UNWEIGHTED MEAN

Saumyojit

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Given in wiki :
the grades in each class on a test were:Morning class = 62, 67, 71, 74, 76, 77, 78, 79, 79, 80, 80, 81, 81, 82, 83, 84, 86, 89, 93, 98
Afternoon class = 81, 82, 83, 84, 85, 86, 87, 87, 88, 88, 89, 89, 89, 90, 90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95, 96, 97, 98, 99
The mean for the morning class is 80 and the mean of the afternoon class is 90. The unweighted mean of the 80 and 90 is 85, so the unweighted mean of the two means is 85. However, this does not account for the difference in number of students in each class (20 versus 30); hence the value of 85 does not reflect the average student grade (independent of class)

this does not account for the difference in number of students in each class---> what does this line mean? i get it that 80 is the avg marks of per student in class A and 90 of per student in class B. so mean of avg marks of A & B= (80+90)/2

That means i am considering the average marks of one student from class A (weight of 80=1), not considering the marks of remaining 19 and same goes for class b considering the average marks of one student from class B, not considering the marks of remaining 29
@Dr.Peterson
 

Dr.Peterson

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You appear to be asking a question and then answering it yourself. Do you still have a question?

The way I would say it is that the individual means tell us that we could replace the morning class with 20 grades of 80, and the afternoon class with 30 grades of 90. If you just average the 80 and 90, you are in effect ignoring the last 10 in the afternoon class, as if the classes were the same size. And that is what "not accounting for the difference" means.
 

Saumyojit

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Please read every line sir whenever u are free dont need to hurry. Many alternatives ringing in my mind @Dr.Peterson @JeffM
If you just average the 80 and 90
if i consider the size was 20 for both the classes; given the avg of class A is 80 and avg of marks of b is 90;
that means the result of this average i.e 85 will represent the avg marks of per student of both class A and B. what if i did it like this : (80*20+90*20)/40=8.5 ...this 8.5 is also a average in some sense right . I know i am wrong.

Why this method will be wrong (80*20+90*20)/40=8.5 as the weightage are same so it useless to use weighted avg where weights are equal; its the same as writing as (80+90)/2 ? but the result is different not the same 8.5 vs 85!

but i want to know that the difference btwen former method{(80*20+90*20)/40} and {(80+90)/2} ; the former considers all the 20 students grades represented by one no 80 and 90 or the sum of marks of 20 student each in a&B in the numerator part(80*20+90*20) & the latter considers only the avg marks per 1 student in class A and per 1 student in class B in the numerator part(80*1+90*1) ?

If you just average the 80 and 90, as if the classes were the same size. And that is what "not accounting for the difference" means.
(80+90)/2 suppose i dont know size of each class but i have only the method (80+90)/2 written on paper and questioner tells me its given in info that 80 & 90 are absolute or avg marks of 2 classes ; I as a answerer will conclude that size of each class will be same as weights are equal; so can i say both the classes have only one student each instead of(20 or any random no)..why i chose 1? that because (80+90)/2 means (80*1+90*1)/2
as implicitly weight is 1 thats why i thought one student got 80(absolute mark not avg in this case) from A & one got 90 from B. This thing came from my intution just now.

you are in effect ignoring the last 10 in the afternoon class
How? Back to the original data. 30 vs 20 students..i am mistakenly finding out the avg of both the averages marks like this ----> (80*1+90*1)/2 90 means inherently it is representing "one" of 30 grades and 80 inherently represents "one" of 20 grades ; so i think that 90 is considering implicitly the extra 10 students grades that why the avg of class b
(sum of all marks /30) came 90 .. i dont get how it is " ignoring the last 10"
 

Dr.Peterson

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Thanks for your patience; I had a lot of work today, doing live online tutoring.

if i consider the size was 20 for both the classes; given the avg of class A is 80 and avg of marks of b is 90;
that means the result of this average i.e 85 will represent the avg marks of per student of both class A and B. what if i did it like this : (80*20+90*20)/40=8.5 ...this 8.5 is also a average in some sense right . I know i am wrong.

Why this method will be wrong (80*20+90*20)/40=8.5 as the weightage are same so it useless to use weighted avg where weights are equal; its the same as writing as (80+90)/2 ? but the result is different not the same 8.5 vs 85!

but i want to know that the difference btwen former method{(80*20+90*20)/40} and {(80+90)/2} ; the former considers all the 20 students grades represented by one no 80 and 90 or the sum of marks of 20 student each in a&B in the numerator part(80*20+90*20) & the latter considers only the avg marks per 1 student in class A and per 1 student in class B in the numerator part(80*1+90*1) ?
You just did the arithmetic wrong! (80*20+90*20)/40=3400/40=85. Did you check your work?

A weighted average is intended to give exactly the same result as the average of the individual items (here, twenty 80's and twenty 90's); and it does so even when the weights are the same.

But when the weights are the same (two classes of the same size), then you get the same result if you just average the averages, (80+90)/2 = 85 as you said.

So these three paragraphs of yours were wasted, due to failure to add carefully.

(80+90)/2 suppose i dont know size of each class but i have only the method (80+90)/2 written on paper and questioner tells me its given in info that 80 & 90 are absolute or avg marks of 2 classes ; I as a answerer will conclude that size of each class will be same as weights are equal; so can i say both the classes have only one student each instead of(20 or any random no)..why i chose 1? that because (80+90)/2 means (80*1+90*1)/2
as implicitly weight is 1 thats why i thought one student got 80(absolute mark not avg in this case) from A & one got 90 from B. This thing came from my intution just now.
All it tells you is that the weights are equal, not what they are. You could be averaging a class of 5000 and another class of 5000. In general, (80n + 90n)/(n+n) = (80+90)/2 = 85, regardless of n.

How? Back to the original data. 30 vs 20 students..i am mistakenly finding out the avg of both the averages marks like this ----> (80*1+90*1)/2 90 means inherently it is representing "one" of 30 grades and 80 inherently represents "one" of 20 grades ; so i think that 90 is considering implicitly the extra 10 students grades that why the avg of class b
(sum of all marks /30) came 90 .. i dont get how it is " ignoring the last 10"
The average of the 20 80's and 30 90's would look like this:

(80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+​
90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+​
90+90+90+90+90+90+90+90+90+90)/50.​

An average with both weights being 20 (or any other equal weights, which I've shown would yield the same result) would be equivalent to this:

(80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+80+​
90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90+90)/40.​

You ignored the last 10 students, by acting as if they weren't there.
 

JeffM

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It seems to me that you make things hard for yourself by trying to make analogies or to find exceptions instead of just following where the mathematics logically leads.

Let's take a very simple example.

Class has three female students with grades of 70, 80, and 90, and five male students with grades of 50, 50, 70, 90, and 90.

The average grade of female students was

\(\displaystyle \dfrac{70 + 80 + 90}{3} = \dfrac{240}{3} = 80.\)

The average grade of male students was

\(\displaystyle \dfrac{50 + 50 + 70 + 90 + 90}{5} = \dfrac{350}{5} = 70.\)

The average grade of all students was

\(\displaystyle \dfrac{70 + 80 + 90 + 50 + 50 + 70 + 90 + 90}{8} =\\

\dfrac{590}{8} = 73.75.\)
Now obviously the mean of 70 and 80 is 75, which is NOT THE AVERAGE FOR THE CLASS.

What causes the error is that averaging the averages gives equal weight to female and male grades even though the numbers of females and males in the class are unequal. It does not make intuitive sense that unequal things should have an equal effect on the outcome.

But we do not need intuition, which may lead us astray because we have now proved by example that averaging means may lead to error.

Let's generalize.

Set W has two discrete non-empty subsets, A and B.

Subset A has a > 0 items. Subset B has b > 0 items. Set W has a + b items.

The sum of the values of the a items in subset A is x. The sum of the values of the b items in subset B is y. The sum of the values of the (a + b) items in set W is (x + y).

The mean value of the a items in subset A is x/a. The mean value of the b items in subset B is y/b. The mean value of the (a + b) items in Set W is (x + y)/(a + b). Any problem seeing that we are expressing the problem in complete generality?

\(\displaystyle \dfrac{x + y}{a + b} \equiv \\

\dfrac{x}{a + b} + \dfrac{y}{a + b} \equiv 1 * \dfrac{x}{a + b} + 1 * \dfrac{y}{a + b} \equiv \\

\dfrac{a}{a} * \dfrac{x}{a + b} + \dfrac{b}{b} * \dfrac{y}{a + b} \equiv \dfrac{a}{a + b} * \dfrac{x}{a} + \dfrac{b}{a + b} * \dfrac{y}{b}.\)
In short,

\(\displaystyle \dfrac{x + y}{a + b} \equiv \dfrac{a}{a + b} * \dfrac{x}{a} + \dfrac{b}{a + b} * \dfrac{y}{b}\) now and forever, Amen.

Only under special circumstances will the unweighted mean of the subsets' means equal the set's mean.
 
Last edited:

JeffM

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It occurs to me that you may be curious about the special circumstances mentioned in my previous post. If you're not curious, just ignore this post because it is far less important than the previous post's general result, WHICH ALWAYS IS CORRECT.

\(\displaystyle \dfrac{x + y}{a + b} = \dfrac{1}{2} * \left ( \dfrac{x}{a} + \dfrac{y}{b} \right ) = \dfrac{bx + ay}{2ab} \implies\)

\(\displaystyle 2ab(x + y) = (a + b)(bx + ay) \implies\)

\(\displaystyle \cancel 2 abx + \cancel 2 aby = \cancel {abx} + a^2y + b^2x + \cancel {aby} \implies\)

\(\displaystyle a^2y - aby + b^2x - abx = 0 \implies\)

\(\displaystyle ay(a - b) + bx(a - b) = 0 \implies (a - b)(ay + bx) = 0.\)

Now the zero product property comes into play. If x and y are both non-negative, the only way that you can get the special case is when a = b. If, however, the product of x and y is negative, then the special case will also arise if x/a = - y/b. So we get this theorem

\(\displaystyle \dfrac{x + y}{a + b} = \dfrac{1}{2} * \left ( \dfrac{x}{a} + \dfrac{y}{b} \right ) = \dfrac{bx + ay}{2ab} \iff a = b \text { or } \dfrac{x}{a} = - \dfrac{y}{b}.\)
 

HallsofIvy

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Please read every line sir whenever u are free dont need to hurry. Many alternatives ringing in my mind @Dr.Peterson @JeffM
if i consider the size was 20 for both the classes; given the avg of class A is 80 and avg of marks of b is 90;
that means the result of this average i.e 85 will represent the avg marks of per student of both class A and B. what if i did it like this : (80*20+90*20)/40=8.5 ...this 8.5 is also a average in some sense right . I know i am wrong.

Why this method will be wrong (80*20+90*20)/40=8.5
That's just bad arithmetic! \(\displaystyle (80*20+ 90*20)/40= (1600+ 1800)/40= 3400/2= 85\), not 8.5!

as the weightage are same so it useless to use weighted avg where weights are equal; its the same as writing as (80+90)/2 ? but the result is different not the same 8.5 vs 85!

but i want to know that the difference btwen former method{(80*20+90*20)/40} and {(80+90)/2} ; the former considers all the 20 students grades represented by one no 80 and 90 or the sum of marks of 20 student each in a&B in the numerator part(80*20+90*20) & the latter considers only the avg marks per 1 student in class A and per 1 student in class B in the numerator part(80*1+90*1) ?

(80+90)/2 suppose i dont know size of each class but i have only the method (80+90)/2 written on paper and questioner tells me its given in info that 80 & 90 are absolute or avg marks of 2 classes ; I as a answerer will conclude that size of each class will be same as weights are equal; so can i say both the classes have only one student each instead of(20 or any random no)..why i chose 1? that because (80+90)/2 means (80*1+90*1)/2
as implicitly weight is 1 thats why i thought one student got 80(absolute mark not avg in this case) from A & one got 90 from B. This thing came from my intution just now.

How? Back to the original data. 30 vs 20 students..i am mistakenly finding out the avg of both the averages marks like this ----> (80*1+90*1)/2 90 means inherently it is representing "one" of 30 grades and 80 inherently represents "one" of 20 grades ; so i think that 90 is considering implicitly the extra 10 students grades that why the avg of class b
(sum of all marks /30) came 90 .. i dont get how it is " ignoring the last 10"
 
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