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What is the median number of employees assigned per project

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What is the median number of employees assigned per project [#permalink] New post 22 Mar 2005, 13:27
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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 assigned to each project.
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 [#permalink] New post 22 Mar 2005, 13:31
I go with E because we have no upper range in (1). It could be 4-1000 per project so we have no way of calculating a value for the median number.
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 [#permalink] New post 22 Mar 2005, 19:31
E....

a,b individually are insufficient, together also they do'nt provide enough informations as we still do'nt kow about the remaining 40% of projects, as to what are they uptoo!!
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 [#permalink] New post 22 Mar 2005, 20:18
I would go with C.
35% of the projects have 2 or less members, 25% of the projects have 4 or more members. Remaining 40% of the projects will have 3 members. Therefore using the information provided by 1 and 2 I can say that median will be 3.
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 [#permalink] New post 22 Mar 2005, 21:52
E.
we cannot say that the rest 40% companies have 3 emplyoees because we do not have any information about that. these 40% companies could have more than 3 employees.
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 [#permalink] New post 23 Mar 2005, 01:30
kapslock wrote:
Guys,

Would anyone post on theory of mean, median and mode? Or maybe provide a link to any such discussion that has happened before?

Thanks


Hello Kapslock,

Please check the following link:

http://www.wtamu.edu/academic/anns/mps/ ... 4_cent.htm

Hope this helps!
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 [#permalink] New post 23 Mar 2005, 15:12
Hi MA,
I belive that we can say that rest of the projects(40%) will have 3 members. Can any value other than 3 be assigned for the number of pepole taking place in the remaining projects?
My answer is no.
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 [#permalink] New post 23 Mar 2005, 15:44
"25% of the projects have 4 or more members", for me construct an upper limit. It doesn't matter #of people participating to the project is 100, 200, they are eventually one number. We are not taking the average. We should be mostly concerned with how many observations will be on the left side and on the right side. This way we can figure the observation takes place in the middle.
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 [#permalink] New post 28 Mar 2005, 08:22
I would go for option E because:-

Lets assume we have 100 projects:

1. 25 projects would have > 4 employees
2. 35 projects would have < 2 employees

Neither 1 nor 2 specifies that they are the upper or lower limits.

The rest of the 40 projects could have
a. > 4 employees, i.e. maybe 5, 6,7 ..or any other greater value
b. <2 employees, i.e. maybe 1 employee each (0 would not make sense anyway)
c. between 2 & 4 employees, i.e. 3 employees each

In the above three cases a, b & c, the medians would differ and hence both together are not sufficient to answer the question. :)
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 [#permalink] New post 28 Mar 2005, 08:32
Let's take a look at your case a:

Quote:
The rest of the 40 projects could have
a. > 4 employees, i.e. maybe 5, 6,7 ..or any other greater value


Now stem (1) says:
Quote:
25 percent of the projects at Company Z have 4 or more employees assigned to each project


In your case you have more than 40 projects that have more than 4 employees. It is directly contrary to what (1) says. If (1) is true, your case a cannot be true. Same thing with your case b. Only c remains.
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 [#permalink] New post 28 Mar 2005, 09:05
HongHu,

Thanks for the explanation. You hit the nail on the head.
C should be the right one.

Cheers,
Paddy.
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 [#permalink] New post 03 Apr 2005, 23:20
Another E here.

I'll need a better explanation for the choice of C.
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Re: DS stat [#permalink] New post 25 Nov 2009, 21:02
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It is C. The question asked for median not mean and therefore you do not need to know the actual numbers of employee working on the project.

Assume you have 100 projects, 25 projects are done by 4 or more employees, 35 projects are done by 2 or less employees. That means the remaining 40 projects are done by 3 employees exactly.

Since you have 100 projects, you median will be project number 50 which is worked on by 3 employees.

Hence, C
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Re: DS stat [#permalink] New post 26 Nov 2009, 00:45
It's clearly C.


MA wrote:
E.
we cannot say that the rest 40% companies have 3 emplyoees because we do not have any information about that. these 40% companies could have more than 3 employees.


Of course there are 3 people @ each of remaining projects. Other possibilities are covered by statements (1) & (2)
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Re: DS stat [#permalink] New post 15 Apr 2010, 04:36
Guys, what if:
1. There are some employees who are involved with both kind of projects (such as Delivery Managers, Directors, HR etc)
2. There are certain freshers/new employee/people on bench :lol:, because it is not mentioned all employee are involved in some projects :?:

So, E.
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Re: DS stat [#permalink] New post 19 Jun 2010, 22:28
lonewolf wrote:
It is C. The question asked for median not mean and therefore you do not need to know the actual numbers of employee working on the project.

Assume you have 100 projects, 25 projects are done by 4 or more employees, 35 projects are done by 2 or less employees. That means the remaining 40 projects are done by 3 employees exactly.

Since you have 100 projects, you median will be project number 50 which is worked on by 3 employees.

Hence, C


IMO, this is the correct explanation. I will go with C.
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Re: DS stat [#permalink] New post 11 Sep 2010, 05:01
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 assigned to each project.

Question asks for "median number of employees per project"

suppose there are 100 projects
as per (1), 25 projects have 4 or more employees i.e. 4, 5, 6, 25 etc.
as per (2), 35 projects have 2 or fewer i.e. 2, 1

Combining 1 & 2, 40 project have exactly 3 employees.

If you combine above 2, how can you derive the median number of employees per project?

Please clarify, as I feel the answer should be E.
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Re: DS stat [#permalink] New post 11 Sep 2010, 05:04
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Orange08 wrote:
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 assigned to each project.

Question asks for "median number of employees per project"

suppose there are 100 projects
as per (1), 25 projects have 4 or more employees i.e. 4, 5, 6, 25 etc.
as per (2), 35 projects have 2 or fewer i.e. 2, 1

Combining 1 & 2, 40 project have exactly 3 employees.

If you combine above 2, how can you derive the median number of employees per project?

Please clarify, as I feel the answer should be E.


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 assigned to each
project.


(1)+(2) 40% of the projects have exactly 3 employees assigned to each of them, as no other option is left for 100%-(25%+35%)=40% of the projects.

Consider there are 100 projects: \{p_1, p_2, ... ,p_{100}\}. The values of p_1 to p_{35} will be 0, 1, or 2; the values of p_{36} to p_{75} will be exactly 3; the values of p_{76} to p_{100} will be 4 or more. Median=\frac{p_{50}+p_{51}}{2}=\frac{3+3}{2}=3.

For example list can be: \{2, 2, 2, ...,(p_{35}=2), (p_{36}=3), 3, ...,(p_{75}=3), (p_{76}=4), 4, ...,(p_{100}=4)\};
OR:
\{0, 0, 1,1,1,2, 2, ...,(p_{35}=2), (p_{36}=3), 3, ...,(p_{75}=3), (p_{76}=4), 5,7, 27, ...,(p_{100}=10000)\} (of course there are a lot of other breakdowns).

In any case median=3.

Answer: C.

Hope it's clear.
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Re: DS stat [#permalink] New post 11 Sep 2010, 05:23
Bunuel, many thanks for your explanation.

One query : while considering median, we need to arrange the numbers in sequence, i guess.

If there are 20 projects,
1) 35% i.e. 7 projects have value as 4,5,6,7,8,8,8
2) 25% i.e. 5 projects have value as 1,2,2,2,0
3) 40% i.e. 8 projects have value as 3,3,3,3,3,3,3,3

So to derive the median, are we suppose to arrange the number in sequence i.e. ascending ?

0,1,2,2,2,3,3,3,3,3,3,3,3,4,5,6,7,8,8

Is my understanding correct Bunuel?
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Re: DS stat [#permalink] New post 11 Sep 2010, 05:40
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Orange08 wrote:
Bunuel, many thanks for your explanation.

One query : while considering median, we need to arrange the numbers in sequence, i guess.

If there are 20 projects,
1) 35% i.e. 7 projects have value as 4,5,6,7,8,8,8
2) 25% i.e. 5 projects have value as 1,2,2,2,0

3) 40% i.e. 8 projects have value as 3,3,3,3,3,3,3,3

So to derive the median, are we suppose to arrange the number in sequence i.e. ascending ?

0,1,2,2,2,3,3,3,3,3,3,3,3,4,5,6,7,8,8

Is my understanding correct Bunuel?


If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.


So if there are 20 projects,
35% i.e. 7 projects have 2 or fewer employees assigned to each project, for example: 0, 0, 1, 1, 1, 2, 2;
40% i.e. 8 projects have exactly 3 employees assigned to each project: 3, 3, 3, 3, 3, 3, 3, 3;
25% i.e. 5 projects have 4 or more employees assigned to each project, for example: 4, 5, 6, 7, 8;

0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5, 6, 7, 8.

So, as 20 is even, then the median number of employees per project would be the average of two middle terms: 10th and 11th --> (3+3)/2=3.

Hope it helps.
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Re: DS stat   [#permalink] 11 Sep 2010, 05:40
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