问题如下图:
选项:
A.
B.
C.
解释:
请问题目如果改成will lie within 10%的话,是不是答案就要选C了?
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?
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?
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。
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?
请问为什么k=2啊,为什么不是用10—4K=8来算k呢