Solution Stats
Problem Comments
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23 Comments
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20 older comments
Israel
on 27 Jan 2012
weighted mean should really be sum(x.*w)/sum(w), and not as defined in the problem.
Davide Ferraro
on 27 Jan 2012
I agree with the definition above. You need to define the sum of all the weights and not the number of weights.
John D'Errico
on 17 Feb 2012
As the others have said, the problem title is flat out wrong. What is required is simply not a weighted average in any standard form.
William Smith
on 25 Sep 2012
Come on Mathworks, delete this problem!
Eilert
on 14 Jan 2014
The test suits for this problem should have tests with different vector length.
There are "solutions" that just divide by 3.
Sneha
on 30 Oct 2014
I think the program title is wrong as in weighted average the denominator should contain sum of all weights
Vittorio
on 9 Jan 2015
Hahaha, who wrote this problem? This is not a weighted average! Just look at the example provided: the weighted average of 1, 2 and 3 is 33.3333, hahahahaha!
Sergej Pauli
on 24 Apr 2015
this is not really the weighted average, because of 10 15 and 20 the average can't never be 33.333 here is a bug
John D'Errico
on 20 Aug 2016
Yes. The description is completely wrong. To start with, a weighted average should have the property that
weighted_average(x,w) == weighted_average(x,k*w)
So a simple re-scaling of the weights should not impact the result.
Benjamin Re
on 8 Dec 2016
Good problem
Muhammad M Sahari
on 20 Dec 2016
kinda misleading, but i get the gist of it
Gan Wee Ting
on 20 Dec 2016
good question
Adi Zabidi
on 27 Dec 2016
yay
Mohan Shanmugam
on 30 Dec 2016
tricky
Limon
on 28 Apr 2017
easy
Informaton
on 29 Apr 2017
Is it possible to remove this problem or at least change the problem name so it does not say "weighted average". As others have pointed out, the definition, via example, is inaccurate.
An average, weighted or otherwise, will not be greater than the largest number in the set (which is the case here).
Informaton
on 8 May 2017
In case you actually want to try a weighted average problem, you can check out the one here
https://www.mathworks.com/matlabcentral/cody/problems/44156-weighted-average
Devin Lyons
on 7 Dec 2017
very good
Shreyash Singh
on 5 May 2018
nice
Mehmed Saad
on 13 Mar 2020
why leading size is always 7? and how :(
Rafael S.T. Vieira
on 19 Jul 2020
On a weighted average we should divide the weighted values by the sum of the weights and not by the number of terms. https://en.wikipedia.org/wiki/Weighted_arithmetic_mean
goc3
on 28 Sep 2020
@Mehmed Saad: some of the many cheating solutions have been deleted.
Benjamin Farris
on 2 Nov 2020
This problem definitely needs to be fixed to use the correct weighted average formula.
Solution Comments
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1 Comment
Daniele Certini
on 18 Oct 2020
Nice problem!
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2 Comments
Daniel Bergman
on 19 Jan 2019
This is NOT the weighted average. The denominator should be the sum of the weights. PLEASE FIX THIS!!!
Rafael S.T. Vieira
on 19 Jul 2020
Daniel is correct.
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1 Comment
Alexander Oleschuk
on 12 May 2018
Nice function!
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1 Comment
Dustin Jones
on 12 Jul 2017
This WAS the leading solution (not mine), but I removed the regexp function so you can see that it is a terrible solution.
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1 Comment
Jess Stuart
on 7 Feb 2017
Your weight vector should be a unit vector. W=w/norm(w) to get a true weighted average. Dividing by the max is dumb.
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1 Comment
fahad aljabal
on 29 Dec 2016
.
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1 Comment
TAYALAN SIVALINGAM
on 29 Dec 2016
keep it up
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2 Comments
Matteo Raso
on 26 Dec 2016
Nice problem
Limon
on 28 Apr 2017
easy
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1 Comment
Yoshimasa Nakatani
on 10 Dec 2016
(^o^)v
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1 Comment
David Verrelli
on 3 Sep 2018
This user (mohamed elbesealy) appears to have gamed the system, with fraudulent "likes" of this unremarkable solution submitted by 'sock-puppet' accounts. —DIV
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2 Comments
John D'Errico
on 20 Aug 2016
This is a seriously flawed problem. I'm sorry, but it is. It uses an incorrect definition of a weighted average. As such, students might solve this problem, then try to use that same expression in code for some future job. A correct solution might better be:
y = x*w'/sum(w);
Li Ding
on 24 Aug 2017
Agree the definition of "weighted average" is flawed. There's still a mistake in your line; it should be y = sum(x.*w)/sum(w), or y = dot(x,w)/sum(w).
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1 Comment
Siva Madugula
on 3 Nov 2015
That's good.
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1 Comment
Kevin Lamb
on 30 Sep 2015
how the heck can you get smaller than?
y = x*w'/3
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1 Comment
Alexander Wickstrom
on 28 May 2015
Sweet.
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1 Comment
Hugo
on 2 Apr 2015
obviously misuse of poor testing...
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1 Comment
Christian
on 4 Mar 2015
Easy, but not the weighted average. Even assuming that the first vector contains the weighting factors (for example the sizes of different sample groups) and the second vector contains the means of the sample groups, than the divisor has to be the sum of the weighting factors instead of the number.
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1 Comment
Piotr Balik
on 28 Jun 2019
This is beautiful! Smart move!
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1 Comment
Giorgos Papakonstantinou
on 5 Mar 2014
This is not the weighted average
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1 Comment
!ntel
on 8 Oct 2013
Good one
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1 Comment
david hansen
on 6 Sep 2013
odd name
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1 Comment
Randall Stace Romero Aguilar
on 23 Aug 2013
I think the problem is not well formulated: in a weighted average, your weights should add up to one
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1 Comment
Frequency Domain
on 24 Jul 2013
FP operations and isequal(), sometimes yield some bitter results
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1 Comment
sea knowledge
on 13 Dec 2012
l doubt on this definition of the problem
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1 Comment
J-G van der Toorn
on 11 Sep 2012
This is not the Weighted average. One should divide by the sum of the weighs, not by the number of them.
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1 Comment
Dimosthenis
on 20 Jul 2012
This is not the weighted average.
The weighted average has the sum of w in the denominator
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2 Comments
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1 Comment
Andrew Davis
on 2 Mar 2012
This probably shouldn't pass...
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1 Comment
Jan
on 28 Jan 2012
I like the commented "y =(x*w.')/length(x)" more than the "shorter" solution.
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