to a lower tail probability:
2. $$P(X < x) = 1 - P(X > x) = 1 - 0.10 = 0.90$$
3. Extract parameters:
4. $$\mu = 20, \quad \sigma = 5$$
5. Find the z-score corresponding to cumulative probability 0.90: $z = 1.282$
6. Rearrange the z-score formula to solve for $x$:
7. $$x = \mu + z\sigma = 20 + (1.282)(5) = 26.41$$
8. Final answer (3 significant figures): $x \approx 26.4$

> **Exam tip:** Always confirm which tail the question is asking for. For example, the upper 10% is the 90th percentile, not the 10th percentile.

## Normal Approximation to the Binomial Distribution

When the number of trials $n$ in a binomial distribution is large, calculating binomial probabilities term by term is impractical. Provided the distribution is not too skewed, $X \sim B(n, p)$ can be approximated by a normal distribution. This is a required CIE Statistics 1 outcome (5.5) and a frequent exam question.

**Normal Approximation to the Binomial** — If $X \sim B(n, p)$ with $q = 1 - p$, and $n$ is large enough that both $np > 5$ and $nq > 5$, then $X$ is approximately normal with mean $\mu = np$ and variance $\sigma^2 = npq$. A continuity correction must be applied because a discrete variable is being approximated by a continuous one.

*Notation:* $X \sim B(n, p) \approx N(np,\ npq)$

*Example:* $X \sim B(100, 0.4)$ has $np = 40 > 5$ and $nq = 60 > 5$, so $X \approx N(40, 24)$.

Because the binomial is discrete but the normal is continuous, apply a continuity correction: replace each integer boundary by the half-integer that keeps the required values inside the region. For $X \geq 5$ use $P(X > 4.5)$; for $X > 5$ use $P(X > 5.5)$; for $X \leq 5$ use $P(X < 5.5)$; for $X < 5$ use $P(X < 4.5)$; and for the single value $P(X = 5)$ use $P(4.5 < X < 5.5)$.

**Worked example:** In a large batch of components, 40% are premium grade. A random sample of 100 components is taken. Use a suitable approximation to find the probability that at least 45 are premium grade.

1. Let $X$ be the number of premium components, so $X \sim B(100, 0.4)$. Check the approximation conditions with $q = 1 - 0.4 = 0.6$:
2. $$np = 100 \times 0.4 = 40 > 5, \quad nq = 100 \times 0.6 = 60 > 5$$
3. Both conditions hold, so approximate with mean $np$ and variance $npq$:
4. $$X \approx N(40,\ 24), \quad \sigma = \sqrt{24} = 4.899$$
5. Apply the continuity correction. 'At least 45' means $X \geq 45$, which becomes $X > 44.5$ for the continuous variable:
6. $$P(X \geq 45) \approx P(X > 44.5)$$
7. Standardize and use the table:
8. $$z = \frac{44.5 - 40}{\sqrt{24}} = \frac{4.5}{4.899} = 0.919$$
9. $$P(X > 44.5) = 1 - \Phi(0.919) = 1 - 0.821 = 0.179$$
10. So the probability that at least 45 components are premium is approximately 0.179 (3 s.f.).

> **Exam tip:** The most-missed step is the continuity correction and its direction: 'at least 45' ($X \geq 45$) uses 44.5, but 'more than 45' ($X > 45$) uses 45.5. Always write $B(n, p)$ and verify $np > 5$ and $nq > 5$ before approximating.

## Common pitfalls

- **Wrong:** Confusing variance and standard deviation when calculating z-scores
  - Why it fails: The notation $N(\mu, \sigma^2)$ gives variance as the second parameter, so students often use variance instead of standard deviation
  - Correct: Always take the square root of the second parameter to get $\sigma$ before starting any calculations
- **Wrong:** Incorrectly calculating probabilities for negative z-scores
  - Why it fails: Standard normal tables only give probabilities for positive z, so students often forget to apply the symmetry rule
  - Correct: Always use the identity $\Phi(-z) = 1 - \Phi(z)$ for negative z-scores
- **Wrong:** Using the upper tail probability directly for inverse normal problems
  - Why it fails: Students use the given upper tail probability to find z, instead of converting to a lower tail probability
  - Correct: Always convert to a lower cumulative probability $p = P(X < x)$ before looking up z in the table
- **Wrong:** Forgetting the continuity correction in the normal approximation to the binomial
  - Why it fails: When approximating $X \sim B(n, p)$ (valid when $np > 5$ and $nq > 5$) by a normal distribution, students forget that the discrete count must be shifted by 0.5
  - Correct: After confirming $np > 5$ and $nq > 5$, add or subtract 0.5 to the binomial value (e.g. $P(X \geq 45) \to P(X > 44.5)$) before standardizing

## Cheatsheet

| Concept | Key Formula / Rule |
| --- | --- |
| General Normal Notation | $X \sim N(\mu, \sigma^2)$ |
| Standard Normal Distribution | $Z \sim N(0, 1)$ |
| Z-score formula | $z = \frac{X - \mu}{\sigma}$ |
| Negative z probability | $\Phi(-z) = 1 - \Phi(z)$ |
| Inverse normal x | $x = \mu + z\sigma$ |
| Empirical Rule (approximate) | $\approx$ 68.3% within 1$\sigma$, $\approx$ 95.4% within 2$\sigma$, $\approx$ 99.7% within 3$\sigma$ (use tables for exact values) |

## What's next

Within Statistics 1, the key application of the normal distribution is the normal approximation to the binomial distribution, covered in the section above. Looking beyond this topic, the normal distribution also underpins sampling and estimation, confidence intervals, and hypothesis testing — but note that these belong to Statistics 2 (Paper 6), a later course, not Statistics 1 (Paper 5). Master the table-based probability and inverse-normal techniques here first.

- [Binomial & Discrete Random Variables (Statistics 1)](https://www.owlsprep.com/study/cie-9709-u4-discrete-random-variables/)
- [Hypothesis Testing (Statistics 2 / Paper 6)](https://www.owlsprep.com/study/cie-9709-u4-hypothesis-testing/)

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