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Concept: Avg and Median

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Concept: Avg and Median  [#permalink]

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New post 24 Oct 2007, 01:19
Can someone explain the concept from MGMAT page 85?

Given the ascending set [X, X, Y, Y, Y, Y] what is greater, median or mean?

median is Y.

Avg = sum/N
Avg = (2x + 4y) / 6

"When solving average problems, it is better to deal with the sum of the terms than their average. If the mean were greater than the median, the sum would be greater than 6y. So the question is really, is (2x + 4y) < 6y?

Since X is less than Y, (2x + 4y) is less than 6y. Therefore, mean < Y.

Given that the median is Y, the mean is less than Y, the median is greater than the mean."

Where did they get 6y from?

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Re: Concept: Avg and Median  [#permalink]

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New post 24 Oct 2007, 05:40
1
bmwhype2 wrote:
Can someone explain the concept from MGMAT page 85?

Given the ascending set [X, X, Y, Y, Y, Y] what is greater, median or mean?

median is Y.

Avg = sum/N
Avg = (2x + 4y) / 6

"When solving average problems, it is better to deal with the sum of the terms than their average. If the mean were greater than the median, the sum would be greater than 6y. So the question is really, is (2x + 4y) < 6y?

Since X is less than Y, (2x + 4y) is less than 6y. Therefore, mean < Y.

Given that the median is Y, the mean is less than Y, the median is greater than the mean."

Where did they get 6y from?


They are checking whether AVG < Y
since AVG = (2x + 4y) / 6 we have to check whether (2x + 4y) / 6 < y
which is equivalent to (2x + 4y) < 6y

Hope I helped ...
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New post 24 Oct 2007, 09:11
this little picture of family income is very good for understanding difference between average (= mean), median and mode...
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USA family income.jpg
USA family income.jpg [ 229.08 KiB | Viewed 2731 times ]

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New post 24 Oct 2007, 09:20
yea. The outliers skew the mean.
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New post 25 Oct 2007, 13:21
ronneyc wrote:
bmwhype2 wrote:
Can someone explain the concept from MGMAT page 85?

Given the ascending set [X, X, Y, Y, Y, Y] what is greater, median or mean?

median is Y.

Avg = sum/N
Avg = (2x + 4y) / 6

"When solving average problems, it is better to deal with the sum of the terms than their average. If the mean were greater than the median, the sum would be greater than 6y. So the question is really, is (2x + 4y) < 6y?

Since X is less than Y, (2x + 4y) is less than 6y. Therefore, mean < Y.

Given that the median is Y, the mean is less than Y, the median is greater than the mean."

Where did they get 6y from?


They are checking whether AVG < Y
since AVG = (2x + 4y) / 6 we have to check whether (2x + 4y) / 6 < y
which is equivalent to (2x + 4y) < 6y

Hope I helped ...


all right. i got it now.
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New post 25 Oct 2007, 16:19
Guys, the set is given with the numbers ascending, so how can we have 2x and 4y? the mean would be (x1+x2+y1+y2+y3+y4)/6 and the median is (y1 +y2)/2

I plugged in numbers and the mean equals the median if and only if the numbers increase by the same amoun, hence same standard deviation, if the numbers increase at random, then we get that the mean is highere than the median.

If this would be a DS question, i would def. go with E

is my judgement wrong?
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New post 25 Oct 2007, 21:51
u are overthinking it.

it is given that

avg vs. median
(2x+4y)/ 6 vs. Y
(2x+4y) vs. 6y
given that x is less than Y, it is clear that 2x+4y is not more than 6y

the sum is smaller than median*N
therefore the avg is smaller than the median
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New post 20 Oct 2017, 05:17
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Concept: Avg and Median &nbs [#permalink] 20 Oct 2017, 05:17
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