一个正态总体的抽样分布
统计量:X‾=1n∑i=1nXi,其中Xi~N(μ,σ2)\overline{X}= \cfrac{1}{n}\sum_{i=1}^nX_{i},其中X_{i}\text{\textasciitilde} N(\mu,{\sigma^{2}} )X=n1i=1∑nXi,其中Xi~N(μ,σ2)S2=1n−1∑i=1n(Xi−X‾)2S^2= \cfrac{1}{n-1}\sum_{i=1}^n(X_{i}-\overline{X})^2 S2=n−11i=1∑n(Xi−X)2推论·:
- X‾~N(μ,σ2n)证明:X‾=1n∑i=1nXi~N(1n∑i=1nμ,∑i=1nσ2n2)=N(μ,σ2n)\overline{X} \text{\textasciitilde} N(\mu,\cfrac{\sigma^{2}}{n})\\ \begin{aligned} 证明: \overline{X}&= \cfrac{1}{n}\sum_{i=1}^nX_{i} \\& \text{\textasciitilde} N( \cfrac{1}{n}\sum_{i=1}^n\mu, \sum_{i=1}^n\cfrac{\sigma^{2}}{n^2})\\ &=N(\mu,\cfrac{\sigma^{2}}{n})\end{aligned} X~N(μ,nσ2)证明:X=n1i=1∑nXi~N(