Data Sufficiency

sorry, i just posted in PB forum..

DS question with median .. ??


What is the median number of employee assigned per project for the projects at company z?

1) 25% of the project at company Z have 4 or more employees assigned to each project.

2) 35% of the project at company z have 2 or fewer employee assigned to each project




pls help, many thanks.

answer is C

Good question - please comment on my logic

35% is less or equal to 2 and 25% is greater or equal to 4. Thus 40% is equal to 3.

So if it is a even number of projects the median will be (3+3)/2 or if it's odd, the answer will be simply 3.

Saffa wrote:Good question - please comment on my logic

35% is less or equal to 2 and 25% is greater or equal to 4. Thus 40% is equal to 3.

So if it is a even number of projects the median will be (3+3)/2 or if it's odd, the answer will be simply 3.

Good logic and right answer!

Statement (1) by itself is insufficient because we have no clue how the rest of the employees are assigned... same problem with statement (2).

When we combine the statement, we know that all the other projects have 3 employees assigned to each of them. If we were to write out the list, "3" would cover the middle terms, so it will end up being the median of the set.

I don't understand why do you know that 40% is 3.
Can you please explain more..

many thanks

Thephu wrote:I don't understand why do you know that 40% is 3.
Can you please explain more..

many thanks

(1) 25% of the project at company Z have 4 or more employees assigned to each project.

(2) 35% of the project at company z have 2 or fewer employee assigned to each project

Statement (1) covers all of the cases where there are 4 or more employees per project. Statement (2) covers all of the cases where there are 2 or fewer employees per project. The only possible # of employees per project that hasn't been covered is exactly 3, which must account for all the remaining scenarios.

So, 100% - (25% + 35%) = 40% for exactly 3 employees/project.

Got it this time....

Thank you very much

Sorry for opening an old thread !!

I understand the first part but how are we adding up 3 + 3 / 2 in the case of even number of projects.

I am not able to understand how do we get 3 + 3?

Stuart Kovinsky wrote:

Saffa wrote:Good question - please comment on my logic

35% is less or equal to 2 and 25% is greater or equal to 4. Thus 40% is equal to 3.

So if it is a even number of projects the median will be (3+3)/2 or if it's odd, the answer will be simply 3.

Good logic and right answer!

Statement (1) by itself is insufficient because we have no clue how the rest of the employees are assigned... same problem with statement (2).

When we combine the statement, we know that all the other projects have 3 employees assigned to each of them. If we were to write out the list, "3" would cover the middle terms, so it will end up being the median of the set.

mehravikas wrote:Sorry for opening an old thread !!

I understand the first part but how are we adding up 3 + 3 / 2 in the case of even number of projects.

I am not able to understand how do we get 3 + 3?

When there are an even number of terms in a set, the median is the average (mean) of the two middle terms.

We know that the first 35% of the projects have 2 members, the middle 40% of the projects have 3 members and the top 25% of the projects have 4 members. If we think about it in a row:

1%-----2 members-----35%36%------------3 members-----------75%76%--------4 members--------100%

we can see that the projects on either side of the 50% mark will both have 3 members in them.

So, we use the average formula:

Avg = sum of terms / # of terms = (3+3)/2 = 3

giving us our median of 3.

thank you for the detailed explanation Stuart. it was extremely helpful

(1) 25% of the project at company Z have 4 or more employees assigned to each project.

(2) 35% of the project at company z have 2 or fewer employee assigned to each project

hence 40% have exactly 3 employeees , so can we say safely that bulk of projects involve 3 employees. Hence the Median is 3.

Hope the logic is correct ?[/url]

The remaining 40% of projects have 3 ppl,why? It can be possible that they have 10ppl.
Please explain me.
I understand that the least 25% are <=2
Max 35% are >=4
Then the rest 40% could also be <=2 or >=4 or 3.Am I correct?
Please Clarify.

skprocks wrote:The remaining 40% of projects have 3 ppl,why? It can be possible that they have 10ppl.
Please explain me.
I understand that the least 25% are <=2
Max 35% are >=4
Then the rest 40% could also be <=2 or >=4 or 3.Am I correct?
Please Clarify.

25% are <=2 that means it could have 0,1 or 2 number of employees/ project

Max 35% are >=4 that means it could have 4,5,6,......any number(we don't know) per project

rest 40% could also be <=2 or >=4 or 3.Am I correct?
No, Since the range <=2 and >=4 is already covered with above scenario. So, we are left with only one possibility i.e. 3 employee/project.

Hi,

I'm so sorry, but I just don't get it!
I wanted to open a new thread for this question but searched and found this one.

I understand that 40% of the projects will have 3 employees.
35% have 2 or less : ok.

But what bothers me is the 25% that have 4 or more. It could be any number above 4...
So how could we say that median is 3 ?

Of course 40>25 and 40>35 but that doesn't mean the median will be the number of employees in the 40% (3)!
Help please!

bowen380 wrote:Hi,

I'm so sorry, but I just don't get it!
I wanted to open a new thread for this question but searched and found this one.

I understand that 40% of the projects will have 3 employees.
35% have 2 or less : ok.

But what bothers me is the 25% that have 4 or more. It could be any number above 4...
So how could we say that median is 3 ?

Of course 40>25 and 40>35 but that doesn't mean the median will be the number of employees in the 40% (3)!
Help please!

You may be confusing median with arithmetic mean.

The median is the middle term of an ordered set. The values of the other terms is irrelevant when calculating the median.

For example:

{2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4}

and

{2, 2, 2, 2, 3, 3, 3, 3, 3, 100, 1000, 10000, 100000}

both have medians of 3.

* * *

(As a small aside: if there's an odd number of terms, the median is the middle term; if there's an even number of terms, the median is the arithmetic mean of the two middle terms.)

Well thank you very much! It's clear in my mind now. As a french person, it has been quite difficult to differentiate all these new terms. And I kept confusing mean and median. Anyway thanks, see you in another topic!