A certain list of 100 data has an average of 6 and a standard deviation of d where d is positive. Which of the following pairs of data, when added to the list, must result in a list of 102 data with a standard deviation less than d?
1- -6 and 0
2- 0 and 0
3- 0 and 6
4- 0 and 12
5 - 6 and 6
What is the best way to go about this? I figured if you add a negative number, naturally the average will decrease and the standard deviation will decrease. Or by adding a 12, the standard deviaiton and average both increase. But OA is E ( 6 and 6), what is the difference between this and 0 and 12?
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- jayhawk2001
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Nice question.
Let the terms be a1, a2 ... a100
d = sqrt { [ (a1-6)^2 + (a2-6)^2 + ... + (a100-6)^2 ] / 100 }
To reduce standard deviation - difference between terms and
average needs to reduce [OR] denominator i.e. number of terms
should increase [OR] both
The terms will likely change the value of the average but given that we
are adding only 2 terms in the range -6 to 12, the difference between
the new average and the old average will be very less and hence in the noise.
Now, lets look at the options --
1- -6 and 0
Numerator increases by 144+0 = 144
2- 0 and 0
Numerator increases by 36+36 = 72
3- 0 and 6
Numerator increases by 36+0 = 36
4- 0 and 12
Numerator increases by 36+36 = 72
5 - 6 and 6
Numerator increases by 0+0 = 0
As you can see, all values except for 6 and 6 tip the standard deviation
to a higher value. Only 6 and 6 will cause the numerator to remain
constant while increasing the denominator to 102
Let the terms be a1, a2 ... a100
d = sqrt { [ (a1-6)^2 + (a2-6)^2 + ... + (a100-6)^2 ] / 100 }
To reduce standard deviation - difference between terms and
average needs to reduce [OR] denominator i.e. number of terms
should increase [OR] both
The terms will likely change the value of the average but given that we
are adding only 2 terms in the range -6 to 12, the difference between
the new average and the old average will be very less and hence in the noise.
Now, lets look at the options --
1- -6 and 0
Numerator increases by 144+0 = 144
2- 0 and 0
Numerator increases by 36+36 = 72
3- 0 and 6
Numerator increases by 36+0 = 36
4- 0 and 12
Numerator increases by 36+36 = 72
5 - 6 and 6
Numerator increases by 0+0 = 0
As you can see, all values except for 6 and 6 tip the standard deviation
to a higher value. Only 6 and 6 will cause the numerator to remain
constant while increasing the denominator to 102
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thanks for the in depth reply, that taught me well lol.
This was the first question of my test. Would that ever happen on the actual GMAT?
This was the first question of my test. Would that ever happen on the actual GMAT?
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- jayhawk2001
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You have seen it on the GMATprep which is as close as it gets to the realmaolivie wrote:thanks for the in depth reply, that taught me well lol.
This was the first question of my test. Would that ever happen on the actual GMAT?
exam. So, in summary, yes, you can expect such questions on the real
deal.
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what I meant to say was can you expect a question at that level for the first question? I thought the first question was usually medium?
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Just out of curiosity how did you get the added values for the numerator increases?
-6 and 0 means a 144 by 0 increase which = 144?
and so on for the rest...
any help with this?
-6 and 0 means a 144 by 0 increase which = 144?
and so on for the rest...
any help with this?
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- Sadowski
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Actually, I think he made a typo. Remember the rules for calculating standard deviation. You find the difference between the number and the average (in this case, our numbers in one of the options are -6 and 0) then square the differences and find the average of those differences.
So, the difference between -6 and 6 is 12. If you square that you get 144. Now, with 0, you're still 6 away from the average, so the square of that is 36.
In the end, you take the non-negative root of the averages = standard deviation.
So, if you want to minimize that number, your goal is to reduce the spread. In other words, you want to add numbers to your population that are as close to the average as possible! So, since our average is 6, we can add extra 6's to the population and reduce the standard deviation (because there's now a larger population and the same spread as before).
Hopefully this helps.
So, the difference between -6 and 6 is 12. If you square that you get 144. Now, with 0, you're still 6 away from the average, so the square of that is 36.
In the end, you take the non-negative root of the averages = standard deviation.
So, if you want to minimize that number, your goal is to reduce the spread. In other words, you want to add numbers to your population that are as close to the average as possible! So, since our average is 6, we can add extra 6's to the population and reduce the standard deviation (because there's now a larger population and the same spread as before).
Hopefully this helps.
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So why won't 0 and 0 be an answer? the difference is just the same as 6 and 6 would be
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- Sadowski
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Because you have to consider the average of the population. It's not the difference between the two numbers you're adding, but the difference between each of the numbers and the average.
Ex: If the average of the population were 0, then adding 0 and 0 would be the correct answer - it would reduce the overall spread by increasing the population, but allowing the average to remain the same.
However, in our example, the average is 6. The difference between 6 (the number we're adding to the population) and 6 (the average of the population) is zero. Standard deviation, thus, goes down while the average stays the same.
Is that more clear?
Ex: If the average of the population were 0, then adding 0 and 0 would be the correct answer - it would reduce the overall spread by increasing the population, but allowing the average to remain the same.
However, in our example, the average is 6. The difference between 6 (the number we're adding to the population) and 6 (the average of the population) is zero. Standard deviation, thus, goes down while the average stays the same.
Is that more clear?