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Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8

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Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 03 Mar 2015, 06:59
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A
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C
D
E

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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 03 Mar 2015, 14:29
2
N: no variance
L: sum of absolute variance = 1+2+0+0+2+1
L: sum of absolute variance = 3 +3+3+3+3+3

answer : N L M
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 03 Mar 2015, 21:55
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1
Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


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Square of Standard deviation is sum of squares of deviation from the mean divided by the number of terms.
If deviation from the mean is higher for the same number of elements, standard deviation will be higher.

L = {3, 4, 5, 5, 6, 7}
Mean is 5 and deviation of 2 elements from mean is 0, of 2 elements is 1 and of 2 elements is 2.

M = {2, 2, 2, 8, 8, 8}
Mean is 5 and deviation of all elements from the mean is 3. So SD here will be higher than SD of L.

N = {15, 15, 15, 15, 15, 15}
Mean is 15 here and all elements have 0 deviation from mean so SD = 0 here. This is the minimum possible SD.

Answer N, L, M
Answer (D)
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 03 Mar 2015, 23:53
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Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


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Looking at the sets, the first thing to do is to finalize N as the set with least Std. dev.(0) as all the terms have the same value.

That leaves us option D & E

By looking at L & M, we see that M has elements which are far from the mean 5 while the set L has elements closer to mean 5 and also two terms have the same value 5. Thus L will have lesser Std. dev than M.

Thus the order is N<L<M

Option D
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 06 Mar 2015, 07:25
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N = {15, 15, 15, 15, 15, 15}
If deviation from the mean is higher standard deviation will be higher.
Hence N is lowest ,then we have 2 options D or E.



Let us start with M as we can observe distribution is symetric

For M = {2, 2, 2, 8, 8, 8}
The median is 5, and as distribution is symetric ,deviation from mean is higher at 3
L = {3, 4, 5, 5, 6, 7}
The median is 5, and as distribution is symetric , highest deviation from mean is higher at 2

M = {2, 2, 2, 8, 8, 8}

Answer (D)
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 06 Mar 2015, 08:57
Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


Kudos for a correct solution.


+1 for D. N,L,M is the correct answer
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 06 Mar 2015, 09:29
1
Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


Kudos for a correct solution.


wider the spread of the numbers , more is the SD..
here the least spread numbers are in N, then L and finally M..
so ans D
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Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 08 Mar 2015, 14:53
Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


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MAGOOSH OFFICIAL SOLUTION:

OK, well first of all, set N has six numbers that are all the same. When all the members of a set are identical, the standard deviation is zero, which is the smallest possible standard deviation. So, automatically, N, must have the lowest. Right away, we can eliminate (A) & (B) & (C). In fact, even if we could do nothing else in this problem, we could guess randomly from the remaining two answers, and the odds would be in our favor. See this post for more on that strategy.

Now we have to compare the standard deviations of Set L and Set M. In Set L, the mean is clearly 5: two of the entries equal 5, so they have a deviation from the mean of zero, and no entry is more than two units from the mean. By contrast, in Set M, the mean is also 5, and here, every number is 3 units away from the mean, so the standard deviation of M is 3. No number in Set L is as much as 3 units away from the mean, so whatever the standard deviation of L is, it absolutely must be less than 3. That means, Set L has the second largest standard deviation, and Set M has the largest of the three. N, L, M in increasing order.

Answer = D.
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8  [#permalink]

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New post 06 Mar 2019, 20:06
Bunuel wrote:
Consider the following sets:
L = {3, 4, 5, 5, 6, 7}
M = {2, 2, 2, 8, 8, 8}
N = {15, 15, 15, 15, 15, 15}
Rank those three sets from least standard deviation to greatest standard deviation.

A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M
E. N, M, L


Kudos for a correct solution.



Since all the numbers in set N are the same, the standard deviation (SD) of set N is 0, and its SD is the smallest since the SD of any set can’t be less than 0.

Since the other two sets have at least two distinct numbers, their SDs are greater than 0. Now, let’s delve deeper into these two sets. We see that they both have a mean of 5. However, the numbers in set L are much closer to 5 (all within 2 units of 5) than those in set M (all are 3 units from 5); therefore, set L has a smaller SD than set M. So the order of the sets ranked by their SDs from least to greatest is: N, L, M.

Answer: D
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Re: Consider the following sets: L = {3, 4, 5, 5, 6, 7} M = {2, 2, 2, 8, 8   [#permalink] 06 Mar 2019, 20:06
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