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The standard deviation of the set 1, 5, 7, 19

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sandy
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The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   03 Jul 2016, 16:03
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Quantity A
Quantity B
The standard deviation of the set 1, 5, 7, 19
The standard deviation of the set 0, 5, 7, 20


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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B
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   04 Jul 2016, 21:05
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Both sets have same mean, but second set has more dispersion from the mean, hence SD of the second set will be more. hence B
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   01 Oct 2017, 06:14
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Just to be a little more formal I provide my solution.

Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.

Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.

Answer B!
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   01 Oct 2017, 07:36
IlCreatore wrote:
Just to be a little more formal I provide my solution.

Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.

Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.

Answer B!


COuld you plz explain, i failed to understand

1. how you got the mean of setA = mean of set B = 10?

2. 9+9/2 = 9? how we get?
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   01 Oct 2017, 07:47
pranab01 wrote:
IlCreatore wrote:
Just to be a little more formal I provide my solution.

Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.

Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.

Answer B!


COuld you plz explain, i failed to understand

1. how you got the mean of setA = mean of set B = 10?

2. 9+9/2 = 9? how we get?


The results are that because I got rid of the numbers who are equal in the two columns, i.e. 5 and 7. In that way we get that the mean is 1+19/2 = 10 and 0+20/2 = 10. Than the sd are (10-1)+(19-10)/2 = 9 and (10-0)(20-10)/2 = 10. This is kind of an approximation that is faster to compute.

However, you could have worked with the two full list as well. Getting the same mean of 8 for the two lists and sd = 5.5. for quantity A and sd = 6 for quantity B. Leading to the same answer, B!
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   01 Oct 2017, 08:59
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IlCreatore wrote:
pranab01 wrote:
IlCreatore wrote:
Just to be a little more formal I provide my solution.

Since the two groups have the same mean (= 10), we can get rid of the common numbers on the two lists, i.e. 5 and 7, since they would provide the same difference from the mean for the two columns. Then, we are left with 1 and 19 on column A and 0 and 20 on column B.

Instead of using the real standard deviation formula, we can get an idea using an approximate formula that is computed as the mean of the differences of the numbers from the mean. Thus, in column A we get 9+9/2 = 9 and on column B, 10+10/2 = 10. Thus, the standard deviation is higher in column B.

Answer B!


COuld you plz explain, i failed to understand

1. how you got the mean of setA = mean of set B = 10?

2. 9+9/2 = 9? how we get?


The results are that because I got rid of the numbers who are equal in the two columns, i.e. 5 and 7. In that way we get that the mean is 1+19/2 = 10 and 0+20/2 = 10. Than the sd are (10-1)+(19-10)/2 = 9 and (10-0)(20-10)/2 = 10. This is kind of an approximation that is faster to compute.

However, you could have worked with the two full list as well. Getting the same mean of 8 for the two lists and sd = 5.5. for quantity A and sd = 6 for quantity B. Leading to the same answer, B!


From where have you got the formula for SD : Im not sure of that formula

Secondly if we look at both qty's

we can see that the smaller number in qty B < qty A and bigger number in qty B> qty A.

So definitely QTY B is more spread out than QTY A. and so the option is B
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The standard deviation of the set 1, 5, 7, 19 vs 0, 5, 7, 20 [#permalink] New post   01 Sep 2020, 17:11
[quantity]The standard deviation of set 1, 5, 7, 19[/quantity]

[quantity]The standard deviation of the set 0, 5, 7, 20[/quantity]




A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.



Source: Manhattan prep 5 lb. Chapter 33, section 2, Question 4

Show: ::
B. Textbook explanation: (B). Standard deviation measures the variance from the mean; the more spread out a set is, the higher the deviation. The set in Quantity B is the same as the one in Quantity A, but with the smallest number even smaller and the largest number even larger, so the set in Quantity B is more spread out and thus has a greater standard deviation.



Can someone check my understanding of the standard deviation for the GRE? Given the explanation of the textbook, it looks like we are measuring the distance from the mean regardless of the sign. So going from 1 to 0 adds a distance of 1 from the mean. Similarly, going from 19 to 20 adds a distance of 1 to the mean. Thus the standard deviation of quantity B is +2 STD units higher than quantity A?

What if quantity B was 1, 5, 7, 18? We gained a distance of 1 by going from 0 to 1, but we lost a distance of 1 by going from 19 to 18. Would the answer be C in this case? Thank you for explaining this!
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   02 Sep 2020, 01:25
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Merged similar posts.

Please

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2) Follow the rules to format it https://gre.myprepclub.com/forum/rules-for ... -1083.html

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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink] New post   08 Nov 2024, 17:32
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The range of first series is 18 and 20 in the second. And that is what it is. SD is about the places of numbers and how far they are away each other.
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Re: The standard deviation of the set 1, 5, 7, 19 [#permalink]
08 Nov 2024, 17:32
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