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## Is it possible to determine the mean of the distribution of 10 scores, but only 9 are given.

With a distribution of 10 scores, nine of them are 3, 5, 9, 1, 9, 2, 0, 3, and 9. The last one is greater than 5 but it is not 9.  Is it possible to determine the mean of the distribution?

I like Kerrie's answer, as long as the scores are discrete, integer data. It appears you have discrete, integers. If that is not the case, then read on:

If the missing score can be any number greater than 5 and less than 9, then plug 5 and 9 into Kerrie's equations intead of 6 and 8; also change the ≤ signs into < signs. This only applies if any score between 5 and 9 is possible (continuous data).

If the data are discrete, but not necessarily integers, then use Kerrie's method but use the nearest discrete data points instead of 6 and 8. For example, if the scores can be 5.0, 5.1, 5.2, 5.3..., then plug in 5.1 and 8.9. Keep the ≤ signs.

Well, it kind of depends on what type of discrete data. According to Georg Cantor, the set Q of rational numbers is also considered discrete, but is dense. As a result, if all rationals in the interval 5<x<9 allowed, the answer is the same as for real numbers except that the mean must be rational.

You can determine the possible range of mean values since you know that the missing value is either a 6,7, or 8...

minimum mean = {sum(3, 5, 9, 1, 9, 2, 0, 3, 9) + 6}/10

maximum mean = {sum(3, 5, 9, 1, 9, 2, 0, 3, 9) + 8}/10

range for mean is {4.7<= m <= 4.9}

You don't mention what the measurement represents or anything about the number of significant digits...If these are truly integer values, then you would most likely round to the nearest integer, 5, no matter what the last measurement.

Your measurement error has increased, but because it is small compared to the uncertainty of the measurement (0.2 compared to 1), you are still able to give a final answer of 5.