{"provider_name":"Hatena Blog","author_name":"derwind","type":"rich","blog_url":"https://randommemory.hatenablog.com/","version":"1.0","image_url":null,"html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Frandommemory.hatenablog.com%2Fentry%2F2015%2F11%2F02%2F001109\" title=\"\u4eee\u8aac\u691c\u5b9a - \u3089\u3093\u3060\u3080\u306a\u8a18\u61b6\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","width":"100%","url":"https://randommemory.hatenablog.com/entry/2015/11/02/001109","published":"2015-11-02 00:11:09","height":"190","description":"\u78ba\u7387\u5909\u6570$\\{X_j\\}_{j=1}^n$\u304c\u6b63\u898f\u5206\u5e03\u306b\u5f93\u3063\u3066\u3044\u308b\u3068\u3059\u308b\u3002\u3053\u306e\u6642\u3001\u6b63\u898f\u5316\u3057\u3066 $$P \\left(a \\le \\frac{\\overline{X} - \\mu}{\\sqrt{\\sigma^2/n}} \\le b \\right) = \\frac{1}{\\sqrt{2\\pi}} \\int_a^b \\exp \\left(- \\frac{x^2}{2} \\right) dx \\hspace{5em} (1)$$\u3068\u306a\u308b\u3002(\u5927\u6570\u306e\u6cd5\u5247\u3068\u4e2d\u5fc3\u6975\u9650\u5b9a\u7406(2) - \u3089\u3093\u3060\u3080\u306a\u8a18\u61b6\u3067\u89e6\u308c\u305f\u4e2d\u5fc3\u6975\u9650\u5b9a\u7406\u306e\u5834\u5408\u3001$\\{X_j\\}$\u304c\u6b63\u898f\u5206\u5e03\u306b\u5f93\u3063\u3066\u3044\u306a\u304f\u3066\u3082\u3001\u6ca2\u5c71\u304b\u304d\u96c6\u3081\u308b\u3068\u3001\u3042\u305f\u304b\u3082\u5f93\u3063\u3066\u3044\u308b\u304b\u306e\u3088\u3046\u2026","blog_title":"\u3089\u3093\u3060\u3080\u306a\u8a18\u61b6","author_url":"https://blog.hatena.ne.jp/derwind/","title":"\u4eee\u8aac\u691c\u5b9a","provider_url":"https://hatena.blog","categories":["statistics"]}