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What is the median number of employees assigned per project

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Intern
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What is the median number of employees assigned per project [#permalink] New post 05 Nov 2004, 23:46
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A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
What is the median number of employees assigned per project for the projects at Company Z?
1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project.
2) 35 percent of the projects at Company Z have 2 or fewer employees assgned to each project.

Please explain your answer. Thank you.
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 [#permalink] New post 06 Nov 2004, 04:33
I am going with E on this one. Unsure thugh.

1. We dont know anything about the remaining 75%. So, INsuff.
2. We dont know anything about the remaining 65%. So, INsuff.

Combining both, it will still matter how the remaining 40% is distributed. If they are in the middle of the distribution(with say 3 employee per project in that 40%), then they will be the median. If that 40% has 10 or more employees per project distribution, then then 35% group will be in the middle. So cant say. Hence E.
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 [#permalink] New post 06 Nov 2004, 06:45
I agree with the above poster that individually, the statements are not sufficient.

But the statements when used in combination tells you the median is 3, and so the answer is C.

A simple graphic:


35%
0--------2


25%
4 --------- infinity


Now 40% of the data is unaccounted for. Where can the data be placed?

If you place the data at 4 or above then you violate statement 1, now more than 25% of the data is above 4! You destroy statement 2 if you try to place the data below 2.


It has to be placed between 2 - 4. So the 35th percentile to the 75th pecentile is 3, and since the median is the 50th percentile, the median = 3.
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 [#permalink] New post 06 Nov 2004, 07:06
1. is insufficient
2 is insufficient

combining 1 and 2,

25 % > = 4 , lets represent the group as A
35 % <= 2 , lets represent this grp as B

==> 40 % = 3

so,
B, B , B...... , 3, 3, 3,3, ... , A, A , A, .....

Since median is the middle number, let s consider 100 grps

so

50 th, 51st grp will always have 3 members. This is sufficient.

so C is the answer.
  [#permalink] 06 Nov 2004, 07:06
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