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阿萌酱 · 2020年12月29日

问一道题:NO.PZ2017092702000157

问题如下:

For a distribution of 2,000 observations with finite variance, sample mean of 10.0%, and standard deviation of 4.0%, what is the minimum number of observations that will lie within 8.0% around the mean according to Chebyshev's Inequality?

选项:

A.

720

B.

1500

C.

1680

解释:

B is correct. Observations within 8% of the sample mean will cover an interval of 8/4 or two standard deviations. Chebyshev’s Inequality says the proportion of the observations P within k standard deviations of the arithmetic mean is at least 1 - 1/k2 for all k > 1. So, solving for k = 2: P = 1 – ¼ = 75%. Given 2,000 observations, this implies at least 1,500 will lie within 8.0% of the mean.
A is incorrect because 720 shows P = 720/2,000 = 36.0% of the observations. Using P to solve for
k implies 36.0% = 1 – 1/k
2, where k
= 1.25. This result would cover an interval only 4% × 1.25 or 5% around the mean (i.e. less than two standard deviations).
C is incorrect because 1,680 shows P = 1,680/2,000 = 84.0% of the observations. Using P to solve for
k implies 84.0% = 1 – 1/k
2, where k
= 2.50. This result would cover an interval of 4% × 2.5, or 10% around the mean (i.e., more than two standard deviations).

这道题目跟老师上课讲的75%的例题有什么区别呢?为什么解题思路不一样。

1 个答案

星星_品职助教 · 2020年12月29日

同学你好,

可以把你提到的这个例题截图粘贴上来用于比较。

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NO.PZ2017092702000157 1500 1680 B is correct. Observations within 8% of the sample mewill cover intervof 8/4 or two stanrviations. Chebyshev’s Inequality says the proportion of the observations P within k stanrviations of the arithmetic meis least 1 - 1/k2 for all k > 1. So, solving for k = 2: P = 1 – ¼ = 75%. Given 2,000 observations, this implies least 1,500 will lie within 8.0% of the mean. A is incorrebecause 720 shows P = 720/2,000 = 36.0% of the observations. Using P to solve for k implies 36.0% = 1 – 1/k2, where k = 1.25. This result woulcover intervonly 4% × 1.25 or 5% arounthe me(i.e. less thtwo stanrviations). C is incorrebecause 1,680 shows P = 1,680/2,000 = 84.0% of the observations. Using P to solve for k implies 84.0% = 1 – 1/k2, where k = 2.50. This result woulcover intervof 4% × 2.5, or 10% arounthe me(i.e., more thtwo stanrviations). 为什么标准差是4%,问在8%的k就成了2?

2021-06-01 13:37 1 · 回答

NO.PZ2017092702000157 老师,您好,这道题提到了的均值和方差是样本值,为什么不是K倍的 S/(根号下N),即标准误?切比雪夫不等式是只针对总体的均值和方差吗?

2021-02-18 21:23 1 · 回答

看了老师之前的回答,和本题下面的解答,也了解这个公式但有点困惑为什么Observations within 8% of the sample mewill cover intervof 8/4 or two stanrviations. 为什么8%可以cover2个stanrviations呢?他们的关系是怎么分析出来的呢?

2020-03-03 22:59 1 · 回答

请问为什么k=2啊,为什么不是用10—4K=8来算k呢

2020-01-01 11:20 1 · 回答