WORKSHEET ON WEIGHTED AVERAGES:
Hi welcome visitors to our website:
in this worksheet, I am going to discuss in detail the concept of weighted average and one very interesting short-cut to solve questions on weighted average.
To enrich your learning, you can also visit our you tube on the concept of weighted average.
So, let’s get on with the formula of weighted average, the property of weighted average and finally the short cut to find the ratio of the number of terms in two groups whose individual averages are given.
Let us take up the following problem on weighted average:
SOLVED PROBLEM 1:
In a group of 80 members, the number of males is 45 and the number of females is 35. The average age of the males is m and that of the females is f. compare the average age of the two groups compared to the average of the ages of the two groups i.e. (f + m)/2. f > m.
Solution:
First all, let us set up the data for weighted average as we do normally as follows:
If you have watched our webpage on weighted average and its property, then you will understand that weighted average will be a number nearer to the average of the group of males i.e. ‘m’, which is the average of the group having more number of terms.
(here, terms represent ages of the males and females)
Now, tell me if a number is closer to the average ‘m’ than to the other average ‘f’, then what will it be compared to (f + m)/2 which is the midpoint of the two average numbers m and f?
As per the data written above, you may conclude that since the weighted average age of the two groups males and females lies closer to the average age ‘m’ from the property of weighted average, therefore from the diagram drawn above in the data representation, weighted average age of the two groups will be less than the midpoint (f + m)/2.
Here lies the mistake often committed by students while solving this question on weighted average age of the two groups of males and female in the above question.
My counter to your answer is:
Your conclusion that weighted average age of the two groups is nearer to the average age ‘m’ than to the other average ‘f’ is true; but compared to (f + m)/2 it is less or greater cannot be still determined without knowing the relationship between ‘f’ and ‘m’.
Because there can exist three relationships between ‘f’ and ‘m’:
f < m or f = m or f > m
i.e. the average age of the females group may be less than the average age of the males or
the average age of the group of females may be same as that of the average age of the males or even
the average age of the group of females could be greater than the average of the males.
The exploration of the above three possibilities of relationship between the average age of the females and the average age of the males is triggered by the information given by the question : f > m
Suppose the average age of the two groups of females and the males is same i.e. f = m, then how can you conclude from the property of weighted average that the weighted average age of the two groups will be less than the average of the average ages: (f + m)/2.
What I am trying to drive home the point is that the weighted average property will enable you to tell that weighted average age of the two groups will definitely be closer to the average of the males, but what the property cannot tell you is what it will be compared to the average (f + m)/2.
To be able to determine the second point above i.e. what is weighted average compared to the average (f + m)/2, we need to know the relationship between the average ages ‘m’ and ‘f’, i.e. which of the three relationships between them exists as explored above.
That’s why the question has supplied the additional detail f > m.
We must realize the importance of the information f > m and its purpose is to determine if weighted average age, which is closer to the average age ‘m’ of the males, is less than or equal to or greater than the average of the average ages : f and m i.e. (f + m)/2
Now, after having ascertained the purpose of the relationship between f and m i.e. f > m, we can now give the answer that weighted average age of the two groups will be less than the average (f + m)/2.
If it is not still clear, then we can make the explanation more clear by assuming some numbers for f and m based on the relationship f > m.
Let us take average age of the males’ group, m as 20 and the average age of the females’ group as some 40
So by substituting the above assumed values for f and m, the average (f + m)/2 will be (40 + 20)/2 = 60/2 = 30
Now, you can clearly see that the weighted average age of the two groups males and females will be less than 30.
How? Its not clear still. Ok, we will go as below:
From the weighted average property, what we can say is the weighted average age of the two males and females whose individual average ages have been taken as 20 and 40 will be a number closer to the average age of the males which is 20.
A number which is closer to the average age 20 than to the other average age 40 must by implication be less than the midpoint of 20 and 40 i.e. 30
For example, 28 is closer to 20 than to 40 as 28 is only 8 numbers away from 20 but slightly more away from 40, i.e. 12 away from 40.
Now figure out in your mind what 28 is compared to 30.
Obviously less.
That’s why weighted average age of the two groups by being nearer to the average age of the males i.e. ‘m’ will be also less than the average of the average ages i.e. (f + m)/2 by taking into consideration the information f > m given in the question.
Now I will tell you a short cut to solve a question on weighted average:
Consider the following problem:
The average marks of a group having ‘p’ students in a class is 75 and the average marks of another group having ‘n’ students is 92. Find what value the fraction p/n will be, if the weighted average marks of the two groups is 84
Solution:
First of all, present the data for the weighted average as we do usually:
To find the value of the fraction p/n, move in the direction of the arrows as drawn above in the presentation:
p / n = (92 – 84)/(84 – 75) = 8/9.
That’s it.
Isn’t that a wonderful short-cut.
Well, the above fraction p/n signifies another important property of weighted average, which will discuss in another worksheet.
