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The median of the first group is the lower or first quartile, and is equal to (0 + 1)/2 = 0.5. The median of the second group is the upper or third quartile, and is equal to (27 + 61)/2 = 44. The smallest and largest observations are 0 and 63. So the five-number summary would be 0, 0.5, 7.5, 44, 63.
Any probability distribution on the real number set has at least one median, but in pathological cases there may be more than one median: if F is constant 1/2 on an interval (so that f = 0 there), then any value of that interval is a median.
Upper 1.5*IQR whisker = Q 3 + 1.5 * IQR = 9 + 3 = 12. (If there is no data point at 12, then the highest point less than 12.) Pattern of latter two bullet points: If there are no data points at the true quartiles, use data points slightly "inland" (closer to the median) from the actual quartiles. This means the 1.5*IQR whiskers can be uneven in ...
Although this number has a decimal representation that is an infinite string of ones, Excel displays only the leading 15 figures. In the second line, the number one is added to the fraction, and again Excel displays only 15 figures. In the third line, one is subtracted from the sum using Excel.
The median is the middle number of the group when they are ranked in order. (If there are an even number of numbers, the mean of the middle two is taken.) 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 ...
The bold numbers (36, 39) are used to calculate the median as their average. As there are an even number of data points, the first three methods all give the same results. (The Method 3 is executed such that the median is not chosen as a new data point and the Method 1 started.)
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.
Thus, the probability of observing 3 or fewer plus signs or 7 or more plus signs in the survival data, if the median survival is 200 weeks, is 0.3438. The expected number of plus signs is 5 if the null hypothesis is true. Observing 3 or fewer, or 7 or more pluses is not significantly different from 5. The null hypothesis is not rejected.