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)

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

F(x)

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

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

Mean

$$E[X] = \mu$$

Variance

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

MGF

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

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 $\operatorname{Var}(X) = \sigma^2$.
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 that test run.

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.