8.4 Formulas

Factorial

$n!=n(n-1)(n-2)...\left(1\right)$

$0!=1$

Combinations

$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{(n-r)!r!}$

Binomial distribution

$X~B(n,p)$

$P(X=x)=\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x{q}^{n-x}$ , for $x=0,1,2,...,n$

Geometric distribution

$X~G\left(p\right)$

$P(X=x)=q^{x-1}p$ , for $x=1,2,3,...$

Hypergeometric distribution

$X~H(r,b,n)$

$P(X=x)=\left(\frac{\left(\genfrac{}{}{0ex}{}{r}{x}\right)\left(\genfrac{}{}{0ex}{}{b}{n-x}\right)}{\left(\genfrac{}{}{0ex}{}{r+b}{n}\right)}\right)$

Poisson distribution

$X~P\left(\mu \right)$

$P(X=x)=\frac{{\mu }^x{e}^{-\mu }}{x!}$

Uniform distribution

$X~U(a,b)$

$f\left(X\right)=\frac{1}{b-a}$ , $a<x<b$

Exponential distribution

$X~\mathrm{Exp}\left(m\right)$

$f\left(x\right)=m{e}^{-\mathrm{mx}}$ , $m>0,x\ge 0$

Normal distribution

$X~N(\mu ,{\sigma }^2)$

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

Gamma function

$\Gamma \left(z\right)={\int }_0^{\infty }{x}^{z-1}{e}^{-x}\mathrm{dx}$ $z>0$

$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

$\Gamma (m+1)=m!$ for $m$ , a nonnegative integer

otherwise: $\Gamma (a+1)=\mathrm{a\Gamma }\left(a\right)$

Student-t distribution

$X~t_{\mathrm{df}}$

$f\left(x\right)=\frac{{\left(1+\frac{x^2}{n}\right)}^{\frac{-(n+1)}{2}}\Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\mathrm{n\pi }}\Gamma \left(\frac{n}{2}\right)}$

$X=\frac{Z}{\sqrt{\frac{Y}{n}}}$

$Z~N\left(\mathrm{0,1}\right)$ , $Y~{{\rm X}}_{\mathrm{df}}^2$ , $n$ = degrees of freedom

Chi-square distribution

$X~{{\rm X}}_{\mathrm{df}}^2$

$f\left(x\right)=\frac{x^{\frac{n-2}{2}}{e}^{\frac{-x}{2}}}{2^{\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}$ , $x>0$ , $n$ = positive integer and degrees of freedom

F distribution

$X~F_{\mathrm{df}\left(n\right),\mathrm{df}\left(d\right)}$

$\mathrm{df}\left(n\right)=$ degrees of freedom for the numerator

$\mathrm{df}\left(d\right)=$ degrees of freedom for the denominator

$f\left(x\right)=\frac{\Gamma \left(\frac{u+v}{2}\right)}{\Gamma \left(\frac{u}{2}\right)\Gamma \left(\frac{v}{2}\right)}{\left(\frac{u}{v}\right)}^{\frac{u}{2}}{x}^{(\frac{u}{2}-1)}[1+\left(\frac{u}{v}\right)x^{-.5\left(u+v\right)}]$

$X=\frac{Y_u}{W_v}$ , $Y$ , $W$ are chi-square