Consider the following data of runners in a race. The first data is from a certain year the second one is from the year after.

$Last\space year \space|\space Mean: 525s \space|\space MAD: 25s$

$This \space year \space|\space Mean: 613s \space|\space MAD: 21s$

The question asks to make inference from the measures of variability.

My answer:

The MAD this year is lower than last year, this means that more students were closer to the mean time taken to finish the race. Since the mean this year is greater than the mean last year, more students were slower this year.

I feel like this should be right, but I am not quite sure if it is the right inference from the data given.

  • 1
    $\begingroup$ @Tim Isn't MAD = mean absolute deviation $\endgroup$ – Mhf Jun 11 at 6:18
  • $\begingroup$ Both definitions are used, although Median AD is probably more correct. Which definition is used in your problem? $\endgroup$ – user2974951 Jun 11 at 6:58
  • $\begingroup$ @user2974951 is most likely Mean AD. $\endgroup$ – Mhf Jun 12 at 3:00

It seems to be a question so let me just give you some hints to figure it out by yourself.

  • $\begingroup$ It's confusing that it is used as an acronym for both of these things. In my case, MAD was used as an acronym for mean absolute deviation. Thank you, Tim! $\endgroup$ – Mhf Jun 12 at 2:59

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