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This example calculates the five-number summary for the following set of observations: 0, 0, 1, 2, 63, 61, 27, 13. These are the number of moons of each planet in the Solar System. It helps to put the observations in ascending order: 0, 0, 1, 2, 13, 27, 61, 63. There are eight observations, so the median is the mean of the two middle numbers ...
The median of a power law distribution x −a, with exponent a > 1 is 2 1/(a − 1) x min, where x min is the minimum value for which the power law holds [10] The median of an exponential distribution with rate parameter λ is the natural logarithm of 2 divided by the rate parameter: λ −1 ln 2.
Firstly, computing median of an odd list is faster and simpler; while one could use an even list, this requires taking the average of the two middle elements, which is slower than simply selecting the single exact middle element. Secondly, five is the smallest odd number such that median of medians works.
Consider the data (1, 1, 2, 2, 4, 6, 9). It has a median value of 2. The absolute deviations about 2 are (1, 1, 0, 0, 2, 4, 7) which in turn have a median value of 1 (because the sorted absolute deviations are (0, 0, 1, 1, 2, 4, 7)). So the median absolute deviation for this data is 1.
The mode of a sample is the element that occurs most often in the collection. For example, the mode of the sample [1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17] is 6. Given the list of data [1, 1, 2, 4, 4] its mode is not unique. A dataset, in such a case, is said to be bimodal, while a set with more than two modes may be described as multimodal.
The median of medians method partitions the input into sets of five elements, and uses some other non-recursive method to find the median of each of these sets in constant time per set. It then recursively calls itself to find the median of these n / 5 {\displaystyle n/5} medians.
For example, given binary data, say heads or tails, if a data set consists of 2 heads and 1 tails, then the mode is "heads", but the empirical measure is 2/3 heads, 1/3 tails, which minimizes the cross-entropy (total surprisal) from the data set.
Equal weights should result in a weighted median equal to the median. This median is 2.5 since it is an even set. The lower weighted median is 2 with partition sums of 0.25 and 0.5, and the upper weighted median is 3 with partition sums of 0.5 and 0.25. These partitions each satisfy their respective special condition and the general condition.