Probability density function, expected value and variance of each probability distribution

I’d like to summarize probability density function (f(x)), expected value (E(X)) and variance (V(X)) of following probability distribution.

Supergeometric distribution
$\displaystyle f(x) = \frac{{}_{M}C_x\cdot{}_{N-M}C_{n-x}}{{}_{N}C_{n}}$	$\displaystyle x = Max(0, n - (N - M)), \cdots, Min(n, M)$
$\displaystyle E(X) = np$	$\displaystyle p = \frac{M}{N}$
$\displaystyle V(X) = np(1 - p)\frac{N - n}{N - 1}$
Binomial distribution
$\displaystyle f(x) = {}_{n}C_{x}p^{x}(1 - p)^{n - x}$	$\displaystyle x = 0, 1, \cdots, n$
$\displaystyle E(X) = np$	$\displaystyle 0 < p < 1[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle V(X) = np(1 - p)$
Poisson distribution
$\displaystyle f(x) = \frac{\exp(- \lambda)\cdot\lambda^x}{x!}$	$\displaystyle x = 0, 1, 2, \cdots$
$\displaystyle E(X) = \lambda$	$\lambda = np$
$\displaystyle V(X) = \lambda$
Geometric distribution
$\displaystyle f(x) = p(1 - p)^{x - 1}$	$\displaystyle x = 1, 2, 3, \cdots$
$\displaystyle E(X) = \frac{1}{p}$	$\displaystyle 0 < p < 1[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle V(X) = \frac{1 - p}{p^2}$
Negative binomial distribution
$\displaystyle f(x) = {}_{k + x - 1}C_{x}p^{k}(1-p)^{x}$	$\displaystyle x = 0, 1, 2, \cdots$
$\displaystyle E(X) = \frac{k(1 - p)}{p}$
$\displaystyle V(X) = \frac{k(1 - p)}{p^2}$
Normal distribution
$\displaystyle f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x - m)^2}{2\sigma^2}\right)$	$\displaystyle - \infty < x < \infty[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle E(X) = m$
$\displaystyle V(X) = \sigma^2$	$\displaystyle \sigma > 0$
Exponential distribution
$\displaystyle f(x) = \lambda\exp(-\lambda x)$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle E(X) = \frac{1}{\lambda}$
$\displaystyle V(X) = \frac{1}{\lambda^2}$
Gamma distribution
$\displaystyle f(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha - 1}\exp(- \lambda x)$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle E(X) = \frac{\alpha}{\lambda}$	$\displaystyle \lambda, \alpha > 0$
$\displaystyle V(X) = \frac{\alpha}{\lambda^2}$
Beta distribution
$\displaystyle f(x) = \frac{x^{\alpha - 1}(1 - x)^{\beta - 1}}{B(\alpha, \beta)}$	$\displaystyle 0 < x < 1[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle f(x) = 0$	$\displaystyle x \le 0, 1 \le x$
$\displaystyle E(X) = \frac{\alpha}{\alpha + \beta}$
$\displaystyle V(X) = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$
Cauchy distribution
$\displaystyle f(x) = \frac{\alpha}{\pi(\alpha^2 + (x - \lambda)^2)}$	$\displaystyle \alpha > 0$
Log-normal distribution
$\displaystyle f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp\left(-\frac{(\log(x) - m)^2}{2\sigma^2}\right)$	$\displaystyle x > 0$
$\displaystyle f(x) = 0$	$\displaystyle x \le 0$
$\displaystyle E(X) = \exp\left(m + \frac{\sigma^2}{2}\right)$	$\displaystyle \sigma > 0$
$\displaystyle V(X) = \exp[2m + \sigma^2](\exp[\sigma^2] - 1)$
Pareto distribution
$\displaystyle f(x) = \frac{x_0^\alpha}{x^{\alpha + 1}}$	$\displaystyle x \ge x_0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle E(X) = \frac{ax_0}{a - 1}$	$\displaystyle a > 1$
$\displaystyle V(X) = \frac{ax_0^2}{a - 2} - \left(\frac{ax_0}{a - 1}\right)^2$	$\displaystyle a > 2$
Weibull distribution
$\displaystyle f(x) = \frac{\gamma x^{\gamma - 1}}{\lambda^b}\exp\left[-\left(\frac{x}{y}\right)^\gamma\right]$	$\displaystyle x \ge 0$
$\displaystyle f(x) = 0$	$\displaystyle x < 0[/latex]</td> </tr> <tr> <td><img src='https://s0.wp.com/latex.php?latex&bg=T&fg=000000&s=0' alt='' title='' class='latex' />\displaystyle E(X) = \lambda\Gamma\left(1 + \frac{1}{\gamma}\right)$	$\displaystyle \lambda > 0$
$\displaystyle V(X) = \lambda^2\left[\Gamma\left(2 + \frac{1}{\gamma}\right) - \left(\Gamma\left(1 + \frac{1}{\gamma}\right)\right)^2\right]$	$\displaystyle \gamma > 0$