NO.PZ2018103102000151
问题如下：
Zhang remembers learning about the weighted harmonic mean. She decides to calculate the weighted harmonic mean for the index and makes the following statement to support her decision:
“The harmonic mean can be used to mitigate the effects of both large and small outliers.”
Zhang’s statement to support using the harmonic mean is best described as:
选项：
A.incorrect with respect to large outliers
incorrect with respect to small outliners
correct
解释：
Zhang’s statement is incorrect with respect to small outliers. The harmonic mean tends to mitigate the impact of large outliers. It may aggravate the impact of small outliers, but such outliers are bounded by zero on the downside.
A is incorrect. The harmonic mean may aggravate the impact of small outliers, but such outliers are bounded by zero on the downside.
C is incorrect. The harmonic mean may aggravate the impact of small outliers, but such outliers are bounded by zero on the downside.
首先，调和平均数能够去除非常大的异常值，这个我明白。
PPT 里面这句话我一直没太明白The harmonic mean may aggravate the impact of small outliers, but such outliers are bounded by zero on the downside.
我的理解是这样的比如说
0.01-2-3-4-500这五个数据，第一次取倒数的时候，500的超大异常值被抹去了。但是0.01被放大了，求和之后再次求倒数，意味着0.01被放大的值再一次被倒数效应所抹去。这样来说，是不是属于调和平均数的优势啊？
老师讲课的时候说，中位数既可以去除最大的异常值，也可以去除最小的异常值，这个我明白。
但是调和平均数只能去除最大的异常值，去除不了最小的异常值呢？这个为啥
