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)?



 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.