The Normal Distribution

The normal distribution is the bell-shaped curve that heights, measurement errors and many averages follow. Every calculation with it goes through one step: standardising.

Properties of the normal curve

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

read as: $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$.

• Bell-shaped and symmetric about $\mu$.
• Mean, median and mode all equal $\mu$.
• The total area under the curve is 1.
• The curve approaches the horizontal axis but never touches it.

The empirical rule: approximately

• $68\%$ of values lie within $1$ standard deviation of the mean,
• $95\%$ within $2$,
• $99.7\%$ within $3$.

Standardising

Any normal variable converts to the standard normal $Z \sim N(0,1)$ by

$$Z = \frac{X - \mu}{\sigma}$$

$Z$ measures how many standard deviations a value lies above (or below) the mean. Tables are published for $Z$ alone, which is why every question passes through this step.

For $X \sim N(50, 8^2)$, standardise $X = 60$:

$$z = \frac{60 - 50}{8} = 1.25$$

so 60 is $1.25$ standard deviations above the mean.

Finding probabilities

Tables give $\Phi(z) = P(Z < z)$ — the area to the left.

$P(X < 60)$ where $X \sim N(50, 64)$:

$$P(Z < 1.25) = \Phi(1.25) = 0.8944$$

$P(X > 60)$:

$$1 - 0.8944 = 0.1056$$

Negative $z$. By symmetry,

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

so $\Phi(-1) = 1 - 0.8413 = 0.1587$.

Between two values: subtract.

$$P(45 < X < 60) = \Phi(1.25) - \Phi(-0.625) = 0.8944 - 0.2660 = 0.6284$$

The inverse problem

Sometimes the probability is given and the value is wanted.

For $X \sim N(50, 64)$, find $k$ such that $P(X < k) = 0.90$.

From tables, $\Phi(z) = 0.90$ at $z = 1.2816$. Then reverse the standardisation:

$$X = \mu + z\sigma = 50 + 8(1.2816) = 60.3 \text{ (3 s.f.)}$$

Two values worth memorising, since they recur throughout hypothesis testing:

$$\Phi(1.645) = 0.95 \qquad \Phi(1.96) = 0.975$$

The normal approximation to the binomial

For large $n$, a binomial distribution is close to a normal one:

$$X \sim B(n,p) \ \approx \ N(np,\ npq) \qquad \text{provided } np > 5 \text{ and } nq > 5$$

Because a discrete distribution is being approximated by a continuous one, a continuity correction is required: each whole number is treated as covering the interval half a unit either side.

$X \sim B(100, 0.4)$. Estimate $P(X \le 45)$.

$$np = 40 > 5 \qquad nq = 60 > 5 \ \checkmark$$ $$\mu = 40 \qquad \sigma = \sqrt{100 \times 0.4 \times 0.6} = \sqrt{24} = 4.899$$ $$P(Y < 45.5) = \Phi\!\left(\frac{45.5 - 40}{4.899}\right) = \Phi(1.123) = 0.869$$

The exact binomial value is $0.8689$ — agreement to three decimal places.

Modelling with the normal

The normal distribution suits quantities that are continuous, symmetric about a central value, and unlikely to be extreme — heights, masses, measurement errors, examination marks.

It suits poorly anything strongly skewed (incomes), bounded in a way that matters (times, which cannot be negative, when the mean is close to zero), or naturally discrete with few categories.