83) Median wage

I think the options provided don't suffice the problem well. But, how do we work on such problems?

This will be interesting to watch....

any i think we are going from vauge to concrete

x numbers ->one number answer

which i think is not possible or very diffcult

in this case
more than $4000 and
less than $5000
you have check lots of possibilities
and are 4900 and 4300 included in less than 5000....are they overlapping

so,i feel incomplete info or very long question not suitable for 2 minutes solution

answers choice like greater than.... or less then .... would have made sense ,i guess....but even in that case personally i think it will be quite difficult...

what is the source?

frank1 wrote:This will be interesting to watch....

any i think we are going from vauge to concrete

x numbers ->one number answer

which i think is not possible or very diffcult

in this case
more than $4000 and
less than $5000
you have check lots of possibilities
and are 4900 and 4300 included in less than 5000....are they overlapping

so,i feel incomplete info or very long question not suitable for 2 minutes solution

answers choice like greater than.... or less then .... would have made sense ,i guess....but even in that case personally i think it will be quite difficult...

what is the source?

Agree with you frank1

ern5231 wrote:In 100 companies, 64% people are paid more than $4000, 60% people less than $5000, 11% equal to $4300, 3% equal to 4900. What is the median wage?
a) 4150 b) 4300 c) 4450 d) 4500 e) 4600

wage of
3% = 4900 11% = 4300
64% > 4000 or 36% < 4000
60% < 5000 or 40% > 5000
so 24% have wages between 4000 - 5000

stuck

ern5231 wrote:In 100 companies, 64% people are paid more than $4000, 60% people less than $5000, 11% equal to $4300, 3% equal to 4900. What is the median wage?

a) 4150 b) 4300 c) 4450 d) 4500 e) 4600

First thing to note: the median is the middle term of an ordered set if there is an odd number of terms and the average of the two middle terms if there is an even number of terms. Since we have 100 terms, the median will be the average of the 50th and 51st terms.

Well, that's the theory behind how to answer the question; next, we'd solve for the 50th and 51st terms.

However, in this question that's impossble. As people have noted, the 4000-5000 range is 24% of the total, but we only have info on 14% of that range. My guess is that the 4000-5000 range is supposed to be 14% of the total, or the 4300 and 4900 people should add up to 24%.

With the current wording, there's no way to solve.

Please post the source of all of your questions - where is this one from?