I Chebyshev inquation
$E(X)=\mu,D(X)=\sigma^{2}$
$P\{|X-\mu|>=\varepsilon\}<=\frac{\sigma^{2}}{\varepsilon^{2}}\\ it's\ equation:\\ P\{|X-\mu|<\varepsilon\}>=1-\frac{\sigma^{2}}{\varepsilon^{2}}$
$Prove:$
$X-f(x)$
$P\{|X-\mu|>=\varepsilon\}=\int_{|x-\mu|>=\varepsilon}f(x)dx<=\frac{1}{\sigma^{2}}\int_{-\infty}^{\infty}(x-\mu)^{2}f(x)dx=\frac{\sigma^{2}}{\varepsilon^{2}}$
II Law of large numbers
$\lim_{n->\infty}P\{|\frac{f_{A}}{n}-p|<\varepsilon\}=1$
III Central limit theorem
$\frac{\sum_{k=1}^{n}X_{k}-n\mu}{\sqrt{n}\mu}\ samiler \ equal\ N(0,1)$
IV Simple of clt and it’s application
一船舶在海区航行，每次遭受海浪冲击，摇晃角大于3’的概率为$p=\frac{1}{3}$假设遭受了90 000次海浪冲击，其中有29 500~30 500次摇晃角度大于3’的概率是多少？
$Solve:$
$X~B(90\ 000,\frac{1}{3})$
$P\{X=k\}=C_{90\ 000}^{k}(\frac{1}{3})^{k}(\frac{2}{3})^{90\ 000-k},\ k=0,1,...,90\ 000$
$it's too complex of solving that using X~B,but we can using the 'clt' samiler the probaility$
$P\{29\ 5000 <=X<=30\ 5000\}=P\{\frac{29\ 5000-np}{\sqrt{np(1-p)}}<=\frac{X-np}{\sqrt{np(1-p)}}<=\frac{30\ 500-np}{np(1-p)}\}$
$it's\ samilier\ that:\\ \int_{\frac{29\ 500-np}{np(1-p)}}^{\frac{30\ 500-np}{np(1-p)}}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}dt$
$=\Phi(\frac{30\ 500-np}{\sqrt{np(1-p)}})-\Phi(\frac{30\ 500-np}{\sqrt{np(1-p)}})\\ =\Phi(\frac{5\sqrt{2}}{2})-\Phi(-\frac{5\sqrt{2}}{2})=0.999\ 5$