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Re: Statistics Made Easy - All in One Topic! [#permalink]
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GDT wrote:
nitesh50 wrote:
Bunuel wrote:
A 750 Level GMAT Question on Statistics!

BY KARISHMA, VERITAS PREP


In this post, we have a very interesting statistics question for you. Above, we have already discussed statistics concepts such as mean, median, range.

This question needs you to apply all these concepts but can still be easily done in under two minutes. Now, without further ado, let’s go on to the question – there is a lot to discuss there.

Question: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N?

(A)10
(B)12
(C)14
(D)15
(E)20

Solution: Let’s first assimilate the information we have. We need to find the minimum number of experts that must be there. Why should there be a minimum number of people satisfying these statistics? Let’s try to understand that with some numbers.

Say, N cannot be 1 i.e. there cannot be a single expert in the unit because then you cannot have the range of $10,000. You need at least two people to have a range – the difference of their salaries would be the range in that case.

So there are at least 2 people – say one with salary 0 and the other with 10,000. No salary will lie outside this range.

Median is $5000 – i.e. when all salaries are listed in increasing order, the middle salary (or average of middle two) is $5000. With 2 people, one at 0 and the other at 10,000, the median will be the average of the two i.e. (0 + 10,000)/2 = $5000. Since there are at least 10 people, there is probably someone earning $5000. Let’s put in 5000 there for reference.

0 … 5000 … 10,000

Arithmetic mean of all the salaries is $7000. Now, mean of 0, 5000 and 10,000 is $5000, not $7000 so this means that we need to add some more people. We need to add them more toward 10,000 than toward 0 to get a higher mean. So we will try to get a mean of $7000.

Let’s use deviations from the mean method to find where we need to add more people.

0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right.

Now we have

0 … 5000 … 10000, 10000, 10000

Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000.

5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead.

When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left.

Let’s try one such set of addition:

0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000

The median is not $5000 yet. Let’s try one more set of addition.

0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000

The median now is $5000 and we have maintained the mean at $7000.

This gives us a total of 15 people.

Answer (D) This question is discussed HERE.

Granted, the question is tough but note that it uses very basic concepts and that is the hallmark of a good GMAT question!

Try to come up with some other methods of solving this.



The answer seems WRONG. The set of {5000,5000,5000,5000,5000,5000,7500,7500,10000,15000} solves this in N=10.


VeritasKarishma, MentorTutoring

Can you pls explain how to arrive at n=10. I understood how we arrived at n=15, but as the sequence mentioned by this user suggests, n=10 is possible.

Thanks in advance!


Please check the original post here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... tatistics/

This is the actual question:
Question: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?

The question here is missing some keywords.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
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VeritasKarishma wrote:
Please check the original post here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... tatistics/

This is the actual question:
Question: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?

The question here is missing some keywords.

Thank you for the clarification, Karishma. The phrasing does indeed make all the difference.

- Andrew
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Statistics Made Easy - All in One Topic! [#permalink]
Bunuel wrote:
Let’s use deviations from the mean method to find where we need to add more people.

0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right.

Now we have

0 … 5000 … 10000, 10000, 10000

Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000.

5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead.

When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left.

Let’s try one such set of addition:

0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000

The median is not $5000 yet. Let’s try one more set of addition.

0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000

The median now is $5000 and we have maintained the mean at $7000.

This gives us a total of 15 people.



Hi VeritasKarishma ! I appreciate your alternate math approach. But a simple math trick can make the above process easy and less prone to careless mistake -

Let N = 2x+1

\(\frac{(10000+5000)x}{2x+1} = 7000\), gives N = 15.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
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vineetm wrote:
Bunuel wrote:
Let’s use deviations from the mean method to find where we need to add more people.

0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right.

Now we have

0 … 5000 … 10000, 10000, 10000

Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000.

5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead.

When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left.

Let’s try one such set of addition:

0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000

The median is not $5000 yet. Let’s try one more set of addition.

0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000

The median now is $5000 and we have maintained the mean at $7000.

This gives us a total of 15 people.



Hi VeritasKarishma ! I appreciate your alternate math approach. But a simple math trick can make the above process easy and less prone to careless mistake -

Let N = 2x+1

\(\frac{(10000+5000)x}{2x+1} = 7000\), gives N = 15.


Thanks vineetm. As I wrote above, this is the actual question:

Question: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?

I am not sure how you arrived at the trick. Please do explain.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Bunuel wrote:

Statistics Made Easy - All in One Topic!


CONTENT



The Meaning of Arithmetic Mean

BY KARISHMA, VERITAS PREP


Let’s start today with statistics – mean, median, mode, range and standard deviation. The topics are simple but the fun lies in the questions. Some questions on these topics can be extremely tricky especially those dealing with median, range and standard deviation. Anyway, we will tackle mean today.

So what do you mean by the arithmetic mean of some observations? I guess most of you will reply that it is the ‘Sum of Observations/Total number of observations’. But that is how you calculate mean. My question is ‘what is mean?’ Loosely, arithmetic mean is the number that represents all the observations. Say, if I know that the mean age of a group is 10, I would guess that the age of Robbie, who is a part of that group, is 10. Of course Robbie’s actual age could be anything but the best guess would be 10.

Say, I tell you that the average age of a group of 10 people is 15 yrs. Can you tell me the sum of the ages of all 10 people? I am sure you will say that it is 10*15 = 150. You can think of it in two ways:

Mean = Sum of all ages/No of people

So Sum of all ages = Mean * (No of people) = 15*10

Or

Since there are 10 people and each person’s age is represented by 15, the sum of their ages = 10*15. Basically, the total sum was distributed evenly among the 10 people and each person got 15 yrs.

Now, let’s say you made a mistake. A boy whose age you thought was 20 was actually 30. What is the correct mean? Again, you can think of it in two ways:

New sum = 150 + 10 = 160

New average = 160/10 = 16

Or

You can say that there is an extra 10 that has to be distributed evenly among the 10 people, so each person gets 1 extra. Hence, the average becomes 15 + 1 = 16.

As you might have guessed, we will work on the second interpretation. Let’s look at an example now.

Example 1: The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

Solution: First tell me, if the age of the additional person were 15 yrs, what would have happened to the average? The average would have remained the same since this new person’s age would have been the same as the age that represents the group. But his age is 39 – 15 = 24 more than the average. We know that we need to evenly split the extra among all the people to get the new average. When 24 is split evenly among all the people (including the new guy), everyone gets 2 extra (since average age increased from 15 to 17). There must be 24/2 = 12 people now (including the new guy) i.e. n must be 11 (without including the new guy).

This question is discussed HERE.

Let’s look at another similar example though a little trickier. Try solving it on your own first. If not logically, try using the formula approach. Then see how elegant the solution becomes once you start ‘thinking’ instead of just ‘calculating’.

Example 2: When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Solution: What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

This question is discussed HERE.

If you use the formula instead, it would take you quite a while to manipulate the two variables to get the value of n. I hope you see the beauty of this method. Next week, we will discuss some GMAT questions based on Arithmetic mean!


How can you use the same formula in Example 2?
This doesn't really make much sense to me.

The formula in example 1 was based on the "current" n and some additional number.

But in 2, you are already working with n+1 which results in -1 average and using the same formula to calculate n?

Am I missing something?
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Re: Statistics Made Easy - All in One Topic! [#permalink]
When we divide 45 by 4, we get 11.25. Do we have 4 consecutive integers such that their mean is 11.25? No, because mean of even number of consecutive integers is always of the form x.5. In this what is meant by ' even number of consecutive integers is always of the form x.5'?
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Statistics Made Easy - All in One Topic! [#permalink]
D4kshGargas wrote:
Bunuel wrote:

Statistics Made Easy - All in One Topic!




The Meaning of Arithmetic Mean

BY KARISHMA, VERITAS PREP


Let’s start today with statistics – mean, median, mode, range and standard deviation. The topics are simple but the fun lies in the questions. Some questions on these topics can be extremely tricky especially those dealing with median, range and standard deviation. Anyway, we will tackle mean today.

So what do you mean by the arithmetic mean of some observations? I guess most of you will reply that it is the ‘Sum of Observations/Total number of observations’. But that is how you calculate mean. My question is ‘what is mean?’ Loosely, arithmetic mean is the number that represents all the observations. Say, if I know that the mean age of a group is 10, I would guess that the age of Robbie, who is a part of that group, is 10. Of course Robbie’s actual age could be anything but the best guess would be 10.

Say, I tell you that the average age of a group of 10 people is 15 yrs. Can you tell me the sum of the ages of all 10 people? I am sure you will say that it is 10*15 = 150. You can think of it in two ways:

Mean = Sum of all ages/No of people

So Sum of all ages = Mean * (No of people) = 15*10

Or

Since there are 10 people and each person’s age is represented by 15, the sum of their ages = 10*15. Basically, the total sum was distributed evenly among the 10 people and each person got 15 yrs.

Now, let’s say you made a mistake. A boy whose age you thought was 20 was actually 30. What is the correct mean? Again, you can think of it in two ways:

New sum = 150 + 10 = 160

New average = 160/10 = 16

Or

You can say that there is an extra 10 that has to be distributed evenly among the 10 people, so each person gets 1 extra. Hence, the average becomes 15 + 1 = 16.

As you might have guessed, we will work on the second interpretation. Let’s look at an example now.

Example 1: The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?

(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

Solution: First tell me, if the age of the additional person were 15 yrs, what would have happened to the average? The average would have remained the same since this new person’s age would have been the same as the age that represents the group. But his age is 39 – 15 = 24 more than the average. We know that we need to evenly split the extra among all the people to get the new average. When 24 is split evenly among all the people (including the new guy), everyone gets 2 extra (since average age increased from 15 to 17). There must be 24/2 = 12 people now (including the new guy) i.e. n must be 11 (without including the new guy).

This question is discussed HERE.

Let’s look at another similar example though a little trickier. Try solving it on your own first. If not logically, try using the formula approach. Then see how elegant the solution becomes once you start ‘thinking’ instead of just ‘calculating’.

Example 2: When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Solution: What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

This question is discussed HERE.

If you use the formula instead, it would take you quite a while to manipulate the two variables to get the value of n. I hope you see the beauty of this method. Next week, we will discuss some GMAT questions based on Arithmetic mean!


How can you use the same formula in Example 2?
This doesn't really make much sense to me.

The formula in example 1 was based on the "current" n and some additional number.

But in 2, you are already working with n+1 which results in -1 average and using the same formula to calculate n?

Am I missing something?




1st scenario- Sum + 39/ (n+1)= Average+2
2nd Scenario- sum + 15/ (n+1)= average - 1

Combining both 1 and 2-
difference between the two added observations, i.e 39-15= 24 is creating a difference of '3' in the average.

To understand the difference of '3' in the average, let's use number line. As addition of 15 and 39 decreases the average by 1 and increases the average by 2 respectively.
(-1, 0, 1, 2)

from -1 to 0= 1 unit
from 0 to 1= 1 unit
from 1 to 2= 1 unit
so the change in the average made by the extra 24 units(years) is of total 3 units(years).

- extra value that is more than the average/ number of observations = change in the average
(as we say that the extra additional value will be evenly split among the total number of observations, leading to increase/decrease in the new average)

so 24/ (n+1)= 3
24/3= n+1
8= n+1
n=7
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Re: Statistics Made Easy - All in One Topic! [#permalink]
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duttan wrote:
When we divide 45 by 4, we get 11.25. Do we have 4 consecutive integers such that their mean is 11.25? No, because mean of even number of consecutive integers is always of the form x.5. In this what is meant by ' even number of consecutive integers is always of the form x.5'?


Yes. The mean of even number of consecutive integers is always of the form ?.5. For example, the mean of 2, 3, 4, 5 is 3.5.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Bunuel wrote:
duttan wrote:
When we divide 45 by 4, we get 11.25. Do we have 4 consecutive integers such that their mean is 11.25? No, because mean of even number of consecutive integers is always of the form x.5. In this what is meant by ' even number of consecutive integers is always of the form x.5'?


Yes. The mean of even number of consecutive integers is always of the form ?.5. For example, the mean of 2, 3, 4, 5 is 3.5.


ahh got it.. You meant average of even number of consecutive integers in never an integer (x.5 was used to represent decimal form). Thank you.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
In the 4 example of SD, It is written that

Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5}

T has higher SD. It has two extra numbers far from the mean. (There is a caveat here too!)

What do you think about cases 5, 6, and 7? I will give you the answers to these three cases in the next post!

I just tried this in the SD calculator and it seems since 2 nos 1,1, 5,5 were added without any deviation from the mean, It stays the same.


Can someone confirm it please?
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Bunuel wrote:
A 750+ Level Question on SD

BY KARISHMA, VERITAS PREP


Above, we looked at a 750+ level question on mean, median and range concepts of Statistics. Here we have a 750+ level question on standard deviation concept of Statistics. We do hope you enjoy checking it out.

Before you begin, you might want to review the post that discusses standard deviation: Dealing With Standard Deviation

So here goes the question.

Question: Given that set S has four odd integers and their range is 4, how many distinct values can the standard deviation of S take?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Solution: Recall what standard deviation is. It measures the dispersion of all the elements from the mean. It doesn’t matter what the actual elements are and what the arithmetic mean is – the standard deviation of set {1, 3, 5} will be the same as the standard deviation of set {6, 8, 10} since in each set there are 3 elements such that one is at mean, one is 2 below the mean and one is 2 above the mean. So when we calculate the standard deviation, it will give us exactly the same value for both sets. Similarly, standard deviation of set {1, 3, 3, 5, 6} will be the same as standard deviation of {10, 12, 12, 14, 15} and so on. But note that the standard deviation of set {25, 27, 29, 29, 30} will be different because it represents a different arrangement on the number line.

Let’s look at the given question now.

Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5

x and y could be same or different but x would always be smaller than or equal to y.

- If x and y were same, we could select the values of x and y in 3 different ways: both could be 1; both could be 3; both could be 5

- If x and y were different, we could select the values of x and y in 3C2 ways: x could be 1 and y could be 3; x could be 1 and y could be 5; x could be 3 and y could be 5.

For clarification, let’s enumerate the different ways in which we can write set S:

{1, 1, 1, 5}, {1, 3, 3, 5}, {1, 5, 5, 5}, {1, 1, 3, 5}, {1, 1, 5, 5}, {1, 3, 5, 5}

These are the 6 ways in which we can choose the numbers in our example.

Will all of them have unique standard deviations? Do all of them represent different distributions on the number line? Actually, no!

Standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are the same. Why?

Standard deviation measures distance from mean. It has nothing to do with the actual value of mean and actual value of numbers. Note that the distribution of numbers on the number line is the same in both cases. The two sets are just mirror images of each other.

For the set {1, 1, 1, 5}, mean is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

For the set {1, 5, 5, 5}, mean is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

The deviations in both cases are the same -> 1, 1, 1 and 3. So when we square the deviations, add them up, divide by 4 and then find the square root, the figure we will get will be the same.

Similarly, {1, 1, 3, 5} and {1, 3, 5, 5} will have the same SD. Again, they are mirror images of each other on the number line.

The rest of the two sets: {1, 3, 3, 5} and {1, 1, 5, 5} will have distinct standard deviations since their distributions on the number line are unique.

In all, there are 4 different values that standard deviation can take in such a case.

Answer (B) This question is discussed HERE.





Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5
in {11,x,y,15} x and y can be 13 or x=11 y=13
so why only 3 values
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Rabab36 wrote:
Bunuel wrote:
A 750+ Level Question on SD

BY KARISHMA, VERITAS PREP


Above, we looked at a 750+ level question on mean, median and range concepts of Statistics. Here we have a 750+ level question on standard deviation concept of Statistics. We do hope you enjoy checking it out.

Before you begin, you might want to review the post that discusses standard deviation: Dealing With Standard Deviation

So here goes the question.

Question: Given that set S has four odd integers and their range is 4, how many distinct values can the standard deviation of S take?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Solution: Recall what standard deviation is. It measures the dispersion of all the elements from the mean. It doesn’t matter what the actual elements are and what the arithmetic mean is – the standard deviation of set {1, 3, 5} will be the same as the standard deviation of set {6, 8, 10} since in each set there are 3 elements such that one is at mean, one is 2 below the mean and one is 2 above the mean. So when we calculate the standard deviation, it will give us exactly the same value for both sets. Similarly, standard deviation of set {1, 3, 3, 5, 6} will be the same as standard deviation of {10, 12, 12, 14, 15} and so on. But note that the standard deviation of set {25, 27, 29, 29, 30} will be different because it represents a different arrangement on the number line.

Let’s look at the given question now.

Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5

x and y could be same or different but x would always be smaller than or equal to y.

- If x and y were same, we could select the values of x and y in 3 different ways: both could be 1; both could be 3; both could be 5

- If x and y were different, we could select the values of x and y in 3C2 ways: x could be 1 and y could be 3; x could be 1 and y could be 5; x could be 3 and y could be 5.

For clarification, let’s enumerate the different ways in which we can write set S:

{1, 1, 1, 5}, {1, 3, 3, 5}, {1, 5, 5, 5}, {1, 1, 3, 5}, {1, 1, 5, 5}, {1, 3, 5, 5}

These are the 6 ways in which we can choose the numbers in our example.

Will all of them have unique standard deviations? Do all of them represent different distributions on the number line? Actually, no!

Standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are the same. Why?

Standard deviation measures distance from mean. It has nothing to do with the actual value of mean and actual value of numbers. Note that the distribution of numbers on the number line is the same in both cases. The two sets are just mirror images of each other.

For the set {1, 1, 1, 5}, mean is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

For the set {1, 5, 5, 5}, mean is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

The deviations in both cases are the same -> 1, 1, 1 and 3. So when we square the deviations, add them up, divide by 4 and then find the square root, the figure we will get will be the same.

Similarly, {1, 1, 3, 5} and {1, 3, 5, 5} will have the same SD. Again, they are mirror images of each other on the number line.

The rest of the two sets: {1, 3, 3, 5} and {1, 1, 5, 5} will have distinct standard deviations since their distributions on the number line are unique.

In all, there are 4 different values that standard deviation can take in such a case.

Answer (B) This question is discussed HERE.





Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5
in {11,x,y,15} x and y can be 13 or x=11 y=13
so why only 3 values


I answered this question here. Let me know if it helps.
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5}

T has higher SD. It has two extra numbers far from the mean. (There is a caveat here too!)

how this is possible?

S has SD of 2.3 [sqrt(16/3)]and T has 1.7[sqrt(16/5)]? Then how T has high SD?

please reply
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pudu wrote:
Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5}

T has higher SD. It has two extra numbers far from the mean. (There is a caveat here too!)

how this is possible?

S has SD of 2.3 [sqrt(16/3)]and T has 1.7[sqrt(16/5)]? Then how T has high SD?

please reply


The standard deviation of {1, 3, 5} is \(2 \sqrt{\frac{2}{3}}≈1.63\)

The standard deviation of {1, 1, 3, 5, 5} is \(\frac{4}{\sqrt{5}}≈1.79\)
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Bunuel wrote:
pudu wrote:
Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5}

T has higher SD. It has two extra numbers far from the mean. (There is a caveat here too!)

how this is possible?

S has SD of 2.3 [sqrt(16/3)]and T has 1.7[sqrt(16/5)]? Then how T has high SD?

please reply


The standard deviation of {1, 3, 5} is \(2 \sqrt{\frac{2}{3}}≈1.63\)

The standard deviation of {1, 1, 3, 5, 5} is \(\frac{4}{\sqrt{5}}≈1.79\)


second one is wrong...mean is 3 not 5...1+1+3+5+5=15 and 15/5=3...
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Re: Statistics Made Easy - All in One Topic! [#permalink]
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pudu wrote:
Bunuel wrote:
pudu wrote:
Case 4: S = {1, 3, 5} or T = {1, 1, 3, 5, 5}

T has higher SD. It has two extra numbers far from the mean. (There is a caveat here too!)

how this is possible?

S has SD of 2.3 [sqrt(16/3)]and T has 1.7[sqrt(16/5)]? Then how T has high SD?

please reply


The standard deviation of {1, 3, 5} is \(2 \sqrt{\frac{2}{3}}≈1.63\)

The standard deviation of {1, 1, 3, 5, 5} is \(\frac{4}{\sqrt{5}}≈1.79\)


second one is wrong...mean is 3 not 5...1+1+3+5+5=15 and 15/5=3...


Not sure what you are trying to say there but everything is correct there:
https://www.wolframalpha.com/input?i=Po ... +3%2C+5%7D
https://www.wolframalpha.com/input?i=Po ... 5%2C+5%7D+
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Re: Statistics Made Easy - All in One Topic! [#permalink]
Mean of {1, 1, 1, 5} is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean. Mean of {1, 5, 5, 5} is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean. Sum of the squared deviations will be the same in both the cases and the number of elements is also the same in both the cases. Therefore, both these sets will have the same SD.


How do we conclude that these two sets have the same SD?
Does this mean that if we have the same number of elements in two sets at the same distance from their mean, their SD will be the same??


Can any of the experts please clarify?
KarishmaB Bunuel


Thank you.
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