Doubt regarding median of unordered sets

[ameya]

I am using Manhattan strategy guides and right now studying Statistics from Word translation guide. I came across a statement that states:

A) For the unordered set {x, 2, 5, 11, 11, 12, 33}. No matter whether x is less than 11, equal to 11, or greater than 11, the median of the resulting set will be 11.

B) For the unordered set {x, 2, 5, 11, 12, 12, 33} depends on x. If x is 11 or less, the median is 11. If x is between 11 and 12, the median is x. Finally, if x is 12 or more, the median is 12.

I cannot understand why median is different in both the options if it is said that median depends ONLY on the middle number in case of odd number of terms or avg. of two numbers in case of even number of terms.

Please help me understand this concept better?

Ameya

[ameya]

No replies :| can anyone help me on this?

Ameya

[jnelson0612]

Sure! Please notice that the first set has two 11s and the second has two 12s. That will be relevant!

So, let's consider the possibilities for the first set{x, 2, 5, 11, 11, 12, 33}:
a) x is less than 11. If I line every number up in order, an 11 is still the middle number. 11 is the median.
b) x is 11. Again, if I line up every number in order, an 11 is the middle number, or median.
c) x is greater than 11. After ordering the numbers, an 11 is the middle number or median.

Let's consider the second set {x, 2, 5, 11, 12, 12, 33}:
a) x is less than 11. The middle number is 11, which is the median.
b) x is 11. The middle number is still 11, which is the median.
c) x is greater than 11. Let's say it is 15, for example. Our new set is {2, 5, 11, 12, 12, 15, 33}. Now the middle number is 12, which is the new median. In this case, if x is a number that is 12 or larger this x pushes a 12 into the middle place.

Please let us know if you have further questions. :-)