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Victoria522 · 2021年05月17日

请问这个题目是考的哪个知识点?完全找不到思路…谢谢老师!

NO.PZ2015120604000178

问题如下:

BIM is a well-developed public company and its market capitalization growth rates over past few decades are normally distributed.The population has a mean of 10% and a standard deviation of 5.4%. An investor wants to figure out the probability that BIM's average capitalization growth rate will over 12.32% in the next three years (all the data are monthly statistics).Which of the following option is most accurate?

选项:

A.

0.5%.

B.

1%.

C.

5%.

解释:

A is correct.

It is appropriate to use the central limit theroy to calculate.

Standard error=5.4%/36\sqrt {\rm{36}}

P(X>12.32%)=P(Z> (12.32%-10%)/0.009)=P(Z>2.58). So the corresponding probability is 0.5%.

请问这个题目考的哪个知识点?为什么要先计算标准误。谢谢!!

4 个答案

星星_品职助教 · 2021年11月19日

@欢欢

总体均值是确定的数字,不存在标准差。

样本均值是随机变量,如果要对其标准化。需要样本均值自身的均值;和样本均值的标准差,也就是标准误。标准误的求法为总体标准差σ除以根号n。

星星_品职助教 · 2021年11月13日

@欢欢

1)本题不涉及到检验,2.58对应的是99%的置信区间,即两侧尾部面积之和为1%,单侧尾部面积为0.5%。所以P(Z>2.58)就是0.5%;

2)要看是谁的标准差。由于本题计算的是样本均值的标准差,所以就是标准误。

欢欢 · 2021年11月19日

也就是说如果题目给出的总体均值的标准差,就可以在分母直接用;但如果是样本的,则需要根据中心极限定理,转变成标准误比较准确是吗。

星星_品职助教 · 2021年09月12日

@𝒜𝒩𝒥𝒜 安雅🎃

同学你好,根据题干中的“.....in the next three years (all the data are monthly statistics)”可知一共用了36个数据

星星_品职助教 · 2021年05月17日

同学你好,

这道题求的是the probability that BIM's average capitalization growth rate will over 12.32% in the next three years (all the data are monthly statistics),关键词是average。

根据这个描述可以看出,要求的是X的“均值”,即X bar大于12.32%的概率(X为BIM 's capitalization growth rate)。

X bar大于12.32%的概率无法直接得出,但根据中心极限定理可知,X bar服从正态分布,所可以通过正态分布标准化的公式,将P(X>12.32%)转化成可以查表的标准正态分布,即z分布的形式,进而查表求出概率。

P(X bar>12.32%)=P(Z>(12.32%-X bar的均值)/X bar的标准差)

同样根据中心极限定理,X bar的均值=总体均值=10%; X bar的标准差(也就是standard error)=总体标准差/ √n=5.4%/ √6=0.9%,

所以P(X bar>12.32%)=P(Z>(12.32%-10%)/0.9%)= P(Z>2.58)

可以通过标准正态分布表查出2.58对应的左侧概率是0.9951,所以P(Z>2.58)=0.0049≈0.5%。

也可以通过上课要求背诵的正态分布置信区间关键值,直接得出2.58对应99%的置信区间,所以右侧尾部面积为1%/2=0.5%。

𝒜𝒩𝒥𝒜 安雅🎃 · 2021年09月12日

老师,为什么算standard error的时候,样本数量n是36?

欢欢 · 2021年11月12日

P(Z>2.58)这不单尾检验吗?但是2.58对应99%的置信区间是双尾的。 另外,这边计算Z统计量的时候为什么而不能直接用总体标准差5.4%,而是要用标准误?谢谢

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