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Statistics Made Easy - All in One Topic!

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New post 14 Sep 2019, 03:54
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Hi, is there a PDF version of the content on this thread?


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Statistics Made Easy - All in One Topic!  [#permalink]

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New post 28 Apr 2020, 07:03
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!
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Re: Statistics Made Easy - All in One Topic!  [#permalink]

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New post 28 Apr 2020, 07:50
1
GDT wrote:
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!

I agree with nitesh50. I am guessing that the person simply started with a fixed condition, looking to justify the smallest value for N, and played around with different values within such constraints. The following is an illustration of how I might approach a similar task.

Step 1: Assume a value for N, say, 10. Given that the average monthly salary is $7,000, the sum of the salaries must be $70,000.

Step 2: Set the median at $5,000 to conform with the given information. This leaves nine more salaries that must sum to $65,000. and the range of salaries must also span $10,000.

Step 3: Rather than get creative, hold the low-end salaries stable, with an eye on setting the range.

{5,000; 5,000; 5,000; 5,000; 5,000; salary; salary; salary; salary; 15,000}

The sum is now $40,000, so $30,000 remains. As long as we can maintain the median salary and ensure that the remaining four salaries sum to $30,000, we can carve up that $30,000 any way we want.

Step 4: Set the sixth salary at $5,000 to preserve the necessary median value.

{5,000; 5,000; 5,000; 5,000; 5,000; 5,000; salary; salary; salary; 15,000}

The sum of the salaries is now $45,000. We can fill in the remaining three salaries with any values between $5,000 and $15,000, inclusive, as long as the three salaries sum to the missing $25,000. This particular user chose $7,500, $7,500, and $10,000, but the salaries could just as easily have been $5,000, $5,000, and $15,000 or any other values that conform to the given conditions.

See if you can apply the same critical reasoning skills to justify another answer. I often teach my GRE® clients to use "forbidden" values that are provided in the problem itself (e.g., if x < 3, then what happens if you try 3 itself? It might be a pivotal point.). Keeping matters as straightforward as possible by holding certain values steady can help you progress on other fronts.

I hope that helps. Thank you for tagging me.

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Re: Statistics Made Easy - All in One Topic!  [#permalink]

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New post 28 Apr 2020, 21:43
1
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.veritasprep.com/blog/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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New post 29 Apr 2020, 04:48
VeritasKarishma wrote:
Please check the original post here: https://www.veritasprep.com/blog/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.

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Statistics Made Easy - All in One Topic!  [#permalink]

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New post 09 May 2020, 22:25
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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New post 12 May 2020, 03:18
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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]

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New post 16 Jun 2020, 07:15
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?
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Re: Statistics Made Easy - All in One Topic!   [#permalink] 16 Jun 2020, 07:15

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