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Does anyone knows, how to change (or which ONE VALUE of data do I have to change), so that the mean of this data are: 1. lower than the median, and 2. as equal as the median. For example if I have the data: 1, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3.

I've read some on the internet which says that:

median + (new value - old value)/number of value (or we called n)= mean

then I've tried to make a new mean value which is lower than the old mean. But it's always give a negative result for the ONE NEW VALUE (of the data). Is it true? So I can't have any ONE NEW POSITIVE VALUE (to make the mean lower/equal than the median)?

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Original:

 x= c( 1, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3)
 mean(x); median(x)
 ## 1.981818
 ## 1.8

Change first element to -10 (other choices below -1 are possible). Mean smaller than median.

 y= c(-10, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3)
 mean(y); median(y)
 ## 0.9818182
 ## 1.8

Change first element to -1 (only value that works). Mean equals median.

 z= c(-1, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3)
 mean(z); median(z)
 ## 1.8
 ## 1.8

I have changed only the first element, but (if I understand the rules) any element below the median (1.8) could be reduced by the same amount. Because a reduction of the total by 2 is required, I think you are correct that the revised sample must have a negative element.

Comment: This is an interesting mathematical puzzle, but I don't see that it lends much to an understanding of the properties or practical uses of the mean and median as measures of centrality.

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  • $\begingroup$ Thank you so much for your answer! This really helps me for a better understanding :) $\endgroup$ – hello Jan 17 '17 at 19:11
  • $\begingroup$ Glad it helped. After thinking about this over lunch, I've revised the sentence before the Comment, (I hope) to make it clearer. $\endgroup$ – BruceET Jan 17 '17 at 20:44

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