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26 May 2017, 04:43
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Are the standard deviations of set A and set B the same?
(1) \(n = 6\)
(2) Set \(A = \{n, 2, 2, 2\}\) and set \(B = \{n, n, n, 2\}\)
(1) \(n = 6\)
(2) Set \(A = \{n, 2, 2, 2\}\) and set \(B = \{n, n, n, 2\}\)
26 May 2017, 04:43
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Official Solution:
In the original condition, there is 1 variable(n). Since you need to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In case of 1), since you do not know which set you substitute n=6, it is not sufficient.
In case of 2), the average (arithmetic mean) of set \(A=\{n, 2, 2, 2\}\) is \(\frac{n+6}{4}\).
If you look at the distance between the average and each element, one "n" is about \(\frac{n+6}{4}-n = \frac{6-3n}{4}\) away from the average and each of the three "2"s is about \(\frac{n+6}{4}-2 = \frac{n-2}{4}\) away from the average.
However, the average (arithmetic mean) of set \(B = \{n, n, n, 2\}\) becomes \(\frac{3n+2}{4}\), and if you look at the distance between the average and each element, one "2" is about \(2- \frac{3n+2}{4} = \frac{6-3n}{4}\) away from the average and
Each of the three "n"s is about \(n- \frac{3n+2}{4} = \frac{n-2}{4}\) away from the average.
In other words, since set A is comprised of \(\frac{6-3n}{4}, \frac{n-2}{4}, \frac{n-2}{4}, \frac{n-2}{4}\)
and set B is also comprised of \(\frac{6-3n}{4}, \frac{n-2}{4}\), \(\frac{n-2}{4}, \frac{n-2}{4}\), the standard deviation of set A and that of set B become the same, hence yes, it is sufficient. Therefore, the answer is B. This is a "CMT 4(A: when you get C too easily, consider A or B)" question.
Answer: B
In the original condition, there is 1 variable(n). Since you need to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In case of 1), since you do not know which set you substitute n=6, it is not sufficient.
In case of 2), the average (arithmetic mean) of set \(A=\{n, 2, 2, 2\}\) is \(\frac{n+6}{4}\).
If you look at the distance between the average and each element, one "n" is about \(\frac{n+6}{4}-n = \frac{6-3n}{4}\) away from the average and each of the three "2"s is about \(\frac{n+6}{4}-2 = \frac{n-2}{4}\) away from the average.
However, the average (arithmetic mean) of set \(B = \{n, n, n, 2\}\) becomes \(\frac{3n+2}{4}\), and if you look at the distance between the average and each element, one "2" is about \(2- \frac{3n+2}{4} = \frac{6-3n}{4}\) away from the average and
Each of the three "n"s is about \(n- \frac{3n+2}{4} = \frac{n-2}{4}\) away from the average.
In other words, since set A is comprised of \(\frac{6-3n}{4}, \frac{n-2}{4}, \frac{n-2}{4}, \frac{n-2}{4}\)
and set B is also comprised of \(\frac{6-3n}{4}, \frac{n-2}{4}\), \(\frac{n-2}{4}, \frac{n-2}{4}\), the standard deviation of set A and that of set B become the same, hence yes, it is sufficient. Therefore, the answer is B. This is a "CMT 4(A: when you get C too easily, consider A or B)" question.
Answer: B
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