Is it true that when the number of terms in a set is even, then it can't be concluded that 50% of terms are greater/less than or equal to the median?
But then, the I found the following statement in an official answer: The average of a set of consecutive integers is also the middle of the set, with an equal number of terms greater and smaller than it.
Is that statement only ALWAYS true for when the number of consecutive terms are NOT even?
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Median of an even set
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OptimusPrep
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Median of a set with even number of terms = average of middle two terms when the set is in ascending/descending orderuniyal01 wrote:Is it true that when the number of terms in a set is even, then it can't be concluded that 50% of terms are greater/less than or equal to the median?
But then, the I found the following statement in an official answer: The average of a set of consecutive integers is also the middle of the set, with an equal number of terms greater and smaller than it.
Is that statement only ALWAYS true for when the number of consecutive terms are NOT even?
Median of a set with odd number of terms = middle term when the set is in ascending/descending order
In an equidistant set such as a set of consecutive integers, the median will always be the average of all the terms.
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Hi uniyal01,
When dealing with GMAT questions, you really have to pay careful attention to the information that you're given. The MEDIAN of a group and the AVERAGE of that group are not necessarily the same thing.
However, when dealing with CONSECUTIVE INTEGERS, the median and the average will be the SAME (and the same number of terms will be above the median as below it).
For example, a group with an odd number of terms....
1, 2, 3, 4, 5
Average = 15/5 = 3
Median = 3
And a group with an even number of terms...
2, 3, 4, 5
Average = 14/4 = 3.5
Median = (3+4)/2 = 3.5
If you're NOT dealing with CONSECUTIVE INTEGERS though, these rules don't necessarily hold true.
GMAT assassins aren't born, they're made,
Rich
When dealing with GMAT questions, you really have to pay careful attention to the information that you're given. The MEDIAN of a group and the AVERAGE of that group are not necessarily the same thing.
However, when dealing with CONSECUTIVE INTEGERS, the median and the average will be the SAME (and the same number of terms will be above the median as below it).
For example, a group with an odd number of terms....
1, 2, 3, 4, 5
Average = 15/5 = 3
Median = 3
And a group with an even number of terms...
2, 3, 4, 5
Average = 14/4 = 3.5
Median = (3+4)/2 = 3.5
If you're NOT dealing with CONSECUTIVE INTEGERS though, these rules don't necessarily hold true.
GMAT assassins aren't born, they're made,
Rich
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800_or_bust
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I'm confused. You can NEVER conclude that 50% of the terms are greater than or equal to the median, and 50% are less than or equal to the median. In an extreme case (i.e. a set with SD=0) all 100% of the terms could be equal to the median, in which case 100% of the terms are greater than or equal to the median and 100% of the terms are less than or equal to the median. This would be true regardless of whether the set had an even or odd number of terms.uniyal01 wrote:Is it true that when the number of terms in a set is even, then it can't be concluded that 50% of terms are greater/less than or equal to the median?
But then, the I found the following statement in an official answer: The average of a set of consecutive integers is also the middle of the set, with an equal number of terms greater and smaller than it.
Is that statement only ALWAYS true for when the number of consecutive terms are NOT even?
800 or bust!
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Matt@VeritasPrep
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We certainly can.800_or_bust wrote:You can NEVER conclude that 50% of the terms are greater than or equal to the median, and 50% are less than or equal to the median.
{1, 2, 3, 4}
Only 1 and 2 satisfy x ≤ median, and only 3 and 4 satisfy x ≥ median, so again we'd have 50% ≤ and 50% ≥.
To satisfy x ≤ y, you don't need to have some x = y.
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ceilidh.erickson
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I think the fundamental thing that you're missing here is that when you have an EVEN number of terms in a consecutive set, the median/average is NOT one of your terms. Look at Rich's example:uniyal01 wrote:Is it true that when the number of terms in a set is even, then it can't be concluded that 50% of terms are greater/less than or equal to the median?
But then, the I found the following statement in an official answer: The average of a set of consecutive integers is also the middle of the set, with an equal number of terms greater and smaller than it.
Is that statement only ALWAYS true for when the number of consecutive terms are NOT even?
2, 3, 4, 5
Average = 14/4 = 3.5
Median = (3+4)/2 = 3.5
3.5 is NOT a term in the set. Thus, half the terms are less than 3.5, and half are greater.
When you have an ODD number of terms in a consecutive set, the median/average IS one of the terms in the set. Again look at Rich's example:
1, 2, 3, 4, 5
Average = 15/5 = 3
Median = 3
3 IS a term in the set. Two terms are less, two terms are greater. (That's not the same as 50% less, 50% greater, because the 3 accounts for 20% of the terms in this particular set).
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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ceilidh.erickson
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We're specifically talking about consecutive / evenly spaced sets here, so the example of all terms being equal does not apply.800_or_bust wrote: I'm confused. You can NEVER conclude that 50% of the terms are greater than or equal to the median, and 50% are less than or equal to the median. In an extreme case (i.e. a set with SD=0) all 100% of the terms could be equal to the median, in which case 100% of the terms are greater than or equal to the median and 100% of the terms are less than or equal to the median. This would be true regardless of whether the set had an even or odd number of terms.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
