Normal distribution

The normal distribution is the symmetric bell-shaped curve fully described by its mean μ and standard deviation σ.

f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\!\Big(-\frac{(x-\mu)^2}{2\sigma^2}\Big)

pdf

$\mu$: mean (centre of symmetry, also the median)
$\sigma$: standard deviation ( $\sigma > 0$ )

Z = \frac{X - \mu}{\sigma} \sim N(0, 1), \qquad P(X \le x) = \Phi\Big(\frac{x-\mu}{\sigma}\Big)

Standardising

$Z$: standard normal variable (mean 0, SD 1)
$\Phi$: standard normal cdf
$z_{1-\alpha/2}$: standard normal quantile, e.g. $z_{0.975} = 1.96$

When to use

Measurements that cluster symmetrically around a centre (heights, measurement error, many process outputs).
As the approximate sampling distribution of means and proportions (see the central limit theorem).

Assumptions

Before using it to model raw data, check the shape with a histogram or Q–Q plot. Skewed or heavy-tailed data (incomes, waiting times, claims) are not normal.

How to interpret

Empirical rule: about 68% of values lie within 1σ of μ, 95% within 2σ (exactly 1.96σ), 99.7% within 3σ.
Forward problem: value → z → probability. Inverse problem: probability → z quantile → value $x = \mu + z\sigma$ .

Common mistakes

Using the variance instead of the SD when standardising.
Assuming data are normal because the sample is large (the CLT is about means, not individual values).
Forgetting that $P(X > x) = 1 - \Phi(z)$ , not $\Phi(z)$ .

Example

Scores are $N(70, 8^2)$ . $P(X > 82)$ : $z = 1.5$ , probability $0.0668$ . $P(62 < X < 78) = 0.6827$ (within 1 SD). The 90th percentile is $70 + 1.2816 \times 8 = 80.25$ .