I have the following question:

Given a mean of 11 and a standard deviation of 2, with a non-normal distribution, can you determine the % of numbers that are between 8 and 12?

My guess is no, because the distribution could be weighted on the left and right... but I'm getting thrown off by the sd of 2. Not sure what that implies for this question.

  • $\begingroup$ Are you asking about the possibility of 70% of numbers are between 8 and 12? $\endgroup$ – vinux Sep 23 '13 at 15:49
  • $\begingroup$ I'm asking if one can determine the probability of the numbers 8-12 (inclusive) existing within the conditions given above. $\endgroup$ – kruton Sep 23 '13 at 15:53
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    $\begingroup$ Notice that if $4/5$ of the numbers are $11-\sqrt{12}\approx 7.54$ and $1/5$ are $11+\sqrt{4/3}\approx 12.15$, the mean is $11$ and the SD is $2$, so none lie between $8$ and $12$. On the other hand, let $p\ge 4/13$ and suppose $100p\%$ of the numbers are $a=(11p-2\sqrt{p-p^2})/p$ and the remaining numbers equal $11 + 2p/\sqrt{p-p^2}$. The mean is $11$ and the SD is $2$. Because $8\le a\lt 11$, at least $100p\%$ of the values lie between $8$ and $12$. Thus the possible answers range from a low of $0\%$ all the way up to (but perhaps not including) $100\%$. $\endgroup$ – whuber Sep 23 '13 at 16:54

You can use the asymmetric, two sided variant of the Chebychev inequality to show that at least m% of the data has to lie between these two bounds.

As an illustration, in the context of the info given alongside the question by the OP, one can further state (quiet confidently) that the tightest bounds appear to be

$$P(8<X<12)\in [0,1)$$

  • $\begingroup$ Interesting... that does require the variance, which I don't have. Under a normal distribution variance would be 4. Do I just use that? $\endgroup$ – kruton Sep 23 '13 at 16:02
  • $\begingroup$ @kruton Yes, the standard deviation is just the square root of the variance. $\endgroup$ – COOLSerdash Sep 23 '13 at 16:02
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    $\begingroup$ Right, so this answer concludes that the probability exceeds $4((11-8)(12-11)-2^2)/(12-8)^2$ = $-1/4$! $\endgroup$ – whuber Sep 23 '13 at 16:36
  • $\begingroup$ @whuber: not exactly: $\ge-1/4$. Do you know of a sharper bound, given the info we have? $\endgroup$ – user603 Sep 23 '13 at 17:03
  • $\begingroup$ The tightest bounds are the half-open interval $[0, 1)$. We have a thread somewhere that establishes the proportion cannot equal $1$. $\endgroup$ – whuber Sep 23 '13 at 17:05

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