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The table above shows the distribution of the number [#permalink]

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27 Mar 2014, 16:56

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The table above shows the distribution of the number of absences from certain English class during the spring semester. For those students who had at least 1 absence, what was the median number of absences?

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.

The table above shows the distribution of the number of absences from certain English class during the spring semester. For those students who had at least 1 absence, what was the median number of absences?

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.

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.
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Re: The table above shows the distribution of the number [#permalink]

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03 Jun 2014, 21:12

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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
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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

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.
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Re: The table above shows the distribution of the number [#permalink]

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06 Jun 2016, 22:04

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