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Yesterday each of the 35 members of a certain task force spent some ti

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Yesterday each of the 35 members of a certain task force spent some ti [#permalink] New post 29 Mar 2009, 09:05
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E

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Question Stats:

64% (02:08) correct 36% (01:58) wrong based on 82 sessions
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Yesterday each of the 35 members of a certain task force spent some time working on project P. The graph shows the number of hours and the number of members who spent that number of hours working on project P yesterday. What was the median number of hours that the members of the task force spent working on project P yesterday?

A. 2
B. 3
C. 4
D. 5
E. 6

OPEN DISCUSSION OF THIS QUESTION IS HERE: yesterday-each-of-the-35-members-of-a-certain-task-force-85576.html
[Reveal] Spoiler: OA

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Re: Yesterday each of the 35 members of a certain task force spent some ti [#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: Yesterday each of the 35 members of a certain task force spent some ti [#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: Yesterday each of the 35 members of a certain task force spent some ti [#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: Yesterday each of the 35 members of a certain task force spent some ti [#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).
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Re: Yesterday each of the 35 members of a certain task force spent some ti [#permalink] New post 06 Nov 2014, 19:05
Can someone please explain the solution ? I'm not getting it at all :(

update :
I got it . Never mind :)
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Re: Yesterday each of the 35 members of a certain task force spent some ti [#permalink] New post 07 Nov 2014, 03:04
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ammara1988 wrote:
Can someone please explain the solution ? I'm not getting it at all :(

update :
I got it . Never mind :)


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Yesterday each of the 35 members of a certain task force spent some time working on project P. The graph shows the number of hours and the number of members who spent that number of hours working on project P yesterday. What was the median number of hours that the members of the task force spent working on project P yesterday?

A. 2
B. 3
C. 4
D. 5
E. 6

We have set consisting of 35 terms (terms=number of members): 9+4+1+2+1+8+10=35. Median of a set, with odd number of terms, would be the middle term, so 18th term. Values of terms: 9 terms=1, 4 terms=2, 1 term=3, 2 terms=4 , 1 term=5, 8 terms=6 and 10 terms=7 --> 18th term is 6.

To illustrate:
1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: yesterday-each-of-the-35-members-of-a-certain-task-force-85576.html
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Re: Yesterday each of the 35 members of a certain task force spent some ti   [#permalink] 07 Nov 2014, 03:04
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