切比不等式

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：

720

1500

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/k², 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/k², 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).