Exam PContinuousFree to read

Normal distribution

A symmetric bell-shaped variable; the limiting distribution of standardised sums by the Central Limit Theorem.

Parameters and support

$\mu$

mean (and median and mode) — $\mu \in \mathbb{R}$

$\sigma$

standard deviation — $\sigma >0$

Support

$-\infty <x<\infty$

The formulas

f(x)

$$f(x)=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{(x-\mu {)}^2}{2{\sigma }^2}\right)$$

F(x)

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

No elementary antiderivative — Φ is read from the standard normal table.

Mean

$$E[X]=\mu$$

Variance

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

MGF

$$M(t)=\exp \left(\mu t+\frac{{\sigma }^2{t}^2}{2}\right)$$

Where the moments come from

1. Standardise: $Z=(X-\mu )/\sigma$ has density $\frac{1}{\sqrt{2\pi }}{e}^{-z^2/2}$.
2. $E[Z]=0$ by symmetry (the integrand is odd), so $E[X]=\mu$.
3. $E[Z^2]=1$ by integrating by parts, so $\mathrm{Var}(X)={\sigma }^2$.
4. Completing the square in the MGF integral gives $M(t)=e^{\mu t+{\sigma }^2{t}^2/2}$.
5. Because the MGF is exponential in $t$, products of normal MGFs are normal MGFs — that is why sums of independent normals stay normal.

Worked example

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

• Adding standard deviations instead of variances when combining independent normals.
• Forgetting the ±0.5 continuity correction when approximating a DISCRETE count.
• Reading σ² off a question that quoted σ, or the reverse.

Related

• Lognormal distribution
• Chi-square distribution
• Gamma distribution

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.