Data Sufficiency

Thephu wrote: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

Awesome Question...............

Interesting question.

What if the first condition was -

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

So will also include the 2nd. condition by default?


P.S. - I couldn't figure out the range at first. It said 4 or more, so that implies any no. above or equal to 4. Being careless.

sumanr84 wrote:

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.

Hey sumanr84
your explanation was the best
even better than Stuart
Cheersss

I drew a diagram. That is how question immediately formed in my head. Draw a line to represent your range and divide it into 2. The halfway point is your median.

1)Now on your line mark approximately where, from the left, the 35% marker would be (about a third of the line. Everything this to the left is 2 and below . From this step you do not get anything other than that 1 is INSUFF

2)In accordance with 2 put a marker at around the 75% mark so that you can conclude that the rest of the line past this marker is above 4.

Now look at your median or middle marker. It is clearly between 2 and 4. Since we cannot have fractions of people, anything between 2 and 4 has to be 3. It is also obvious from this that you need both pieces of info to derive your answer. So you answer choice should be C. This solution took me 35 seconds. I am usually much slower but this is the type of problem I am good at.

What level is this question?

straight C is the answer

I agree with the odd number such as: 1, 1, 2, 2, 3, 3, 3, 3, 3, 4, 100, 1000, 10000 -> median is 3
But how's about even number such as: 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 4, 100, 1000, 10000. How it can be 3 ?

scholardream wrote:But how's about even number such as: 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 4, 100, 1000, 10000. How it can be 3 ?

Hi!

When you have an even number of terms, the median is the average of the two middle terms.

Since there are 14 terms in your set, we want the average of the 7th and 8th terms. Since T7 and T8 are both 3, the average is:

(3+3)/2 = 6/2 = 3

and, accordingly, the median is still 3.

This can be easily solved using diagrammatic approach

I had identified that this can be a typical number line question. I made my numbers equal 100 and thus, the median would be the 50th and 51st number divided by 2. At this point, being a DS question, I dont care about what answer I get, as long as I get an answer that is unique.

1) Dont know aything about the next 75% of employee projects-- Not Suffient
2) Dont know about the remaining 65% of employee projects. Not Sufficient

1+2= We know about the 35% being less that 2 AND the 25% being over 4.

35% have less than 2..... 40% have 3........ 25% have more than 4.--> 3 lands on the 50th and 51st number!!!!!!!

That means 3+3/2= 3.. We got the answer! Jump up for joy and go to the next question!

Okaaaaay. I undertand the logic used but it still doesn't make sense. At best, we know the median for 40%. We don't know the median number of employees for the other 60% of the company. Median for 25% could 5, 6, 10, 100.

Sorry yet again for re-opening this old thread. What confused me was this: how can we know for sure that the remaining 40% isn't split into a 30/10 or 20/20 percentage with a number other than 3 assigned to each sector? The problem doesn't put restrictions on each percentage by stating specifically that each sector must have a distinct number of employees...this is why I chose E. Can someone please explain why my logic here may be flawed or where I am going wrong with this line of thinking...? Thanks a bunch.

Nanners007 wrote:Sorry yet again for re-opening this old thread. What confused me was this: how can we know for sure that the remaining 40% isn't split into a 30/10 or 20/20 percentage with a number other than 3 assigned to each sector? The problem doesn't put restrictions on each percentage by stating specifically that each sector must have a distinct number of employees...this is why I chose E. Can someone please explain why my logic here may be flawed or where I am going wrong with this line of thinking...? Thanks a bunch.

Hi!

We're told that 25% of projects have 4+ employees and 35% have 2- employees. Since the only number that doesn't fall into "4 or more" and "2 or fewer" is 3, all of the remaining projects must have exactly 3 employees.

gr8 Q

i thinking making a number line here helps

C the answer.

Median = 3