
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
April 9th, 2018, 10:44 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 294 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. 
April 10th, 2018, 01:46 AM  #2  
Senior Member Joined: Jun 2015 From: England Posts: 831 Thanks: 244 
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:
$\displaystyle \frac{{A + B}}{{C + D}} \ne \frac{A}{C} + \frac{B}{D}$ Last edited by studiot; April 10th, 2018 at 01:58 AM.  
April 10th, 2018, 09:18 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 294 Thanks: 1 
This is different question.

April 10th, 2018, 10:10 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,174 Thanks: 1644 
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.

April 10th, 2018, 01:34 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 
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.

April 10th, 2018, 01:49 PM  #6  
Newbie Joined: Nov 2013 Posts: 26 Thanks: 8  Quote:
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.  

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 