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12 Oct 2015, 21:58
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (01:34) correct
29%
(01:43)
wrong
based on 695
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Set J consists of the terms {a, b, c, d, e}, where e > d > c > b > a > 1. Which of the following operations would decrease the standard deviation of Set J?
A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e
A. Multiply each term by e/d
B. Divide each term by b/c
C. Multiply each term by −1
D. Divide each term by d/e
E. Multiply each term by c/e
12 Oct 2015, 23:20
Kudos
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Bunuel
CONCEPT: Standard Deviation is Defined as Average Deviation of Terms in the set from the Mean value of the set. i.e.
1) It depends on the separation between the successive terms of the set
2) If a Constant Value is Added/Subtracted in every terms of set then the Separation between successive terms does NOT change Hence S.D. remains Constant
3) If a Constant Value is Multiplied in every terms then the Separation between succesive terms gets multiplied by the constant Hence S.D. remains gets multiplied by same Number
A. Multiply each term by e/d is equivalent to multiplying each term with some value>1 which will Increase the SD
B. Divide each term by b/c is equivalent to Dividing each term with some value<1 which will Increase the SD because b/c<1
C. Multiply each term by −1 will keep the deviation of the terms same only the signs of the terms in set will change hence, No change in SD
D. Divide each term by d/e is equivalent to Dividing each term with some value<1 which will Increase the SD because d/e<1
E. Multiply each term by c/e is equivalent to multiplying each term with some value<1 which will DECREASE the SD by the coefficient c/e
Answer: option E
General Discussion
12 Oct 2015, 23:58
Kudos
Bookmarks
Bunuel
In these type of questions my suggestion would be to consider nos which are easy to multiply or divide
In our case lets consider (5,10,15,20,25) for (a,b,c,d,e) which takes care of the above condition 1<a<b<c<d<e
The (mean-no) differences are (10,5,0,5,10). With this lets get into the options
A. Multiply each term by e/d - 25/20 = 5/4, which is > 1. This will make the STD.dev wider.
B. Divide each term by b/c - 10/15 = 2/3 = 0.66
C. Multiply each term by −1 = No change in Std.dev
D. Divide each term by d/e = sames as multiplying by e/d
E. Multiply each term by c/e - 15/25 = 3/5 = 0.6
As you clearly see, only options B & E bring the difference closer (Multiply the fraction with the original nos to see the gap being reduced). Out of which, 0.6<0.666. Thus Option E
