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Median

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Senior Manager
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Median [#permalink] New post 29 Mar 2009, 09:05
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

67% (01:28) correct 33% (00:57) wrong based on 8 sessions
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pls discuss !!
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Re: Median [#permalink] New post 03 Apr 2009, 00:53
Weighted median needs to be calculated, can anyone plz explain the solution.
I think it is based on standard deviation basics.
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Re: Median [#permalink] New post 03 Apr 2009, 01:31
E. 6

Since I so recently taught middle school, I took a basic approach. I listed 1 nine times, 2 four times, 3 one time, 4 two times, 5 one time, 6 eight times, and 7 ten times. You can then see that 6 is the middle number.

However, you didn't even have to list them all. You can see that the 18th number will be the middle number, so you could stop with the first 6.
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Re: Median [#permalink] New post 03 Apr 2009, 01:49
tenaman10 wrote:
pls discuss !!


E - 6... for the first... for reasons explained above
B - 1/6... for the second

assume mean of 20 nos as x. So n=4x.
sum of 21 nos is 20x + 4x = 24x

n/24x = 20x/24x = 1/6
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Re: Median [#permalink] New post 17 Apr 2009, 19:22
thegr81 wrote:
tenaman10 wrote:
pls discuss !!


E - 6... for the first... for reasons explained above
B - 1/6... for the second

assume mean of 20 nos as x. So n=4x.
sum of 21 nos is 20x + 4x = 24x

n/24x = 20x/24x = 1/6


i agree with b for the second question.
i solved by picking numbers. to make it easy i used 1-20 for the first 20 numbers. given that, n came out to be 42. then i took n/sum of all the numbers, including n (sum of consecutive numbers is avg (middle number) * # of terms).
Re: Median   [#permalink] 17 Apr 2009, 19:22
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