Average + Median

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carllecat
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Topic: Average + Median
PostTue Nov 17, 2009 10:09 am
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The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?

1) The average (arithmetic mean) of the integers in S is less than the average (arithmetic mean) of the integers in T.

2) The median of the integers in S is greater than the median of the integers in T.

Let's say S is composed of 2,2,2,2 &2 , then the sum of these numbers is 10 and then mean is 2 and T is composed of 5 & 5 for a total sum of 10 and a mean of 5.

How can I assess 2)?
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Kalvin
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PostWed Nov 18, 2009 6:59 pm
1, 1, 1, 1, 1, 100

vs

1, 26, 26, 26, 26

or

1, 1, 103

vs

1, 26, 26, 26, 26
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PostWed Nov 18, 2009 10:12 pm
Whats the OA ? E ?
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PostThu Nov 19, 2009 4:23 am
IMO A
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PostThu Nov 19, 2009 5:17 pm
The simple average formula can be expressed: Avg = SUM/# , or as SUM = Avg*#

The stimulus tells us that SUM(t) = SUM(s), therefore we know the products of their respective Avgs and #s must be equal

(1) Avg(s) < Avg(t), therefore #(t) >#(s) to make Avg*# = Avg*#. Sufficient.

(2) The stimulus does not say that the group of numbers is evenly distributed. Therefore, outliers can skew the mean in either direction while keeping the median the same. Regardless, the relationship of the median doesn't provide information about the relative averages or number of integers. Insufficient.
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carllecat
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PostThu Nov 19, 2009 6:56 pm
A Statement 1 alone is sufficient.
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