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What is the median number of employees assigned per project [#permalink]
 10 Nov 2008, 06:41
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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.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient
(E) seems so straightforward....
am i missing something here?
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This question is already posted several times in the forum.
Answer has to be C as the remaining 40% projects will have 3 employees and hence 3 will be the median.
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scthakur wrote: This question is already posted several times in the forum.
Answer has to be C as the remaining 40% projects will have 3 employees and hence 3 will be the median. why the remaining 40% projects will have 3 employees and hence 3 will be the median. Median doesnt have anything to do with dispersion
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lylya4 wrote: scthakur wrote: This question is already posted several times in the forum.
Answer has to be C as the remaining 40% projects will have 3 employees and hence 3 will be the median. why the remaining 40% projects will have 3 employees and hence 3 will be the median. Median doesnt have anything to do with dispersion 25% of the projects have >= 4 employees. 35% of the projects have <= 2 employees. Hence, remaining projects will have only 3 employees. And if you rank them, 3 will be the median.
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scthakur wrote: lylya4 wrote: scthakur wrote: This question is already posted several times in the forum.
Answer has to be C as the remaining 40% projects will have 3 employees and hence 3 will be the median. why the remaining 40% projects will have 3 employees and hence 3 will be the median. Median doesnt have anything to do with dispersion 25% of the projects have >= 4 employees. 35% of the projects have <= 2 employees. Hence, remaining projects will have only 3 employees. And if you rank them, 3 will be the median. e.g. If there are 100 projects, within 25 projects, 1 projects has 4 employees, 1 projects has 5 employees, 1 projects has 6 employess.... 1 projects has 24 employees 40 projects has 3 employees 35 projects has 2 employees 2, 3, 4, 5,6 ,7 ,8 9, 10..... 25 So the median is not 3. If you know the number of employees assigned in each project, you can calculate the median, otherwise its insuff.
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lylya4 wrote: e.g. If there are 100 projects,
within 25 projects, 1 projects has 4 employees, 1 projects has 5 employees, 1 projects has 6 employess.... 1 projects has 24 employees
40 projects has 3 employees
35 projects has 2 employees
2, 3, 4, 5,6 ,7 ,8 9, 10..... 25
So the median is not 3.
If you know the number of employees assigned in each project, you can calculate the median, otherwise its insuff. The important thing to note is that it is the mean employee for the projects, hence, in your example, the number of employees per project will be like the following (for say 100 projects) 1,1,1,1,1,1,1,2,2,2,2,....(till 35 such numbers), 3,3,3,3,3........(till 40 such numbers), 4,5,6,7,8,20, 100, ....(till 25 such numbers) and the median will be 3.
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scthakur wrote: lylya4 wrote: e.g. If there are 100 projects,
within 25 projects, 1 projects has 4 employees, 1 projects has 5 employees, 1 projects has 6 employess.... 1 projects has 24 employees
40 projects has 3 employees
35 projects has 2 employees
2, 3, 4, 5,6 ,7 ,8 9, 10..... 25
So the median is not 3.
If you know the number of employees assigned in each project, you can calculate the median, otherwise its insuff. The important thing to note is that it is the mean employee for the projects, hence, in your example, the number of employees per project will be like the following (for say 100 projects) 1,1,1,1,1,1,1,2,2,2,2,....(till 35 such numbers), 3,3,3,3,3........(till 40 such numbers), 4,5,6,7,8,20, 100, ....(till 25 such numbers) and the median will be 3. You are right, i igot it. +kudo for you
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