The normal distribution as a model for continuous data, its mean and standard deviation, calculating probabilities, the standard normal distribution and standardising, finding values from probabilities, and using the normal approximation to the binomial.

What this dot point is asking

AQA wants you to use the normal distribution to model continuous data, work with its mean and standard deviation, calculate probabilities directly from your calculator, standardise values to the standard normal distribution, solve inverse problems (finding a value given a probability), and use the normal distribution to approximate the binomial. This is a calculator-active topic on Paper 3, so you are expected to use the normal cumulative distribution function and inverse normal function fluently.

The model and its shape

The standard deviation $\sigma$ controls the spread: a larger $\sigma$ gives a flatter, wider curve, while a smaller $\sigma$ gives a tall, narrow one. The points of inflection of the curve occur exactly at $x = \mu \pm \sigma$, which is a useful sketching landmark and a frequent AQA marking point. As a rough guide for sketches and sanity checks, the empirical proportions are about $68$ percent within $\mu \pm \sigma$, about $95$ percent within $\mu \pm 2\sigma$, and about $99.7$ percent within $\mu \pm 3\sigma$.

Calculating probabilities

On Paper 3 you find probabilities directly with the normal cdf, entering the lower bound, upper bound, $\mu$ and $\sigma$. Use $-10^{99}$ (or a very large negative) for "less than" and $10^{99}$ for "greater than". For example, with $X \sim N(50, 16)$ (so $\sigma = 4$), $P(X < 56) = P\left(Z < \frac{56 - 50}{4}\right) = P(Z < 1.5) \approx 0.933$.

Always draw and shade a sketch of the bell curve. It stops sign errors, makes "greater than" versus "less than" obvious, and is often worth an explicit mark. Remember the area to the right is one minus the area to the left, and by symmetry $P(Z > a) = P(Z < -a)$.

Standardising

Inverse problems

To find the value with a given probability below it, look up the z value for that probability (using the inverse normal) and rearrange to $x = \mu + z\sigma$. For example, the value below which $95$ percent of the data lie has $z = 1.6449$, so $x = \mu + 1.6449\sigma$. If two percentiles are given, set up simultaneous equations in $\mu$ and $\sigma$ and solve, as in the bolt question above.

Normal approximation to the binomial

For a binomial $X \sim B(n, p)$ with large $n$ (and $p$ not too close to $0$ or $1$, so that $np$ and $n(1-p)$ are both reasonably large), $X$ is approximately $N(np, np(1-p))$. Because the binomial is discrete and the normal continuous, apply a continuity correction: replace $P(X \le k)$ with $P(Y < k + 0.5)$ and $P(X \ge k)$ with $P(Y > k - 0.5)$, where $Y$ is the approximating normal variable.