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# If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and

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If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and [#permalink]

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17 Jan 2008, 08:51
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If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and 21 are picked out, is the deviation of the remain numbers same to that of ten numbers?
1). The median of remain numbers is 10
2). The mean of remain numbers is same to that of 10 numbers.
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17 Jan 2008, 09:31
I guess D.
I think it is very difficult to keep deviation constant after subtraction numbers.
good question, +1
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17 Jan 2008, 11:38
Good question, +1 too. I picked E. I am not sure how to solve it. I just think that if you remove two numbers, the whole std must change.
Not sure, I would like to see the solution.

young_gun wrote:
If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and 21 are picked out, is the deviation of the remain numbers same to that of ten numbers?
1). The median of remain numbers is 10
2). The mean of remain numbers is same to that of 10 numbers.
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Joined: 29 Mar 2007
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17 Jan 2008, 14:15
young_gun wrote:
If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and 21 are picked out, is the deviation of the remain numbers same to that of ten numbers?
1). The median of remain numbers is 10
2). The mean of remain numbers is same to that of 10 numbers.

I say B. This is def a good question.

1: this is insuff. essentially this says that 9 and 11 were not picked. if we pick 1 or 21 then the std is the same. HOWEVER if we pick say 19. Then you can see the std will change just a bit b/c of the gap created from 17 to 21.

2: This appears to be suff. The mean of the numbers now is 11.

This means that 1 or 21 was NOT picked. for example if you take 1 out of the pick you have 21+3/2 = 12 which is NOT 11.

Same with taking out 21: 19+1/2 which is 10.

I have no idea which number was picked, but that is irrelevant and its a waste of time to find it. because we know the integer picked was NOT 1 or 21, then STD will change because there will be a gap among the consecutive odd integers. Ex: say we take out 5. Now we have a gap of 3 and 7. This gap changes the STD from the uniform std of the consecutive integers to something else.
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17 Jan 2008, 14:35
GMATBLACKBELT wrote:
1: this is insuff. essentially this says that 9 and 11 were not picked. if we pick 1 or 21 then the std is the same. HOWEVER if we pick say 19. Then you can see the std will change just a bit b/c of the gap created from 17 to 21.

Can you prove this?

I think we have something like this:

$$\sigma_{old}=\sqrt{\frac{\sum}{10}}$$

$$\sigma_{new}=\sqrt{\frac{\sum-(1-x_{av})^2-(21-x_{av})^2}{8}}$$

and $$\sigma_{old}<>\sigma_{new}$$
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17 Jan 2008, 14:45
walker wrote:
GMATBLACKBELT wrote:
1: this is insuff. essentially this says that 9 and 11 were not picked. if we pick 1 or 21 then the std is the same. HOWEVER if we pick say 19. Then you can see the std will change just a bit b/c of the gap created from 17 to 21.

Can you prove this?

I think we have something like this:

$$\sigma_{old}=\sqrt{\frac{\sum}{10}}$$

$$\sigma_{new}=\sqrt{\frac{\sum-(1-x_{av})^2-(21-x_{av})^2}{8}}$$

and $$\sigma_{old}<>\sigma_{new}$$

Lets look at the scenarios that produce a median of 10. 9, 11. 7, 13, 3, 17 and 1, 19. well if you remove any of the numbers you can never have 7,13, 3,17 and 1,19 in the middle its just not going to happen when you only remove ONE integer.
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17 Jan 2008, 14:52
Very interesting question

Give me two numbers for which $$\sigma_{new}=\sigma_{old}$$. Otherwise, 1 is sufficient
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17 Jan 2008, 14:55
My guess is D. If two numbers are taken out and the median is still ten, then you know the mean will stay the same. Correct?

The def of Standard deviation is the deviation from the mean[u][/u]. Since they give the mean you can calculate the standard deviation.
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18 Jan 2008, 00:19
GMATBLACKBELT wrote:
Ok I agree S1 is suff.

I really don't know how we can prove this for 2min.
eschn3am, what do you think?
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18 Jan 2008, 03:26
[quote="jingy77"]My guess is D. If two numbers are taken out and the median is still ten, then you know the mean will stay the same. Correct?

The def of Standard deviation is the deviation from the [u]mean[u][/u][/u]. Since they give the mean you can calculate the standard deviation.[/quote]

[quote="jingy77"]My guess is D. If two numbers are taken out and the median is still ten, then you know the mean will stay the same. Correct?

The def of Standard deviation is the deviation from the [u]mean[u][/u][/u]. Since they give the mean you can calculate the standard deviation.[/quote]

I'm not so sure about that. You could choose several sets of two numbers to take out and still leave behind a median of 10 (i.e. as long as 9 and 11 are untouched and you take out one below ten and one above). This means you could pull out 1 and 13, which would skew the whole thing towards a much higher mean, and therefore give you a different dev. I don't think S1 is sufficient.

As for S2, this means you must be pulling out a matching pair (1 and 21) or (3 and 19). I guess the question is how does this affect the sum of squared residuals. I don't really see how it could maintain the same deviation as the original set. But I'm not sure. I'd be guessing between either B or E.
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18 Jan 2008, 07:02
young_gun wrote:
If two of the ten numbers 1, 3, 5, 7, 9, 11, 13, 17, 19, and 21 are picked out, is the deviation of the remain numbers same to that of ten numbers?
1). The median of remain numbers is 10
2). The mean of remain numbers is same to that of 10 numbers.

Mean: $formdata=\frac{1+3+5+7+9+11+13+17+19+21}{10}=\frac{106}{10}=10.6$

Find the difference between each term and mean and square it:

$formdata=(10.6-1)^2=92.16$
$formdata=(10.6-3)^2=57.76$
$formdata=(10.6-5)^2=31.36$
$formdata=(10.6-7)^2=12.96$
$formdata=(10.6-9)^2=2.56$
$formdata=(10.6-11)^2=.16$
$formdata=(10.6-13)^2=7.29$
$formdata=(10.6-17)^2=40.96$
$formdata=(10.6-19)^2=70.56$
$formdata=(10.6-21)^2=108.16$

Now sum up the squares and divide it by the number of terms:

$formdata=\frac{92.16+57.76+31.36+12.96+2.56+.16+7.29+40.96+70.56+108.16}{10}=\frac{423.93}{10}=42.393$

Now find the square root of the new number:

$formdata=sqrt{42.393}=6.51099$

Now we have our standard deviation of 6.51099. For us to be able to pull ANY two numbers out and maintain our standard deviation we need to have specific numbers available to us. The total divided by 10 is equal to 42.393. We need a total of 8 numbers that, when divided by 8, equal 43.393. SO...

$formdata=42.393*8=339.144$

and

$formdata=432.93-339.144=84.786$

The only way we could pull two numbers out of this and maintain our standard deviation is by pulling two numbers whose difference of squares sums up to 84.786.

I don't believe those two numbers exist in our list, so any numbers you pull out will change the standard deviation.

This is a strange problem because there aren't any numbers you can remove and still keep the same standard deviation. Does that make the answer D or E? You know none of the numbers will work even without looking at the statements...bizarre.

HOWEVER...if you run the same problem with 11 numbers and include 15 again you have a real problem. The standard deviation works out so that you can pull a couple numbers that have a difference of squares = to 80 and it will all work out. Then we can use the statements to see if those are the numbers being pulled. Where did this problem come from? I'm almost positive you wouldn't see something like this on the actual test.
Re: DS std dev   [#permalink] 18 Jan 2008, 07:02
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