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Statistics - Mean, Median, Range Question

Today, we have a very interesting statistics question for you. It is based on statistics concepts such as mean, median, range etc.

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 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?



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)


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.




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