Normal Distribution (PDF/CDF)
Statistics
Intro: Compute normal PDF or CDF at x; standardize to Z if needed.
Worked example
- mu=0, sigma=1, x=1.96, mode=CDF
- Goal: compute the normal CDF using $\Phi(z)=\tfrac12\,[1+\mathrm{erf}(z/\sqrt{2})]$.
- Inputs: $\mu=0$, $\sigma=1$, $x=1.96$ (CDF mode).
- Standardize: $z=\dfrac{x-\mu}{\sigma}=\dfrac{1.96-0}{1}=1.96$.
- Reduce to erf: $x_e=z/\sqrt{2}=1.96/1.41421356\approx1.385929$.
- Approximation (Abramowitz–Stegun 7.1.26): $\mathrm{erf}(x)\approx 1 - t\,e^{-x^2}\,[a_1+a_2 t+a_3 t^2+a_4 t^3+a_5 t^4]$, with $p=0.3275911$, $a_1=0.254829592$, $a_2=-0.284496736$, $a_3=1.421413741$, $a_4=-1.453152027$, $a_5=1.061405429$.
- Helper: $t = 1/(1+p x_e) = 1/(1+0.3275911\cdot1.385929) \approx 0.687749$.
- Square & exponential: $x_e^2\approx1.385929^2=1.9208$ so $e^{-x_e^2}\approx e^{-1.9208}=0.1464897$.
- Powers: $t^2\approx0.472999$, $t^3\approx0.325305$, $t^4\approx0.223728$.
- Polynomial: $a_1+a_2 t+a_3 t^2+a_4 t^3+a_5 t^4\approx0.254829592-0.195662+0.672328-0.472717+0.237466=0.496244$.
- Product 1: $t\,e^{-x_e^2}\approx0.687749\cdot0.1464897=0.100654$.
- Product 2: $t\,e^{-x_e^2}\,\text{poly}\approx0.100654\cdot0.496244=0.049996$.
- Approximate erf: $\mathrm{erf}(x_e)\approx1-0.049996=0.950004$.
- Convert back: $\Phi(1.96)=\tfrac12(1+0.950004)=0.975002$.
- Sanity check (two-tailed 5\%): $2(1-0.975002)=0.049996\approx0.0500$.
- Answer: $\boxed{0.9750}$ (rounded to four decimals).
FAQs
Two-tailed areas?
Yes—use symmetry or compute $P(|Z|>z)=2(1−Φ(z))$.
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