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It is possible to estimate the median of the underlying variable. If, say, 22% of the observations are of value 2 or below and 55.0% are of 3 or below (so 33% have the value 3), then the median is 3 since the median is the smallest value of for which () is greater than a half. But the interpolated median is somewhere between 2.50 and 3.50.
For n = 1 or 2, the midrange and the mean are equal (and coincide with the median), and are most efficient for all distributions. For n = 3, the modified mean is the median, and instead the mean is the most efficient measure of central tendency for values of γ 2 from 2.0 to 6.0 as well as from −0.8 to 2.0.
the weighted arithmetic mean of the median and two quartiles. Winsorized mean an arithmetic mean in which extreme values are replaced by values closer to the median. Any of the above may be applied to each dimension of multi-dimensional data, but the results may not be invariant to rotations of the multi-dimensional space. Geometric median
The IQR of a set of values is calculated as the difference between the upper and lower quartiles, Q 3 and Q 1. Each quartile is a median [8] calculated as follows. Given an even 2n or odd 2n+1 number of values first quartile Q 1 = median of the n smallest values third quartile Q 3 = median of the n largest values [8]
The five-number summary gives information about the location (from the median), spread (from the quartiles) and range (from the sample minimum and maximum) of the observations. Since it reports order statistics (rather than, say, the mean) the five-number summary is appropriate for ordinal measurements, as well as interval and ratio measurements.
Thus to find the median, order the list according to its elements' magnitude and then repeatedly remove the pair consisting of the highest and lowest values until either one or two values are left. If exactly one value is left, it is the median; if two values, the median is the arithmetic mean of these two. This method takes the list 1, 7, 3 ...
If you do not choose the median as the new data point, then continue the Method 1 or 2 where you have started. If there are (4n+1) data points, then the lower quartile is 25% of the nth data value plus 75% of the (n+1)th data value; the upper quartile is 75% of the (3n+1)th data point plus 25% of the (3n+2)th data point.
The one-sided variant can be used to prove the proposition that for probability distributions having an expected value and a median, the mean and the median can never differ from each other by more than one standard deviation. To express this in symbols let μ, ν, and σ be respectively the mean, the median, and the standard deviation. Then