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07 Jun 2017, 21:46
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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:
65%
(hard)
Question Stats:
41% (01:15) correct
59%
(01:34)
wrong
based on 29
sessions
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If a, b, c and d are distinct positive integers such that a < b < c < d, which of the following sets has the lowest standard deviation?
a. {a, b, c, d}
b. {-a, -b, -c, -d}
c. {100 – a, 100 – b, 100 – c, 100 - d}
d. {a +1, b + 2, c + 3, d + 4}
e. {a -1, b - 2, c - 3, d - 4}
a. {a, b, c, d}
b. {-a, -b, -c, -d}
c. {100 – a, 100 – b, 100 – c, 100 - d}
d. {a +1, b + 2, c + 3, d + 4}
e. {a -1, b - 2, c - 3, d - 4}
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07 Jun 2017, 22:25
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The closer a group of numbers are to each other, the lower is the standard deviation of that group.
We are given that a<b<c<d.
Std Deviation of {a,b,c,d} is going to be same as {-a,-b,-c,-d}, since the respective differences among the numbers in both these sets are going to be the same. So these two options cannot be the answer.
In case of {100-a, 100-b, 100-c, 100-d} we are adding a same number '100' to each of -a, -b, -c, -d. Adding the same number to each of the terms of a set does NOT change its Std Deviation. So Std deviation of {100-a, 100-b, 100-c, 100-d} is going to be exactly the same as {-a, -b, -c, -d} so this also cannot be the answer.
{a+1, b+2, c+3, d+4} is going to further increase the respective differences among the four integers even more, so the Std deviation is going to increase rather than decrease. So this also cannot be the answer.
But {a-1, b-2, c-3, d-4} is such a set where the respective differences among the four integers are going to decrease, hence Std deviation is going to be lowest,
lower than {a, b, c, d}.
Hence E answer
We are given that a<b<c<d.
Std Deviation of {a,b,c,d} is going to be same as {-a,-b,-c,-d}, since the respective differences among the numbers in both these sets are going to be the same. So these two options cannot be the answer.
In case of {100-a, 100-b, 100-c, 100-d} we are adding a same number '100' to each of -a, -b, -c, -d. Adding the same number to each of the terms of a set does NOT change its Std Deviation. So Std deviation of {100-a, 100-b, 100-c, 100-d} is going to be exactly the same as {-a, -b, -c, -d} so this also cannot be the answer.
{a+1, b+2, c+3, d+4} is going to further increase the respective differences among the four integers even more, so the Std deviation is going to increase rather than decrease. So this also cannot be the answer.
But {a-1, b-2, c-3, d-4} is such a set where the respective differences among the four integers are going to decrease, hence Std deviation is going to be lowest,
lower than {a, b, c, d}.
Hence E answer
08 Jun 2017, 00:28
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quantumliner
We need to look at the set that has the least dispersal about the mean. Lets take an example with 1, 2, 3, 4.
The standard deviation for the first three cases (1,2,3,4); (-1,-2,-3,-4); (99,98,97,96) will be the same as they have the same average dispersion.
For option (D), we have to evaluate the average dispersion of 2,4,6,8, which will certainly be more than that of (E) at 0,0,0,0.
Clearly (E).
