Lesson: Range and Standard Deviation

Comment on Range and Standard Deviation

By "mean" you the mean "the average"?
Because I got confused when you solved the last example (at 8:30) of finding the standard deviation for the two sets, the mean for the first one was 10 according to the mean definition ( arrange from the lowest to the greatest and take the middle number)!
greenlight-admin's picture

Average and (arithmetic) mean are the same thing.
What you described (" arrange from the lowest to the greatest and take the middle number") is the MEDIAN

at 9.54 minutes, you say that, (1,1,2,4,7,13,30,34,47,80) here range 89, average age (mean) is 22 and SD is 24.6, but we know that SD is the average distance the data values are away from the mean, Now as its mean is 22, then the distance from the mean is (21,21,19,18,15,9,8,12,25,58) and the sum of this is 206, and the total number is 10 then average is 20.6, but you said that SD is 24.6, how can you get this? Can you explain it? Thanks
greenlight-admin's picture

In the video, we examine 2 different definitions of Standard Deviation: the formal definition and the informal definition.
At 9:45, I say "if we apply the FORMAL definition of Standard Deviation to both sets," we get a standard deviation of the top set is 24.6

When you applied the INFORMAL definition, you got 20.6. which is pretty close to what we found with the formal definition.

As I mention in the video, the informal definition will not yield the exact same standard deviation that you'd get using the formal definition. However, for the purposes of the GRE, it's close enough.

In fact, at 6:30 in the video, we used the informal definition to find the standard deviation, and when we were finished, the standard deviation we calculated was reasonably close to what we got when we used the formal definition.

As I said, for the purposes of the GRE, the informal definition is good enough to answer answer standard deviation question the GRE can throw your way.

I have been experiencing issues with the loading of the video :(
greenlight-admin's picture

Try clearing the cache on your browser.

Here are the instructions for clearing your cache: https://kb.iu.edu/d/ahic

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Let me know if that helps.

Concise and clear explanations. Thanks for the video! It is really helpful in my prep for GRE.


Please answer this.

TO find range i assumed 160 is max hight (since they asked for min possible) and 140 as min hight, i got 20 as range.

There are 4 possible range.

1) if we think height is dispersed from 140 to 164 then range will be 24.
2) 144 and 164 then range will be 20
3) 140 and 160 then range will be 20
4) 144 and 160 then range will be 16

since they have asked for lease possible range then we should assume that 6 students are of 144 hight and 4 students are of 160 hight, then range will yield minimum. so B is answer.

in case, if they had ask for highest possible range then it would 24.

Thank you so much for your response Brent! :)
greenlight-admin's picture

That's correct.


Please answer this question.

instead of finding 1st, 2nd, 3rd, 4th so on till 100th.

will find 1st and 100th numbers of each.
A) y= 1/3 when x =1 and y = 100/3 when x=100 range will be 33

B) y=1/2 when x=1 and y=100/2 when x=100 range will be 49.5

C) y=1 when x=1 and y=100 when x=100 range will be 99

D) y=52 when x =1 and y=250 when x=100 range will be 198

E) y=-17 when x=1 and y=280 when x=100 range will be 297

looking at the range more spread is there at E, therefore SD will also be greatest.
hence E.

Is this a valid approach?
greenlight-admin's picture

In many cases, the set with the greatest range will also have the greatest standard deviation. But this isn't always the case. Consider these two sets:

Set A: {-10, 0, 0, 0, 0, 0, 0, 10}
Set B: {-8, -8, -8, -8, 8, 8, 8, 8}

In both sets, the mean is 0.

Even though set A has the greater range, set B has the greater standard deviation. So, it also helps to examine some of the numbers in the set as well.

yes exactly, but since all given are linear function, hence our assumption should be true.

Anyhow it is always better to test for some data, in your solution for the above question i found its tedious to find mean and SD. what procedure would be best to find mean or SD in the above kind of question?
greenlight-admin's picture

For these kinds of questions, you need not find the mean or SD. You should be able to just examine each set and see which one has the greatest dispersion.

Hi there! can you please provide some clarity to this question?

The numbers in data set S have a standard deviation of 5. If a new data set is formed by adding 3 to each number in S, what is the standard deviation of the numbers in the new data set?

A 2
B 3
C 5
D 8
E 15
greenlight-admin's picture

Great question.

IMPORTANT CONCEPT: Adding the same value to every value in a set has no effect on the standard deviation.

For example, notice that in the set {5, 7, 9, 11} each value is 2 greater than the one before it.

If we add 10 to each value in that set, we get the new set {15, 17, 19, 21}. In this NEW set, each value is 2 greater than the one before it. So, the standard deviation of the NEW set will be equal to the standard deviation of the original set.

Now onto the question that you posed....

Since we're adding the same value (3) to every value in set
S, the standard deviation of the new set will still be 5.

Answer: C


Hi Brent - can you elaborate on Answer Choice A? Specifically on how we'd find the range of the heights of all students.


Hi Brent,

I have solved the above question in the following manner. Is it correct?

Since its given that the six number are in increasing order and let the range be x, we can infer that 2a - 3.7 = x.
Therefore, a = (x + 3.7)/2

Lets plug in each answer choice in place of x.

When x = 4,
a = (4 + 3.7)/2 which equals 3.85.
But since a has to be in between 4.1 and 8.5. Eliminate option A.

Now when x = 5.2,
a = (5.2 + 3.7)/2 which equals 4.45.
In this case a is between 4.1 and 8.5 but 2a = 8.9 which has to be greater than 9.2. Eliminate option B.

Now when x = 7.3,
a = (7.3 + 3.7)/2 which equals 5.5.
This value is in between 4.1 and 8.5 and also 2a = 11 is greater than 9.2. Keep option C.

Likewise if we follow the above methodology, we have C,D,E as answer.

greenlight-admin's picture

Question link: http://www.urch.com/forums/gre-math/152259-problem.html

Great approach, Suraj!!!


Why isn't the answer 20
greenlight-admin's picture

Question link: https://greprepclub.com/forum/qotd-10-the-table-above-shows-the-frequenc...

The range COULD be 20 if the shortest person were 142 cm tall and the tallest person were 162 cm tall.

However, the question asks for the LEAST possible range of the heights.

Notice, if the shortest person were 144 cm tall and the tallest person were 160 cm, then the range would be 16.

In fact, 16 is the LEAST possible range of the heights.

Does that help?


How does the statement when X>2 comes in? I thought it would required the lease distinct value of the set to be greater than 2
greenlight-admin's picture

Question link: https://greprepclub.com/forum/set-n-is-a-set-of-x-distinct-positive-inte...

I think you're reading the given information incorrectly.

GIVEN: Set N is a set of x distinct positive integers where x > 2.

S, x is the NUMBER of integers in set N.
So, the fact that x > 2 is telling us that there are more than two values in set N.
In other words, x does not tell us anything about the VALUES of the numbers in set N.

Does that help?


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