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Difficulty:
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60% (01:42) correct
40%
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I. j; k; m; n; p
II. j - 10; m; m; m; p + 15
III. j + 2; k + 1; m; n - 1; p - 2
If j, k, m, n, and p are consecutive positive integers such that j < k < m < n < p, the data sets I, II, and III above are ordered from greatest standard deviation to least standard deviation in which of the following?
(A) I, III, II
(B) II, I, III
(C) II, III, I
(D) III, I, II
(E) III, II, I
II. j - 10; m; m; m; p + 15
III. j + 2; k + 1; m; n - 1; p - 2
If j, k, m, n, and p are consecutive positive integers such that j < k < m < n < p, the data sets I, II, and III above are ordered from greatest standard deviation to least standard deviation in which of the following?
(A) I, III, II
(B) II, I, III
(C) II, III, I
(D) III, I, II
(E) III, II, I
07 Jun 2014, 08:13
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Gnpth
If j, k, m, n, and p are consecutive positive integers such that j < k < m < n < p, the data sets I, II, and III above are ordered from greatest standard deviation to least standard deviation in which of the following?
(A) I, III, II
(B) II, I, III
(C) II, III, I
(D) III, I, II
(E) III, II, I
(A) I, III, II
(B) II, I, III
(C) II, III, I
(D) III, I, II
(E) III, II, I
I. j; k; m; n; p
II. j - 10; m; m; m; p + 15
III. j + 2; k + 1; m; n - 1; p - 2
If j, k, m, n, and p are consecutive positive integers such that j < k < m < n < p, the data sets I, II, and III above are ordered from greatest standard deviation to least standard deviation in which of the following?
(A) I, III, II
(B) II, I, III
(C) II, III, I
(D) III, I, II
(E) III, II, I
j, k, m, n, and p are consecutive positive integers such that j < k < m < n < p, thus the third set is {m, m, m, m, m}. The standard deviation of a set with all identical elements is 0 (the SD is always \(\geq{0}\)). Therefore the third set has the smallest standard deviation.
Only option B offers that.
Answer: B.
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Hope this helps.
General Discussion
21 Aug 2014, 13:36
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A 30sec Approach
Range of 1st Set = (p-j)
Range of 2nd Set = P+15 - j +10 = (p-j) +25
Range of 3rd Set = P -2 - j - 2 = (p-j) - 4
R2 > R1 > R3 ==> SD2 > SD1 > SD3
Since, Higher the Range -> More Dispersion in the Set -> More Standard Deviation.
Hope it helps.
Range of 1st Set = (p-j)
Range of 2nd Set = P+15 - j +10 = (p-j) +25
Range of 3rd Set = P -2 - j - 2 = (p-j) - 4
R2 > R1 > R3 ==> SD2 > SD1 > SD3
Since, Higher the Range -> More Dispersion in the Set -> More Standard Deviation.
Hope it helps.
