ritikajain1988 wrote:
Source: Manhattan
Consider the unordered set {x,2,5,11,11,12,33}, No matter whether x is less than 11 ,equal to or greater than 11, the median will ALWAYS BE 11.
While in the unordered set {x,2,5,11,12,12,33}, the median depends on x. If x is 11 or less, median is 11 and if x is between 11 and 12, median is x and finally if x is 12 or more, median is 12
Ques: I did not understand how the statement claims ALWAYS BE 11. Is there a fundamental rule I am missing here? (Though when I try substituting different value, I agree to this statement.)
Good question!
Take a look at the values you already know:
2 5 11 11 12 33
You know six of the seven values. Since there are seven values, including the unknown one, the median will always be the 'middle' value - the one that's fourth in line.
No matter where you put 'x' in the list, the number 11 will always be 4th in line. For instance, if x is a small number, it would go towards the beginning of the list:
2
3 5 11 11 12 33
and the number 11 is fourth.
If x is a large number, it would go towards the end of the list:
2 5 11 11 12
30 33
and the number 11 is still fourth.
This only really works because there are two copies of the number 11 in the list, and those two copies are already pretty much in the middle of the list. Adding a new number to the list will only move the median up or down slightly. For instance, the median in a 6-number list is the average of number 3 and number 4. The median in a 7-number list is number 4. That's only a very small change. If number 3 and number 4 were already equal, then the median won't change at all.
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