Standard Deviation Formula

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:

https://gmatclub.com/forum/a-researcher-computed-the-mean-the-median-and-the-standard-134893.html
https://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

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.

Just a note: EMPOWERgmatRichC and EgmatQuantExpert

I was literally asked on question #2 of my official GMAT today, the following question:

!	Posting real GMAT questions is a direct violation of GMAC and GMAT Club rules. Question is removed. MODERATOR.

(answers are from memory, so not sure if they are accurate)
Point is, I was specifically asked about standard deviation.

I was aware that they normally do not ask us to calculate it. Did I misunderstand the concept?

This is earth shattering news. I've never heard of such a case in a decade of working with the GMAT full-time.
That said, it's still possible to solve quickly for the first five positive integers (or for any five consecutive integers) with minimal math - just jot down the five deviations from the mean (the mean in any evenly spaced set with an odd number of terms will be the middle term.

Hi coled90,

To start, it is illegal to discuss specific questions that you have faced on the Official GMAT, so I'm going to address this in general terms.

First, the S.D. of ANY 5 positive, consecutive integers will be the SAME result, so for the question to define the group as the 'FIRST five positive integers' is technically unnecessary (and that's not something that GMAT writers tend to do). When combined with the fact that the GMAT has never required that a Test-taker solve this type of calculation, there's a fairly high chance that this was an 'Experimental' question (meaning that it did NOT count towards your Score). In addition, you can eliminate most of the wrong answers fairly easily. Since S.D. is based on a fairly step-heavy calculation under a square-root sign, it's highly UNLIKELY that the result will be an integer (meaning that you could eliminate 3 of the 5 answers you listed). Even if you did not know that, you should be able to logically eliminate at least two of the Answers (since '4' is the range of the group and '1' is the difference between any two consecutive integers; NEITHER of those would be equal to the S.D.).

GMAT assassins aren't born, they're made,
Rich