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Median

by sumithshah » Wed Nov 12, 2008 9:18 pm
What si the median of the set if
a) More than half of the numbers are >=1000
b) less than half of the numbers are <=1000


OA is C however, I'm not sure how that comes out.
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by vishubn » Wed Nov 12, 2008 9:33 pm
A) >=1000
1000,1001,1002.....

B)<=1000
1000,999,998

A+B together 1000 is the median

OA C

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by sumithshah » Wed Nov 12, 2008 9:36 pm
Doesn't matter if the number of terms are odd or even?
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by vishubn » Wed Nov 12, 2008 9:39 pm
wat ever it is !! the end limit of the either inequality matters !! So given space of time !! lotta ifs and but's might arise ..

So why not wait for some more comments to pour in 8) i am sure the tiger tamer( LOGITECH ) want to join in :)

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by jimmiejaz » Thu Nov 13, 2008 3:07 am
sumithshah wrote:Doesn't matter if the number of terms are odd or even?
i also a bit sceptical about this. we simply cant assume that if more than half are greater than 1000, exactly the other part ie numbers less than 1000. eg. suppose we have 100 numbers.
according to stmt 1.
lets say we have 60 numbers are greater than 1000.
according to stmt 2.
30 numbers are less than 1000.

now combining both. Both the statements hold true for the example.
But, since we don't know about the other 10 numbers, we can't comment on the median.
Unless, there is some logic/rule for this...
So, i will go with E.

Please correct me if am wrong.

Sumith, can u pls tell the source?
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My first post

by Tryingmybest » Thu Nov 13, 2008 4:52 am
This is my first post in this website . Please point out if my logic is wrong.
I agree with Jimmiejaz in this case.

Statement 1:
Let us consider two sets of example for this statement
Example set 1: 987, 990,1000,1001,1002,1003 Median is 1000.5

Example set 2: 987,991,992,999,1000,1001,1002,1003,1004,1006,1007 ,1008,1009,1010Median is 1002.5

Both the cases above hold good for statement one . But we dont have a definitive median.Not enough

Statement 2:

I guess the same two examples hold good for this statement . Not enough.

There is no point in combining here .

Eventually Choice E :D
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by stop@800 » Thu Nov 13, 2008 9:38 am
I am also with E

The set is
999 1002 1003
999 1003 1004
999 1001 1004
A and B both will hold true but no means to find median


Waiting for logitech, cramya :)
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by logitech » Thu Nov 13, 2008 9:51 am
Sorry I am late ;-)

Well

More than half less than half crap!

So lets call the sample size 5 than for fun

Less than half : 2
More than half : 3

1 2 3 4 5

This case 3rd term will be the median. And question does not give any info whether it has odd or even members:

So whatever we choose for the 3rd term which is greater or equal to 1000 will be our MEDIAN

So E is the answer and C is the incorrect OA
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by sumithshah » Thu Nov 13, 2008 9:53 am
How would your answer change if there were
a) Odd number of terms
b) even number of terms
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by stop@800 » Thu Nov 13, 2008 10:05 am
sumithshah wrote:How would your answer change if there were
a) Odd number of terms
b) even number of terms
No change
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by cramya » Thu Nov 13, 2008 10:17 am
Waiting for logitech, cramya
I agree with E). If its C then we are missing something big time :-)

We dont even know if 1000 is in the list(lets say odd number of terms exists after arranging them in ascending order) or adding the middle terms and dividing by 2 gets us to 1000 after again arranging them in ascending order(if there are even number of terms)

Sumit, what's the source?

If the question was half of the numbers are more than 1000 and half of the numbers are less than 1000 and we know 1000 existed on the list then by definition of median it is 1000 which would be D) if it was a DS question. Any takers for this statement?

Median = value with as many other values above as below


Still think its E) and not C) due to the varius if's and but's in this problem!!
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by logitech » Thu Nov 13, 2008 10:24 am
Guys can I assume this ?

a) More than half of the numbers are >=1000

SO the median is will be always on this side , which means Median will be either equal 1000 or greater than 1000 ?
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by lunarpower » Fri Nov 14, 2008 5:20 am
logitech wrote:Guys can I assume this ?

a) More than half of the numbers are >=1000

SO the median is will be always on this side , which means Median will be either equal 1000 or greater than 1000 ?
true.
but this is just an inequality (median > 1000). that's not the same as actually having the median.

this is definitely (e); there are all kinds of ways this could turn out.
here are just two:

* 87, 1000, 1000 (satisfies both conditions) --> median = 1000
* 87, 2000, 4000 (satisfies both conditions) --> median = 2000
insufficient.

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now, IF BOTH STATEMENTS ARE SUPPOSED TO SAY "MORE THAN", THEN THE ANSWER IS ACTUALLY (C) AS ORIGINALLY GIVEN.

in other words, if more than half the numbers are < 1000 and more than half the numbers are > 1000, then the median must be 1000.
i can explain this in more detail if anyone so desires.

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by logitech » Fri Nov 14, 2008 10:30 am
Thanks Ron. You rock!
LGTCH
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by EricLien9122 » Fri Nov 14, 2008 11:49 am
Hey Ron, thanks for the amazing explanation, but I am still not sure if C is sufficient, because the neither statement stated that the "middle" number is equal to 1000.

So...we can have -20,-10, 9999, 10000, 10001

median=9999

we also can have...

-20,-10,1000,1001,1002

median=1000

Since we can't pinpoint a specific value, shouldn't E be the right answer?

Thank you in advance Ron, I appreciate your comments.
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