Normal Distribution

Importance

The normal distribution (Gaussian) is the most important distribution in statistics. It appears naturally as the limit of sums (Central Limit Theorem) and models many physical phenomena.

Climate: heights, measurement errors, IQ scores, physical measurements, and economic data all approximate normal distributions.

Definition

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

$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad x \in \mathbb{R}$

Parameters

• $\mu$: mean (centre)
• $\sigma^2$: variance (spread), $\sigma > 0$

Range

$X \in (-\infty, \infty)$ (all real numbers)

CDF

$F(x) = \Phi\left(\frac{x - \mu}{\sigma}\right)$

where $\Phi$ is the CDF of the standard normal $N(0,1)$. No closed form exists; we use tables or software.

Properties

Symmetry

The normal is symmetric around $\mu$:

$P(X \leq \mu - a) = P(X \geq \mu + a)$

$f(\mu + a) = f(\mu - a)$

Mean equals Median equals Mode

$E[X] = \mu = \text{median} = \text{mode}$

Variance

$\text{Var}(X) = \sigma^2$

Linear Transformations

If $X \sim N(\mu, \sigma^2)$ and $Y = aX + b$ ($a \neq 0$):

$Y \sim N(a\mu + b, a^2\sigma^2)$

Important: Any linear transformation of a normal is still normal.

Standard Normal

The standard normal $Z \sim N(0, 1)$ has:

$\phi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$

CDF: $\Phi(z) = P(Z \leq z)$

Standardisation

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

$Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$

This converts any normal to standard normal, enabling use of standard tables.

Visualisation

The normal PDF is the famous “bell curve”: symmetric, peaked at $\mu$, with spread controlled by $\sigma$.

• $\mu$ shifts the centre
• $\sigma$ controls the width: larger $\sigma$ = more spread

Probability Calculations

Use standardisation and $\Phi$ tables.

$P(X \leq x) = \Phi\left(\frac{x - \mu}{\sigma}\right)$

$P(a \leq X \leq b) = \Phi\left(\frac{b - \mu}{\sigma}\right) - \Phi\left(\frac{a - \mu}{\sigma}\right)$

Example

$X \sim N(100, 15^2)$. Find $P(X \leq 115)$.

Standardise: $Z = \frac{115 - 100}{15} = 1$

$P(X \leq 115) = P(Z \leq 1) = \Phi(1) \approx 0.841$

From tables: $\Phi(1) \approx 0.8413$.

Empirical Rule (68-95-99.7)

For any normal $N(\mu, \sigma^2)$:

Interval	Coverage
$\mu \pm \sigma$	$\approx 68\%$
$\mu \pm 2\sigma$	$\approx 95\%$
$\mu \pm 3\sigma$	$\approx 99.7\%$

This is useful for quick estimates.

Exact Values

$\Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826$

$\Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9544$

$\Phi(3) - \Phi(-3) = 0.9987 - 0.0013 = 0.9974$

Sum of Independent Normals

If $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ are independent:

$X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$

The sum of independent normals is normal. This is not true for arbitrary sums.

Extending to $n$ Variables

For independent $X_i \sim N(\mu, \sigma^2)$:

$\bar{X} = \frac{1}{n}\sum X_i \sim N\left(\mu, \frac{\sigma^2}{n}\right)$

Summary Table

Property	Value
PDF	$\frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$
Mean	$\mu$
Variance	$\sigma^2$
Standard normal	$N(0, 1)$
Standardisation	$Z = \frac{X - \mu}{\sigma}$
Sum of independent normals	Normal with $\mu = \mu_1 + \mu_2$, $\sigma^2 = \sigma_1^2 + \sigma_2^2$