The table above shows the distribution of the number of absences from

Question

https://gmatclub.com/forum/download/file.php?id=22793 The table above shows the distribution of the number of absences from a certain English class during the spring semester. For those students who had at least 1 absence, what was the median number of absences? (A) 1.5 (B) 2 (C) 2.5 (D) 3 (E) 3.5 PS02353 Table1.png OE: When students who did not have at least 1 absence are eliminated, there are 28 - 4 = 24 students left. The numbers of absences are then 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5 or more, 5 or more, 5 or more. Because 24 is even, the median number of absences for the 24 students is the mean of the 12th and 13th highest number of absences. Both of these numbers are 2 and so the median number of absences is 2. Hi, it appears that number of precise absence in " 5 more" absence is not relevant to the median calculation. Can anyone explain this to me, please.

KarishmaB · Accepted Answer

Yes, it isn't and the reason is this:
In the list {2, 4, 8, 10, 11, 13, 17, 20, 10098}, what is the median? It is 11. Since the total number of items is 9 so median is the middle element i.e. the fifth element. 
Does 10098 play any role in the median? No!

Similarly, your list here looks like this:
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5 or more, 5 or more, 5 or more
It doesn't matter what the last three items are. The median will be the average of the 12th and the 13 the term since there are a total of 24 terms (3+10+3+5+3). Both 12th and 13th terms are 2 so median will be 2.

NoHalfMeasures · Answer

The total no of students with at the least 1 absence is 3+10+3+5+3 = 24
The median would be the average of 12th and 13th term: 2
Ans:B

GMatAspirerCA · Answer

Hi
 I am finding it difficult to understand wording of this question. How are these series deducted from the question?

1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5 or more, 5 or more, 5 or more

Thanks

KarishmaB · Answer

The table gives the number of students for each number of absences. There are 4 students with 0 absences, 3 students with 1 absence each, 10 students with 2 absences etc.

Now if you write down the number of absences of all students in the class in ascending order, you will do something like this:
0 (1st student's absences), 0 (2nd student's absences), 0 (3rd student's absences), 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3 ... and so on...
(there are 4 students with 0 absences, 3 students with 1 absence etc)

If you need to write the number of absences of students with at least one absence in ascending order, you will do the following:
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3 ... and so on...

The median number of absences i.e. average of 12th student and 13th student will be average of 2 and 2 which will be 2. So median number of absences will be 2.

Xin Cho · Answer

VeritasKarishma ,  Bunuel

If we have a list of x items, how do we identify what is the position of the median? I know what is median and how to find it but I am a bit unsure about the proper mathematical way. Thank you in advance.

KarishmaB · Answer

Median is the middle of the list when all items are arranged in increasing (or decreasing) order. So if there are 9 items, median is the 5th item (after arranging in order). If there are total 10 items, then median is the avg of 5th and 6th items and so on.

In a question like this, we know there are a total of 4+3+10+3+5+3 = 28 items.

So median is the avg of 14th and 15th items. When all items are arranged in increasing order, 14th and 15th, both will be 2. Hence median will be 2.

Xin Cho · Answer

Thank you very much  VeritasKarishma

ScottTargetTestPrep · Answer

The precise number of absences in “5 or more” is indeed not relevant in the median calculation with these numbers. Since there are only 3 students in the “5 or more” category, you’ll get the same answer whether the precise absence in that category was 5, 5, and 5; or 5, 6, and 7; or 10, 20, and 30. As you can see, in any of these cases, the numbers at the 12th and 13th spots do not change. It would have been a different story if, for instance, there was only one student each with an absence of 1, 2, 3 and 4 days and 20 students with an absence of 5 or more days. If that were the case, you wouldn’t be to determine the median number of absences; you would only be able to say that the median was 5 or more.

Answer: B

pulak1988 · Answer

chrtpmdr · Answer

What has that to do with number properties?
Isn't that solely statistics?

Bunuel · Answer

_______________________
Edited the tags. Thank you.

lynnglenda · Answer

Students who had at least 1 absence - so we can ignore the first row.
Total number of students excluding the ones with 0 absence = 24
The position of the median of a set of even numbers is given by the average of n/2 and (n+2)/2 = 12, 13
The 2nd and 3rd rows add up to 3+10 = 13. Hence our median lies in 2

Answer : B

DanTheGMATMan · Answer

Cross out in bulk from top and bottom:

https://www.youtube.com/watch?v=n4ih4tvOOXo

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The table above shows the distribution of the number of absences from

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