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## Standard Deviation for the GMAT

Standard deviation is a common measure of dispersion of the numbers or values in a list. The more scattered the data away from the mean, the greater the standard deviation.

Standard deviation is usually represented by the Greek letter sigma (σ), and is calculated using the formula

Where X represents each value in the list,  represents the mean of the values, and n is the number values in the list.

Example 1:

Let’s find the standard deviation of the data −5, 0, 15, 20, and 20 in order to understand the concept of standard deviation.

The average or the arithmetic mean is 10.

 X X − 10 (X − 10)2 −5 −15 225 0 −10 100 15 5 25 20 10 100 20 10 100 Total 50 550

Standard deviation = √(550/5) = √110 = 10.488 (approx.)

It is important to note that the standard deviation depends on every value. However, the standard deviation is most affected by the values that are farthest from the mean. Thus, even without calculating the standard deviation of {0, 5, 15, 15, and 15}, we can say that the standard deviation of {0, 5, 15, 15, and 15} will be less than √110 even though the mean is still 10.

Example 2:

Let’s take an example from OG 2017 to understand how standard deviation may be tested on the GMAT.

OG 2017, Question No. 367

4, 6, 8, 10, 12, 14, 16, 18, 20, 22

List M (not shown) consists of 8 different integers, each of which is in the list shown. What is the standard deviation of the numbers in list M?

(1)  The average (arithmetic mean) of the numbers in list M is equal to the average of the numbers in  the list shown.

(2) List M does not contain 22.

Solution:

We are given 10 different numbers, 8 of which are in List M. The question asked is: what is the standard deviation of the numbers in list M?

We know that we can find the standard deviation of a set of numbers if we know the numbers. Thus the question turns out to be: which 8 numbers are in the list or which 2 numbers are not in the list?

Let’s consider Statement (1): The average (arithmetic mean) of the numbers in list M is equal to the average of the numbers in the list shown. It follows that the average of the two numbers excluded should be same as the average of the numbers in the list shown.

The average of the numbers in the list shown is 13, and, therefore, the sum of the two numbers excluded should be 26. You find that the numbers could be 4 and 22, or they could be 12 and 14, or even they could be any three other pairs of that sort (6 and 20, 8 and 18, or 10 and 12).

It’s important to observe that the numbers 2 and 22 are far away from the mean; whereas, 12 and 14 are close to the mean. Therefore, as we exclude different pairs of numbers, the standard deviations will be different.

The information provided in Statement (1) is not sufficient for answering the question.

Let’s consider Statement (2): List M does not contain 22. Obviously, this doesn’t tell us which 8 numbers are there in List M.

The information provided in Statement (2) alone is also not sufficient for answering the question.

However, if we combine the information provided in Statement (1) and the information provided in Statement (2), we know for certain that the 2 numbers that are not in List M are 4 and 22.

Therefore, both statements together are sufficient for answering the question.

Note: Did you notice that in the list shown, the 10 numbers are evenly spaced? If you did, it must have been easy for you to the find the average and figure out the pairs of numbers that add up to 26.

Observations: Note that to solve the question we did not have to calculate the standard deviation. It was enough to know that (i) the standard deviation depends on every number/value in a list and (ii) the standard deviation depends on the extent of dispersion of the numbers/values in the list.

Corollary:

(i)  If every number/value in a list is increased or decreased by a fixed amount, the standard deviation remains unchanged.

(ii) Standard deviation can’t be negative; minimum possible standard deviation of the numbers/values in any list is 0 — when all the numbers/values are the same.

Conclusion: The more closely packed are the values around the mean, the smaller the standard deviation. The greater the standard deviation, the farther away are the values from the mean.

Next time, we will talk about standard deviations of normally distributed data.

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