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16 Sep 2014, 00:41
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Is the range of data set \(S\) greater than its average (arithmetic mean)?
(1) All elements in data set \(S\) are negative.
(2) The median of data set \(S\) is negative.
(1) All elements in data set \(S\) are negative.
(2) The median of data set \(S\) is negative.
16 Sep 2014, 00:41
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Official Solution:
Is the range of data set \(S\) greater than its average (arithmetic mean)?
Note that the range of any set is always greater than or equal to zero.
(1) All elements in data set \(S\) are negative.
Since the mean of a set with all negative elements is also negative, it will always be less than its range (which is always non-negative). Sufficient.
(2) The median of data set \(S\) is negative.
This suggests that the data set contains at least one negative element. Let's examine two possible cases:
A. If all elements in set \(S\) are negative, then we have the same scenario as in the first statement, and therefore, \(Range \gt Mean\);
B. If not all elements in set \(S\) are negative, then since \(Range = Largest - Smallest\), the range will be greater than the largest element (since the smallest element in set \(S\) is negative). For example, consider the set {-1, -1, 2}: the range of this set is \(Range = 2 - (-1) = 3 \gt 2\). Conversely, the mean will be less than the largest element, indicating that \(Range \gt Largest \gt Mean\).
So, in any case \(Range \gt Mean\). Sufficient.
Answer: D
Is the range of data set \(S\) greater than its average (arithmetic mean)?
Note that the range of any set is always greater than or equal to zero.
(1) All elements in data set \(S\) are negative.
Since the mean of a set with all negative elements is also negative, it will always be less than its range (which is always non-negative). Sufficient.
(2) The median of data set \(S\) is negative.
This suggests that the data set contains at least one negative element. Let's examine two possible cases:
A. If all elements in set \(S\) are negative, then we have the same scenario as in the first statement, and therefore, \(Range \gt Mean\);
B. If not all elements in set \(S\) are negative, then since \(Range = Largest - Smallest\), the range will be greater than the largest element (since the smallest element in set \(S\) is negative). For example, consider the set {-1, -1, 2}: the range of this set is \(Range = 2 - (-1) = 3 \gt 2\). Conversely, the mean will be less than the largest element, indicating that \(Range \gt Largest \gt Mean\).
So, in any case \(Range \gt Mean\). Sufficient.
Answer: D
General Discussion
popov
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31 Jul 2015, 00:00
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Bunuel
What if the set has only one number Set: {x}. Can we rule out this option saying that range is not defined for set consisting of one number? Or range is 0 for this set?
