Online Word Problems #4

[Carla]

Online Question Bank : Word Problems : #4

I went over the online solution and noticed that in this solution some statistics rules are presented that had not been in the guide.

I was just wondering if there are other rules that you think it would be good to know related to statistics and standard deviation.

Also I wonder if you might have any suggestions of a method to solve these types of problems.

[Guest]

The table below represents three sets of numbers with their respective medians, means and standard deviations. The third set, Set [A+B], denotes the set that is formed by combining Set A and Set B.

Median Mean StandardDeviation
Set A X Y Z
Set B L M N
Set [A + B] Q R S

If X - Y > 0 and L - M = 0, then which of the following must be true?

I. Z > N
II. R > M
III. Q > R


I only
II only
III only
I and II only
none

[StaceyKoprince]

In general, you should definitely know how to handle both average (mean) and median. Standard deviation is necessary if you want a high score.

Some of the discussion in the explanation for this problem is basic to understanding what median and mean represent. For example, if the median of a set of numbers is greater than the mean of that same set - this just means that the numbers below the median must be farther from the median than the numbers above - eg 1, 3, 4, 16, 17, 18, 19, the median is 16 and the mean is 11.1. If the median is smaller than the mean, then the numbers above the median must be farther from the numbers below - just the opposite of the previous example.

Also know that there are some relationships between mean and median when you have a set of consecutive integers (must be consecutive). If you have a set of consecutive integers, then you can find the average by just averaging the first and last terms. For example, 1, 2, 3, 4, 5. For the average, I can just take (1+5)/2 = 3. If the set has an odd number of consecutive terms, then the median will also equal the mean. In that last example, 3 is both the median and the mean.

The rule given for a "composite set" in this explanation is a more obscure rule, however. Only worry about remembering that if you want a 700+ score. (Notice that this problem is labeled 700-800; it's a very hard problem.)

[smitha.gudapakkam]

Hi ,

So is the answer to the problem that only statement III is true?

Thanks
Smitha

[tim]

correct answer for this one is E..

[krishnan.anju1987]

Hi,

Sorry for continuing this silent thread. Here you mentioned that the answer for this would be none E. However, if I think along the lines mentioned by you in this post, since L=M, the set contains odd and consecutive integers. The distance between the elements is constant. Now since x-y>0, x>y. This implies that the median is greater than mean and hence the numbers above the mean are farther away from the mean when compared to the numbers/elements below the mean. However, when both the sets are combined to form the three sets, the elements above the mean will be farther away from the mean when compared to the elements below the mean and hence the third point must be true. Now having said that, the mean itself will vary since the elements of both sets have been joined, Is this the reason that the third one is not correct. I tried recalculating based on the statements but am now plain confused.

[tim]

there are many mistakes here, so i'll just point out the first one and throw the question back to you: there is nothing in the problem that requires consecutive odd integers, so don't assume they are..

[chetan86]

Why the answer is E??

--------- Median-- Mean--- StandardDeviation
Set A------ X ------Y------ Z
Set B---- --L------ M------ N
Set [A + B]- Q----- R ------S

If I set the value as :

--------- Median-- Mean--- StandardDeviation
Set A------ 2 ------1------ Z
Set B---- --1------ 1------ N
Set [A + B]-3----- 2 ------S

---------------OR-----------------

--------- Median-- Mean--- StandardDeviation
Set A------ 2 ------1------ Z
Set B---- --10------10------ N
Set [A + B]--12----- 11 ------S

So anyways Q would be greater than R.
So can anyone please explain why the answer is E?

[chetan86]

Got it!!!

Condition is given as X-Y>0.
So if I take X= (-1) and Y = (-2)
then (-1)-(-2) = -1+2 i.e. 1>0, condition will be satisfied.

--------- Median-- Mean--- StandardDeviation
Set A------ (-1) ------(-2)------ Z
Set B-------(+10)------(+10)------ N
Set [A + B]--(-9)----- (-8) ------S

Hence it is not always Q>R because -9 is not greater than -8.

Let me know if have done anything wrong here.

Regards,
Chetan

[RonPurewal]

Got it!!!

Condition is given as X-Y>0.
So if I take X= (-1) and Y = (-2)
then (-1)-(-2) = -1+2 i.e. 1>0, condition will be satisfied.

--------- Median-- Mean--- StandardDeviation
Set A------ (-1) ------(-2)------ Z
Set B-------(+10)------(+10)------ N
Set [A + B]--(-9)----- (-8) ------S

Hence it is not always Q>R because -9 is not greater than -8.

Let me know if have done anything wrong here.

Regards,
Chetan

Yeah ... unfortunately, just about everything here seems to be wrong. In fact, it appears that you've forgotten -- or just aren't considering in the first place -- the fact that these quantities are means and medians of SETS of numbers.

Look at the numbers in your first post:

--------- Median-- Mean--- StandardDeviation
Set A------ 2 ------1------ Z
Set B---- --10------10------ N
Set [A + B]--12----- 11 ------S

These numbers are impossible.

* These medians are impossible. If the middle number of one set is 2, and the middle number of another set is 10, then, if you combine the sets, the middle number must be 2, 10, or something in between those. Can't be 12.

* The means are also impossible. If the average of one set is 1 and the average of another set is 10, then the combined average must lie strictly between 1 and 10. Can't be 11.

So, it would appear that at least one, and perhaps even both, of the following is/are happening:
1/ You're just thinking that these are random numbers, and are forgetting that they are means/medians/etc.
2/ You're misunderstanding what the problem means by "set [a + b]". (In this problem, it has nothing whatsoever to do with adding numbers together; it's a matter of taking two lists and combining them into a single list.)

[harika.apu]

In general, you should definitely know how to handle both average (mean) and median. Standard deviation is necessary if you want a high score.

Some of the discussion in the explanation for this problem is basic to understanding what median and mean represent. For example, if the median of a set of numbers is greater than the mean of that same set - this just means that the numbers below the median must be farther from the median than the numbers above - eg 1, 3, 4, 16, 17, 18, 19, the median is 16 and the mean is 11.1. If the median is smaller than the mean, then the numbers above the median must be farther from the numbers below - just the opposite of the previous example.

Also know that there are some relationships between mean and median when you have a set of consecutive integers (must be consecutive). If you have a set of consecutive integers, then you can find the average by just averaging the first and last terms. For example, 1, 2, 3, 4, 5. For the average, I can just take (1+5)/2 = 3. If the set has an odd number of consecutive terms, then the median will also equal the mean. In that last example, 3 is both the median and the mean.

The rule given for a "composite set" in this explanation is a more obscure rule, however. Only worry about remembering that if you want a 700+ score. (Notice that this problem is labeled 700-800; it's a very hard problem.)

Hello Stacey ,
Just need one more confirmation.
i think same behaviour even applies to mean as it does to median .(referring to your discussion in first para).


Thanks.

[RonPurewal]

Hello Stacey ,
Just need one more confirmation.
i think same behaviour even applies to mean as it does to median .(referring to your discussion in first para).


Thanks.

sorry—what, specifically, are you trying to confirm?

presumably you're referring to the second paragraph—but, in that paragraph, stacey's observation is about comparisons between the median and the mean.
that paragraph contains no statements about the mean alone (or about the median alone), so i can't tell what you are saying here.

please clarify, thanks.

[harika.apu]

Hello Stacey ,
Just need one more confirmation.
i think same behaviour even applies to mean as it does to median .(referring to your discussion in first para).


Thanks.

sorry—what, specifically, are you trying to confirm?

presumably you're referring to the second paragraph—but, in that paragraph, stacey's observation is about comparisons between the median and the mean.
that paragraph contains no statements about the mean alone (or about the median alone), so i can't tell what you are saying here.

please clarify, thanks.

Hello Ron ,
Sorry , i was referring to 2nd paragraph.
the point i was trying to clarify is that
when median > mean ,are numbers below the mean more farther from mean than numbers above the mean(from mean) ?

Thanks:)

[tim]

when median > mean ,are numbers below the mean more farther from mean than numbers above the mean(from mean) ?

On average, yes. If median > mean, then more than half the numbers are greater than the mean. This means that on average they must be closer to the mean than the numbers below the mean in order for everything to balance at the mean.

[RonPurewal]

be sure to note that, in the situation described above, it is impossible to conclude anything about any individual numbers in the set.