Problem 106. Weighted average

Created by Will

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Solution Stats

14019 Solutions
7560 Solvers

Last Solution submitted on Sep 25, 2020

Last 200 Solutions

Problem Comments

21 Comments

21 Comments

Show 18 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

Solution Comments

Solution 1710206

2 Comments

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.

Solution 1526738

1 Comment

1 Comment

Alexander Oleschuk on 12 May 2018

Nice function!

Solution 1230064

1 Comment

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.

Solution 1117120

1 Comment

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.

Solution 1094092

1 Comment

1 Comment

fahad aljabal on 29 Dec 2016

Solution 1093871

1 Comment

1 Comment

TAYALAN SIVALINGAM on 29 Dec 2016

keep it up

Solution 1089812

2 Comments

2 Comments

Matteo Raso on 26 Dec 2016

Nice problem

Limon on 28 Apr 2017

easy

Solution 1078502

1 Comment

1 Comment

Yoshimasa Nakatani on 10 Dec 2016

(^o^)v

Solution 1017453

1 Comment

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

Solution 939161

2 Comments

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).

Solution 770817

1 Comment

1 Comment

Siva Madugula on 3 Nov 2015

That's good.

Solution 750303

1 Comment

1 Comment

Kevin Lamb on 30 Sep 2015

how the heck can you get smaller than?
y = x*w'/3

Solution 677180

1 Comment

1 Comment

Alexander Wickstrom on 28 May 2015

Sweet.

Solution 610202

1 Comment

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Hugo on 2 Apr 2015

obviously misuse of poor testing...

Solution 591260

1 Comment

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.

Solution 582669