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For the numbers 95, 96, 97, 98, 99, 100, 101, 102, 103, 104,

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For the numbers 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, [#permalink]

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New post 25 Jul 2008, 03:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For the numbers 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, and 105, if m represents the mean and d represents the standard deviation, how many are greater than m-d and less than m+d?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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New post 25 Jul 2008, 03:45
Rules for normally distributed data (Taken from WiKi)
In practice, one often assumes that the data are from an approximately normally distributed population. This is ideally justified by the classical central limit theorem, which says that sums of many independent, identically distributed random variables tend towards the normal distribution as a limit. If that assumption is justified, then about 68% of the values are within 1 standard deviation of the mean, about 95% of the values are within two standard deviations and about 99.7% lie within 3 standard deviations. This is known as the 68-95-99.7 rule, or the empirical rule.

We have 11 items and 68% will falls within m+d and m-d.
11*68/100 = 7 (approximately)

IMO D.

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New post 25 Jul 2008, 03:55
The set is not normally distributed, sorry.

It is at most "uniformly distributed" (normal distribution is a continous probability function and has nothing to do with the set here).

Good answer is (D) though, but it is not because of what you wrote. ;)

Edit see my answer to the same question posted a few days ago:

Oski wrote:
m is 100

and \(d^2 = \frac{5^2+4^2+3^2+2^2+1^2+0^2+1^2+2^2+3^2+4^2+5^2}{11} = 10\) so \(d = sqrt{10} = 3.X\)

Therfore: m-d = 96.Y and m+d = 103.X

97,98,99,100,101,102,103 are in this range

==> Answer is (D)

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New post 25 Jul 2008, 05:22
Yes, exactly- the distribution is a uniform discrete distribution, and is not at all like a normal distribution. It's actually a bad idea to learn the 68-95-99.7 rule for the GMAT- you won't need it, and if you try to use it, you'll almost certainly be using it in a case where it does not apply. It's actually very easy to make a set where almost no values are within one standard deviation of the mean - the 68% number from normally distributed data is only applicable in special circumstances. If you look at, say:

{-10, -10, -10, 10, 10, 10}

Then the mean is 0 and the standard deviation is 10. That is, 100% of values are within one standard deviation of the mean. Let's make the set huge- we could have one trillion values all equal to 10, and one trillion values all equal to -10. We still have a mean of 0 and a standard deviation of 10: again, 100% of values are within one standard deviation of the mean. Now add 0 to the set. If we add a value to a set that is equal to the mean, the standard deviation will go down (unless the range is zero)- the standard deviation is now (very slightly) less than 10. We have two trillion-and-one values in the set, and two trillion of these values are more than one standard deviation away from the mean.
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New post 25 Jul 2008, 06:19
Interesting set of problems ... I didn't expect to see those on the GMAT.

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New post 25 Jul 2008, 06:53
m = 100
\(d = sqrt{10}=3.1\)

m+d=103.1
m-d=106.9

97, 98, 99, 100, 101, 102, 103 are within the range

7 numbers

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Re: standard deviation.   [#permalink] 25 Jul 2008, 06:53
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For the numbers 95, 96, 97, 98, 99, 100, 101, 102, 103, 104,

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