Calculating mean from 5 number summary

Click For Summary
Calculating the mean from a 5-number summary (min, Q1, median, Q3, max) is not possible with exact precision, but rough estimates can be made. An approximation formula is suggested: (min + max + 2(Q1 + median + Q3))/8. In a specific data set, the mean was found to be significantly different from the estimate, indicating the limitations of this method. The discussion highlights that if the distribution is symmetrical, the mean and median may be close in value. Overall, while estimating the mean from a 5-number summary can provide insights, it is inherently imprecise.
Banaticus
Messages
32
Reaction score
0
It seems like it should be possible to calculate the mean (usual average) from a 5-number summary of a set of numbers (min, first quartile or Q1, median, third quartile or Q3, and max). You should be able to calculate roughly what a percentile is, then by taking each discrete percentile and then taking the average of those hundred numbers... or better yet by using calculus and taking every percentile point, and the average of every point, you should be able to come really close to the mean, if not compute it directly. I, however, don't know math well enough to do that, nor do I remember any calculus.

In one data set that I'm looking at, the min, Q1, median, Q3, and max are: 0, 3900, 18882, 50145.5, 1250000
And the mean is: 46172.04545 or just under Q3.
How can the mean be calculated from those 5 numbers?
 
Mathematics news on Phys.org
You can't exactly. A rough estimate (analogous to trapezoid rule for integral approximation) is:
(min + max + 2(Q1+median+Q3))/8.
 
Banaticus said:
It seems like it should be possible to calculate the mean (usual average) from a 5-number summary of a set of numbers ...
Did you mean estimate? Surely you can see (as mathman pointed out) that you can't get an exact calculation based on just summary numbers.
 
mathman said:
You can't exactly. A rough estimate (analogous to trapezoid rule for integral approximation) is:
(min + max + 2(Q1+median+Q3))/8.
Hmm, ok, thanks. I have one data set where the 5-number summary is: 0, 29496, 68552, 124280, 780575. The mean is 80041.24331 and that approximation comes up with 153153.875, so I guess it's a really rough estimate.
phinds said:
Did you mean estimate?
Sure, why not. If I clearly don't know what I'm talking about, feel free to attempt to fill in the gaps. :)
 
Banaticus said:
It seems like it should be possible to calculate the mean (usual average) from a 5-number summary of a set of numbers (min, first quartile or Q1, median, third quartile or Q3, and max). You should be able to calculate roughly what a percentile is, then by taking each discrete percentile and then taking the average of those hundred numbers... or better yet by using calculus and taking every percentile point, and the average of every point, you should be able to come really close to the mean, if not compute it directly. I, however, don't know math well enough to do that, nor do I remember any calculus.

In one data set that I'm looking at, the min, Q1, median, Q3, and max are: 0, 3900, 18882, 50145.5, 1250000
And the mean is: 46172.04545 or just under Q3.
How can the mean be calculated from those 5 numbers?
You cannot. But if you the distribution is fairly symmetrical (like the familiar bell-curve), the mean and the median are approximately equal.
 
Banaticus said:
Hmm, ok, thanks. I have one data set where the 5-number summary is: 0, 29496, 68552, 124280, 780575. The mean is 80041.24331 and that approximation comes up with 153153.875, so I guess it's a really rough estimate.

Sure, why not. If I clearly don't know what I'm talking about, feel free to attempt to fill in the gaps. :)
The last number (780575) swamps the other four.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
13
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K