My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Thanks Tree2Thanks
  • 2 Post By Jomo
Reply
 
LinkBack Thread Tools Display Modes
April 9th, 2018, 10:44 PM   #1
Senior Member
 
Joined: Aug 2014
From: India

Posts: 310
Thanks: 1

[Weighted average] Is last para saying that first para is wrong?

In my book, in Averages chapter there is a line which tells us:

There are two sections A and B of a class where the average height of section A is 150 cm and that of section B is 160 cm. On the basis of this information alone, we cannot find the average of entire class (of the two sections).

The Average height of the entire class is

$\large\frac {total\,height\,of\,the\,entire\, class}{total\, number\, of\, students\, in\, the\, entire \,class}$

Since we don't have any information regarding the number of students in the two sections, we cannot find the average of the entire class. Now, suppose that we are given that there are 60 students in section A and 40 students in section B, then we can calculate the average height of the entire class which, in this case will be equal to

$\large\frac {60\times150 + 40\times160}{60+40} = \small154 cm$.

The average height 154 cm of the entire class is called "weighted average" of the class.

The above step in calculating the weighted average of the class can be rewritten as below:

$\large\frac{60\times150+40\times160}{60+40} = \frac{60}{100}\small150$ + $\large\frac{40}{100} \small160$

= $\large\frac{3}{5}\small150 + \large\frac{2}{5}\small160 $

It is clear from the above step that we would have been able to calculate the average height of the entire class even if we had not been given the height of students in the individual section but only the ratio of the number of students in the two sections (which in this case is 3:2)

My doubt is: In first para says: we cannot find the average of entire class.

And last para says: we would have been able to calculate the average height of the entire class even if we had not been given the height of students in the individual section.

I didn't understand.

Is last para saying that first para is wrong?

Last edited by skipjack; April 10th, 2018 at 10:06 AM.
Ganesh Ujwal is offline  
 
April 10th, 2018, 01:46 AM   #2
Senior Member
 
Joined: Jun 2015
From: England

Posts: 853
Thanks: 258

Why did you start another thread about the same question?

Why did you not continue in your original thread, where your question has already been answered?

Why is my book shows that we cannot find the average of entire class?

Quote:
Is last para saying that first para is wrong?
No, it is saying

$\displaystyle \frac{{A + B}}{{C + D}} \ne \frac{A}{C} + \frac{B}{D}$

Last edited by studiot; April 10th, 2018 at 01:58 AM.
studiot is offline  
April 10th, 2018, 09:18 AM   #3
Senior Member
 
Joined: Aug 2014
From: India

Posts: 310
Thanks: 1

This is different question.
Ganesh Ujwal is offline  
April 10th, 2018, 10:10 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 19,511
Thanks: 1743

I think you misquoted "we would have been able to calculate the average height of the entire class even if we had not been given the height of students in the individual section", as that is untrue.
skipjack is offline  
April 10th, 2018, 01:34 PM   #5
Global Moderator
 
Joined: May 2007

Posts: 6,582
Thanks: 610

The point is we cannot find the average without weights. These weights can be given as the total numbers in each group or at least the ratio of these numbers.
mathman is online now  
April 10th, 2018, 01:49 PM   #6
Newbie
 
Joined: Nov 2013

Posts: 26
Thanks: 8

Quote:
Originally Posted by Ganesh Ujwal View Post
In my book, in Averages chapter there is a line which tells us:

There are two sections A and B of a class where the average height of section A is 150 cm and that of section B is 160 cm. On the basis of this information alone, we cannot find the average of entire class (of the two sections).

The Average height of the entire class is

$\large\frac {total\,height\,of\,the\,entire\, class}{total\, number\, of\, students\, in\, the\, entire \,class}$

Since we don't have any information regarding the number of students in the two sections, we cannot find the average of the entire class. Now, suppose that we are given that there are 60 students in section A and 40 students in section B, then we can calculate the average height of the entire class which, in this case will be equal to

$\large\frac {60\times150 + 40\times160}{60+40} = \small154 cm$.

The average height 154 cm of the entire class is called "weighted average" of the class.

The above step in calculating the weighted average of the class can be rewritten as below:

$\large\frac{60\times150+40\times160}{60+40} = \frac{60}{100}\small150$ + $\large\frac{40}{100} \small160$

= $\large\frac{3}{5}\small150 + \large\frac{2}{5}\small160 $

It is clear from the above step that we would have been able to calculate the average height of the entire class even if we had not been given the height of students in the individual section but only the ratio of the number of students in the two sections (which in this case is 3:2)

My doubt is: In first para says: we cannot find the average of entire class.

And last para says: we would have been able to calculate the average height of the entire class even if we had not been given the height of students in the individual section.

I didn't understand.

Is last para saying that first para is wrong?
The 1st paragraph says that if you only know that the average height of section A is 150 cm and that of section B is 160 cm you can not find the average of the total class. This statement is true.

Then it went on to say that if you knew the number of students in each section then you can find the average. This statement is true.

Then the math showed after reducing that you don't need to even know the size of each section BUT rather you can also find the average if you know the ratio.

That last paragraph says that you can find the average of the class BUT you need more information then given in 1st para. You either need the size of each section or the ratio or maybe even something else (which was not given).

In either case, you need more than the average of each section.
Thanks from topsquark and JeffM1
Jomo is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
average, para, weighted, wrong



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Weighted average? pdmm Elementary Math 3 June 11th, 2017 01:58 PM
Do magnets\ magnetic fields become weaker when exposed to para magnetic objects? Ganesh Ujwal Physics 1 December 27th, 2014 04:14 AM
Weighted Average hayashiryo Elementary Math 3 September 21st, 2014 05:47 PM
hola soy dominicana y estoy acá para aprender . claudiadominicana New Users 1 April 1st, 2014 06:09 AM
para equation kfarnan Algebra 4 November 23rd, 2010 12:16 PM





Copyright © 2018 My Math Forum. All rights reserved.