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# Standard Deviation Formula

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Intern
Joined: 10 Aug 2013
Posts: 16
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28 Mar 2015, 21:34
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How can we calculate S.D if we know number of terms ,n, and that the series of numbers is consecutive.
I remember there is a direct formula.Can someone pls share it.
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28 Mar 2015, 22:40
Hi AIMGMAT770,

The GMAT is NEVER going to ask you to calculate Standard Deviation, so you don't need to know the formula to do so; instead, you'll be tested on your general knowledge of the concept and the ideas that it is based on. It's not a big subject on Test Day (you'll likely see just 1 question on it). The basic premise in most SD questions is about how "spread out" a group of numbers is, how to raise/lower the SD of the group or some type of general comparison (eg. which group has the higher SD).

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30 Mar 2015, 23:15
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Expert's post
AIMGMAT770

I have attached a snippet on Standard Deviation from the Official Guide to GMAT. You can use this approach to compute standard deviation for any set of numbers. GMAT in general does not ask you to compute standard deviation of a set of numbers, however if they do ask you to compute it then they will give you the formula for standard deviation(they have tested it) and you do need to know how to use it. The example given in the Official Guide should be enough to take on any computation problem on the GMAT.

What is more important is to understand the conceptual meaning of standard deviation. For example, the case of consecutive integers is interesting. What is the standard deviation of five consecutive integers? Does it depend on the specific set of consecutive integers, meaning do these two sets: {1,2,3,4,5} and {13,14,15,16,17}, have different standard deviations. You should be able to answer this question once you conceptually understand what standard deviation is all about. You will also need to know how the average of a set of consecutive integers is related to the numbers in the set.

My advice is not to worry about memorizing special cases, but instead to have the ability to deal with them as needed.
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StdDeviationGMAT.png [ 88.36 KiB | Viewed 1488 times ]

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Joined: 04 Jan 2015
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18 Apr 2015, 06:28
AIMGMAT770 wrote:
How can we calculate S.D if we know number of terms ,n, and that the series of numbers is consecutive.
I remember there is a direct formula.Can someone pls share it.

Dear AIMGMAT770,

As has already been pointed above, you're not going to see a PS question of the type:

'What is the Standard Deviation of the list {2, 3, 4, 5, 6}?
(A) some value
(B) some value . ..
(E) some value "

The GMAT will not test your ability to calculate Standard Deviation. Instead, it'll test your understanding of the idea that Standard Deviation measures how spread out the data in a given set is.

So, the questions that you can get on Standard Deviation will be of the following types:

1.
Set P consists of 5 distinct integers. What is the standard deviation of set P?
(1) The mean of the set P is equal to the median of the set
(2) The 5 integers are consecutive

(Please note that since this is a DS question, you'll not be required to actually solve for the value of Standard Deviation. As you'll notice, this question is built upon the conceptual question posed by dabral in the post above)

2.
Which of the following sets will have the highest Standard Deviation?
For the following sets, which of the set would have the highest standard deviation
(A) {10, 20, 30}
(B) {99, 100, 101}
(C) {195, 200, 205}
(D) {992, 1000, 1008}
(E) {10001, 10002, 10003}

3.
For two sets P = {10, 20, 30} and Q= {15, 20, 25}, which of the following statements are true?
I. Standard Deviation(P) > Standard Deviation(Q)
II. If 5 is added to each term of set P, the standard deviation of set P will become equal to the standard deviation of Set Q
III. If each term of Set Q is multiplied by 2, the standard deviation of Set P will become equal to the standard deviation of Set Q.

Please also find below the links to a couple of Official questions on Standard Deviation:

http://gmatclub.com/forum/a-researcher-computed-the-mean-the-median-and-the-standard-134893.html
http://gmatclub.com/forum/list-m-not-shown-consists-of-8-different-integers-each-140777.html

For the GMAT, the concepts of Average (Mean) and Median are far more important and tested more frequently than the concept of Standard Deviation.

Hope this helped.

- Japinder
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Joined: 22 Apr 2015
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23 Apr 2015, 09:00
Hi AIMGMAT770,

Let's consider a series of n consecutive numbers: $$a, a+1, a+2......a+n-1$$
To find the standard deviation, we first calculate the mean, which in this case is same as the average of first and last number.
Mean = $$(a+a+n-1)/2 = a + (n-1)/2$$

Next we need to find the squared distance of each number of the series from the mean:

Case 1: n is odd
For the 1st number, a -> $$[a - (a + (n-1)/2)]^2 = [(n-1)/2]^2$$
For the 2nd number, a +1 -> $$[a + 1 - (a + (n-1)/2)]^2 = [1-(n-1)/2]^2$$
.
.
.
For the number preceding the middle number, a + (n-1)/2 -1 -> $$[a + (n-1)/2 -1- (a + (n-1)/2)]^2 = 1$$
For the middle number, a + (n-1)/2 (Mean is also the (n-1)/2th term of the series) -> $$[a + (n-1)/2 - (a + (n-1)/2)]^2 = 0$$
For the number succeeding the middle number, a + (n-1)/2 + 1 ->$$[a + (n-1)/2 +1- (a + (n-1)/2)]^2 = 1$$
.
.
.
For the second lat number, a +n -2 -> $$[a + n -2 - (a + (n-1)/2)]^2 = [(n-1)/2-1]^2$$
For the last number, a +n -1 -> $$[a + n -1 - (a + (n-1)/2)]^2 = [(n-1)/2]^2$$

Sum of the squared distance = $$2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]$$
Variance = $$2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]/n$$

Standard deviation = $$\sqrt{[2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]/n]}$$

As sum of the squares of consecutive n numbers is n(n+1)(2n+1)/6

Hence, the Standard deviation simplifies to $$\sqrt{(n^2 -1)/12}$$

Case 2: n is even
For the 1st number, a -> $$[a - (a + (n-1)/2)]^2 = [(n-1)/2]^2$$
For the 2nd number, a +1 -> $$[a + 1 - (a + (n-1)/2)]^2 = [1-(n-1)/2]^2$$
.
.
.
For the number preceding the mean, a + (n-1)/2 - 0.5 -> $$[a + (n-1)/2 -0.5 - (a + (n-1)/2)]^2 = 0.5^2$$
{a + (n-1)/2 (Mean is the average of the two middle terms and hence at a distance of 0.5 from them)}
For the number succeeding the mean, a + (n-1)/2 + 0.5 ->$$[a + (n-1)/2 + 0.5 - (a + (n-1)/2)]^2 = 0.5^2$$
.
.
For the second last number, a +n -2 -> $$[a + n -2 - (a + (n-1)/2)]^2 = [(n-1)/2-1]^2$$
For the last number, a +n -1 -> $$[a + n -1 - (a + (n-1)/2)]^2 = [(n-1)/2]^2$$

Sum of the squared distance = $$2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]$$
Variance = $$2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]/n$$
Standard deviation = $$\sqrt{2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]/n}$$

As you can see, in both the cases you only need to know the value of n to find the Standard deviation.

It is very unlikely that you will be asked to find the standard deviation but you can definitely get a DS question where the question stem asks you to find the standard deviation of a series of n consecutive numbers and one of the statements gives you the value of n.
All you need to understand here that if you know the number of terms, you can find the standard deviation.
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28 Jan 2018, 23:59
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Re: Standard Deviation Formula   [#permalink] 28 Jan 2018, 23:59
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