Normal Distribution PDF⚠ unverified

Understanding the Parameters

• The value $x$ — where on the axis the density is evaluated. The PDF is largest at $x = \mu$ and falls off symmetrically; it never reaches zero but decays extremely fast (as $e^{-x^2}$) in the tails.

• The mean $\mu$ — the location parameter: the centre of symmetry and the peak. Shifting $\mu$ slides the whole bell left or right without changing its shape. For the normal, mean, median, and mode all coincide at $\mu$.

• The standard deviation $\sigma$ — the scale parameter: the width of the bell. Small $\sigma$ gives a tall, narrow peak (values tightly clustered); large $\sigma$ a short, broad one. The famous 68–95–99.7 rule says $68\%$ of the probability lies within $\pm1\sigma$, $95\%$ within $\pm2\sigma$, and $99.7\%$ within $\pm3\sigma$ of the mean — the basis of "sigma" quality metrics (Process Capability Cpk).

• The output $f(x)$ — a probability density, not a probability. It has units of "per unit $x$"; actual probabilities come from integrating $f$ over an interval (the area under the curve). The total area is exactly $1$.

• The normalising constant. The factor $1/(\sigma\sqrt{2\pi})$ exists solely to make the total area integrate to $1$ — without it the exponential bell would enclose the wrong area.

Derivation (Approaching a Proof)

The normal PDF can be motivated several ways; the cleanest is the requirement that the total probability integrate to $1$. Start from the Gaussian kernel $e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ — a bell-shaped function peaked at $\mu$ whose width is controlled by $\sigma$. To be a valid density it must integrate to unity, so we need the normalising constant $C$ such that

$$C\int_{-\infty}^{\infty} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\,dx = 1.$$

The integral is the classic Gaussian integral. Substituting $u = (x-\mu)/(\sigma\sqrt2)$ turns it into $\sigma\sqrt2\int_{-\infty}^{\infty}e^{-u^2}\,du = \sigma\sqrt2\cdot\sqrt\pi = \sigma\sqrt{2\pi}$, using the famous result $\int_{-\infty}^{\infty}e^{-u^2}du = \sqrt\pi$ (proved by squaring the integral and evaluating in polar coordinates). Therefore $C = 1/(\sigma\sqrt{2\pi})$, giving

$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}. \qquad\blacksquare$$

One can further show $\mu$ is the mean (by symmetry) and $\sigma^2$ the variance (by integrating $x^2 f$), so the two parameters are exactly the distribution's centre and spread. The deeper reason the normal appears everywhere is the central limit theorem: the normalised sum of many independent, finite-variance random variables converges to this density regardless of their individual distributions.

Dimensional check. The exponent $(x-\mu)^2/\sigma^2$ is dimensionless; the prefactor $1/\sigma$ carries units of "per unit $x$", so $f(x)$ is a density (probability per unit $x$) and $\int f\,dx$ is dimensionless. $\checkmark$

History and Development

• De Moivre, Laplace, Gauss. The curve arose in Abraham de Moivre's 1733 approximation to the binomial, was developed by Pierre-Simon Laplace, and is named for Carl Friedrich Gauss, who derived it (1809) as the error distribution that makes the arithmetic mean the most likely estimate — the origin of "Gaussian" and of least-squares.

• The central limit theorem. Formalised through the 19th and early 20th centuries (Laplace, Lyapunov, Lindeberg), the CLT explains the normal's ubiquity: aggregate many small independent effects and the result is approximately normal, whatever the parts look like.

• In engineering. The normal underlies measurement uncertainty (Expanded Uncertainty, Coverage Factor), statistical process control (Process Capability Cpk), tolerance analysis, and the reliability of load-vs-strength interference — making it the connective tissue of the whole statistics/uncertainty toolkit.

Related Concepts: Process Capability Cpk, Weibull Distribution PDF, Coverage Factor, Expanded Uncertainty, Sample Normal, Relative Uncertainty

Notes: Registry calculator normal-pdf (unverified). Gaussian PDF — correct as shipped. A probability density (integrate for probabilities); total area $= 1$. 68–95–99.7 rule at $\pm1,2,3\sigma$. Fully specified by $\mu$ (location) and $\sigma$ (scale); ubiquitous via the central limit theorem. Returns $0$ if $\sigma\le0$.