王秋宇 · 2022年01月08日
NO.PZ2015120604000049
问题如下：
The table below shows the monthly stock returns of Ivy Corp.
According to the above table ，Calculate the population variance for Ivy Corp. returns, assuming the population has 6 observations?
选项：
A. 8.01%.
B. 77.1%2.
C. 64.2%2.
解释：
C is correct
=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2
这题目是不是有歧义呢，好像用n或者n-1都说得通啊。
NO.PZ2015120604000049 问题如下 The table below shows the monthly storeturns of Ivy Corp.Accorng to the above table ，Calculate the population varianfor Ivy Corp. returns, assuming the population h6 observations? A.8.01%. B.77.1%2. C.64.2%2. C is correctpopulationvariance=∑(X−μ)2npopulation\quvariance=\fr{ { \sum { (X-\mu ) } }^{ 2 } }{ n } populationvariance=n∑(X−μ)2=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2 用计算器的STAT功能，Sx=8.7788，σx=8.0139，怎么会出现两个标准差的，而且按理说不应该看样本标准差这个值吗？
NO.PZ2015120604000049问题如下The table below shows the monthly storeturns of Ivy Corp.Accorng to the above table ，Calculate the population varianfor Ivy Corp. returns, assuming the population h6 observations?A.8.01%.B.77.1%2.C.64.2%2. C is correctpopulationvariance=∑(X−μ)2npopulation\quvariance=\fr{ { \sum { (X-\mu ) } }^{ 2 } }{ n } populationvariance=n∑(X−μ)2=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2 计算器如果12的Y01=2，算出来 和单独按2次 不一样… 前者N=5，后者N=6。但是我看计算器使用的时候，第一种方法说是可以呀？
NO.PZ2015120604000049 问题如下 The table below shows the monthly storeturns of Ivy Corp.Accorng to the above table ，Calculate the population varianfor Ivy Corp. returns, assuming the population h6 observations? A.8.01%. B.77.1%2. C.64.2%2. C is correctpopulationvariance=∑(X−μ)2npopulation\quvariance=\fr{ { \sum { (X-\mu ) } }^{ 2 } }{ n } populationvariance=n∑(X−μ)2=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2 收益率20% 我X1输入1.2收益率-5%，我X2输入0.95……我似乎算出来的标准差其实是一样的 都是8.01
NO.PZ2015120604000049 问题如下 The table below shows the monthly storeturns of Ivy Corp.Accorng to the above table ，Calculate the population varianfor Ivy Corp. returns, assuming the population h6 observations? A.8.01%. B.77.1%2. C.64.2%2. C is correctpopulationvariance=∑(X−μ)2npopulation\quvariance=\fr{ { \sum { (X-\mu ) } }^{ 2 } }{ n } populationvariance=n∑(X−μ)2=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2 老师，这个问题是我看了您对别的同学的回答后的疑问。这个知识点我可能在之前遗漏了。就是总体方差和总体标准差的概念能不能简单概述一下？为什么是平方的关系？和各自的英文对应是什么？
NO.PZ2015120604000049 问题如下 The table below shows the monthly storeturns of Ivy Corp.Accorng to the above table ，Calculate the population varianfor Ivy Corp. returns, assuming the population h6 observations? A.8.01%. B.77.1%2. C.64.2%2. C is correctpopulationvariance=∑(X−μ)2npopulation\quvariance=\fr{ { \sum { (X-\mu ) } }^{ 2 } }{ n } populationvariance=n∑(X−μ)2=[(20- 7.7)2 + (4 - 7.7)2 + (-5 - 7.7)2 + (12- 7.7)2 + (3 - 7.7)2 + (12- 7.7)2] / 6 = 64.2%2 想请教老师，这道题我将百分数都化为小数来按，为什么出来的结果不对？依次化为小数，0.2、0.4、-0.05、0.12、0.03、0.12，按计算器的时候，我的按键步骤为2NTA——X01 键入0.2，enter，Y01不动，保持为01的默认——X02键入0.04，enter，Y02保持0.1的默认——X03键入-0.05，enter，Y03默认——X04键入0.12，enter，Y04键入2——X05键入0.03，enter，Y05默认为1——2NSTAT——得到的总体数据的标准差为0.085182，8.51%——总体的方差为72.42%的样子，未能得到答案，不知道是哪里出了问题？
