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From the list of 10 consecutive odd integers >

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From the list of 10 consecutive odd integers > [#permalink] New post 18 Aug 2009, 17:48
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A
B
C
D
E

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0% (00:00) correct 100% (01:59) wrong based on 2 sessions
From the list of 10 consecutive odd integers –> 1,3,5,7,9,11,13,15,17 and 19 if 2 integers are removed then does the standard deviation remain unchanged?
(1) Median of the remaining numbers is 10
(2) The Mean of the remaining numbers is 10
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Re: Mean,median and sd GOOD DS [#permalink] New post 18 Aug 2009, 18:08
Answer should be E I think. Here is my explanation.

First of all before evaluating any statements, we must make note that the given list of numbers has a standard deviation of 2. (Since each number differs from other by 2 units)

(1) Median of the remaining numbers is 10

CASE 1 - Lets take out two numbers 9,11. Then the list becomes 1,3,5,7,13,15,17,19 (median is 10) but st. dev is definitely greater than 2 because 7 and 13 are 6 units aparts
CASE 2 - Lets take out two numebrs 1,19. Then the list becomes 3,5,7,9,11,13,15,17 (median is 10) but st. dev is 2 in this case since each number is 2 units apart from its adjoining numbers.

Hence INSUFF.

(2) The Mean of the remaining numbers is 10. Clearly insufficient.

Again same thing. Take the same examples in CASE1 and CASE 2 as above (because both of them have median = mean = 10). But st dev changes in one case while it remains the same in the other case.

Even combining (1) and (2) taking examples from CASE 1 and 2, we cannot say if st dev changes or not. Hence E.

What is the OA?
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Re: Mean,median and sd GOOD DS [#permalink] New post 19 Aug 2009, 08:46
sdrandom1 wrote:

First of all before evaluating any statements, we must make note that the given list of numbers has a standard deviation of 2. (Since each number differs from other by 2 units)


The standard deviation of the original set is not equal to 2. Remember that you calculate standard deviation by first computing the distance from each element to the mean, and in the given set, none of those distances is equal to 2. The mean of the set provided is equal to 10, and the distances from each element to the mean are:
9, 7, 5, 3, 1, 1, 3, 5, 7, 9

To compute the standard deviation, we square these distances, average those squares, and then take the square root, so the standard deviation is equal to:
\sqrt{\frac{9^2 + 7^2 + 5^2 + 3^2 + 1^2 + 1^2 + 3^2 + 5^2 + 7^2 + 9^2}{10}} = \sqrt{33} \approx 5.75

In any case, no matter what two elements are removed from the list 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, the standard deviation will change (though it's a bit of work to prove that), so you do not even need the statements to answer the question - the answer must be yes. I suppose that means the answer is D, but it doesn't seem to be a well-designed question.

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Re: Mean,median and sd GOOD DS [#permalink] New post 19 Sep 2009, 04:40
Tough question. Agree with D. Constructed an Excel table to find the question.

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Re: Mean,median and sd GOOD DS   [#permalink] 19 Sep 2009, 04:40
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