Continuous random variables and the Normal distribution, standardising to the standard Normal, finding probabilities, and the Normal approximation to the binomial.

What this dot point is asking

WJEC wants you to model continuous data with the Normal distribution $N(\mu, \sigma^2)$, to standardise values into z-scores, to find probabilities (using a calculator or tables), to work backwards from a probability to an unknown mean or standard deviation, and to use the Normal approximation to the binomial. The Normal distribution is the basis of the hypothesis test on a Normal mean in the next topic.

The answer

Continuous random variables and the Normal distribution

A continuous random variable can take any value in a range, so probability is the area under its density curve, and $P(X = a) = 0$ for any single value.

Standardising

Any Normal variable is converted to the standard Normal $Z \sim N(0, 1)$ by subtracting the mean and dividing by the standard deviation.

The z-score says how many standard deviations $x$ is from the mean. Modern calculators give Normal probabilities directly, but standardising is still required to find an unknown $\mu$ or $\sigma$.

Finding probabilities

Working backwards and the binomial approximation

To find an unknown parameter, read the z-value corresponding to the given probability (the percentage point), then solve $z = \dfrac{x - \mu}{\sigma}$.

For large $n$ (and $p$ not too close to $0$ or $1$), the binomial $B(n, p)$ is well approximated by $N(np, np(1-p))$, which lets you handle large samples that would be tedious term by term. The approximation matches the binomial mean $np$ and variance $np(1-p)$, and works best when $np$ and $n(1-p)$ are both reasonably large so the binomial histogram is roughly symmetric and bell-shaped.

Examples in context

Example 1. Quality thresholds. A factory rejects components more than two standard deviations from the mean diameter. Since about $95\%$ of a Normal distribution lies within two standard deviations, roughly $5\%$ are rejected, split equally between too large and too small. The empirical rule gives a quick reject rate without a calculator.

Example 2. Exam scaling. Marks are Normally distributed with mean $60$ and standard deviation $12$. A mark of $84$ has $z = \dfrac{84 - 60}{12} = 2$, so it beats about $97.7\%$ of candidates. The z-score converts a raw mark into a percentile.

Try this

Q1. $X \sim N(100, 15^2)$. Standardise the value $130$. [2 marks]

• Cue. $z = \dfrac{130 - 100}{15} = 2$.

Q2. Find $P(Z < -1)$ for the standard Normal. [2 marks]

• Cue. By symmetry $P(Z < -1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$.

Q3. For $X \sim N(20, 5^2)$, find $P(X > 25)$. [2 marks]

• Cue. $z = 1$, so $P(X > 25) = 1 - 0.8413 = 0.1587$.