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The common distributions

Bernoulli, binomial, Poisson, uniform, exponential, normal, and what each models.

Binomial counts successes; Poisson counts rare events; exponential waits for the next one; normal is what sums look like. Picture it: the bell curve and the 95% between ±1.96. Think it: the normal appears because of the central limit theorem, not because nature prefers it.

Worked example · integrate e^(-x^2/2)/sqrt(2*pi) dx from -1.96 to 1.96

Integrate sqrt(2)·e^(-x^2/2)/(2·sqrt(pi)) from -1.96000 to 1.96000

\int_{-1.96}^{1.96} \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}\, dx

Step by step

  1. \int_{-1.96}^{1.96} \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}\, dx = \frac{\sqrt{2}}{2 \sqrt{\pi}} \int e^{- \frac{x^{2}}{2}}\, dx

    Pull the constant \frac{\sqrt{2}}{2 \sqrt{\pi}} out of the integral.

  3. \int e^{- \frac{x^{2}}{2}}\, dx = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}

    Erf rule.

  4. F(1.96) - F(-1.96) = \left(\frac{\operatorname{erf}{\left(0.98 \sqrt{2} \right)}}{2}\right) - \left(- \frac{\operatorname{erf}{\left(0.98 \sqrt{2} \right)}}{2}\right)

    Fundamental theorem of calculus: plug in the limits.

  5. = \operatorname{erf}{\left(0.98 \sqrt{2} \right)} \approx 0.95000

    Simplify.

Reveal the answer
\operatorname{erf}{\left(0.98 \sqrt{2} \right)}

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How to: The common distributions

  1. First find an antiderivative F, then evaluate F(b) − F(a).
  2. Pull the constant \frac{\sqrt{2}}{2 \sqrt{\pi}} out of the integral.
  3. Erf rule.
  4. Fundamental theorem of calculus: plug in the limits.
  5. Simplify.

Questions people ask

What is the difference between probability and statistics?

Probability goes from a known model to what the data should look like; statistics goes from data back to the model. Probability theory is the deductive half.

What does the law of large numbers promise?

That the average of many independent samples converges to the expected value. It says nothing about any single trial.

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