I have a boxplot that is drawn and the whiskers are the same length but the median is closer to the upper quartile than the middle.
Would this be considered skewed or symmetrical?
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In small samples from symmetric distributions the median may frequently be much closer to one hinge (effectively, quartile) than the other.
Even very "nicely" behaved unimodal distributions like the normal can have this boxplot-asymmetry happen. If the sample size is 10, then the median will be 63% or more of the way toward the nearest quartile half the time (i.e. the absolute quartile skewness will half the time exceed 0.26). That is, for sampling from a normal with $n=10$, this is a typical (i.e. median) amount of asymmetry in the box:
(since 25% it will be at least this far up, and 25% of the time it would be equally as far below the midhinge point)
At n=9 it's 67% for samples from a normal distribution. [Equivalently the absolute value of the quartile skewness will exceed 1/3 half the time.]
If the population distribution is non-normal but still symmetric, a typical boxplot might be much more asymmetric (one symmetric example I just came up with had the median-line position about 92.5% of the way toward the nearest quartile at least half the time at n=10).
Which is to say, in small samples, we have to be very cautious about interpreting asymmetry in the boxplot as implying asymmetry of the population from which the sample was drawn.
For a given distribution, as $n$ increases, such a large deviation from symmetry becomes less likely and then we may feel somewhat safer in calling it asymmetric (e.g. for normal samples at n=100 the typical split of the box is 45-55), but even at large samples there would still be symmetric distributions that tended to have very asymmetric-looking boxes (they won't be unimodal, though -- but you generally won't be able to tell that from the boxplot either).