3. Weighted Average
Weighted average method is applied where we need to find the average of groups of different sizes. Where x1,x2,...,xn are the values and w1,w2,...,wn are the weights, the weighted average is
x=w1+w2+w3+...+wnw1x1+w2x2+w3x3+...+wnxn=i=1∑nwii=1∑nwixi
Average is to be found for the values, i.e., the x -terms. Weights, i.e., the w -terms are the number of occurrences of these x-terms.
(Note: To find the weighted average of two x-values, we can use the alligation method detailed in the
Mixtures & Alligations lesson.)
Example 6
The average of the percentage marks scored by Standard X students in four schools - R, S, T and U are 75,80,68 and 90 respectively. If number of students in R, S, T and U are 48,60,84 and 96 respectively, what is the average of the percentage marks scored by students in the 4 schools combined?
Solution
In this question, the averages of the respective schools are provided. However, the number of students (the weights) in each school is different.
Average =48+60+84+9648×75+60×80+84×68+96×90
=2883600+4800+5712+8640 =28822752=79
Alternatively (Recommended Method)
The impact of weights is always relative to one another.
∴ Weights of 48,60,84 and 96 can be simplified as a ratio and then applied.
48:60:84:96=4:5:7:8
Average =4+5+7+84×75+5×80+7×68+8×90=24300+400+476+720=241896=4316=79
Answer: 79
Example 7
In a class, 30% of the students play 1 sport, 30% play 2 sports and 40% play 3 sports. What is the average number of sports played by a student?
Solution
We need to find the average for the number of sports, which are the x-values. The weights here are the number of students, which are given as percentages.
Average =0.3+0.3+0.40.3×1+0.3×2+0.4×3=10.3+0.6+1.2=2.1
Note: 100%=1. As the sum of percentages equals 100%, going forward, we needn't separately add them up in the denominator.
Answer: 2.1
Example 8
The average age of 90 officers in a division is 36 in 2000. In 2004,45 new officers, whose average age was 31, joined the division. Besides this, if no other officer left or joined the division between 2000 and 2010, what was the average age of the officers in the division in 2010?
Solution
As each of these officers get 4 years older in 2004, their average age will also increase by 4.
∴ Average age of the 90 officers in 2004=36+4=40
Average age of the 45 officers joining in 2004=31
Ratio of weights =90:45=2:1
Average age of all the officers in 2004=2+12×40+1×31=37
6 years later, the age of each of these officers would have increased by 6.
∴ Average of all the officers in 2010=37+6=43
Alternatively
This is not the recommended method. This involves applying simple average, i.e. sum of values by the number of items.
Sum of ages of 90 officers in 2004=40×90=3600
Sum of ages of 40 officers in 2004=31×45=1395
Average age in 2004=90+453600+1395=37
Average age in 2010 =37+6=43
Answer: 43
3.1 Weighted Average using Assumed Mean
This is similar to the assumed mean in the previous section. The difference here is the application of weights.
Where
a is the Assumed Mean, xi is a value and wi is a weight,
Weighted Average =a+i=1∑nwii=1∑n(xi−a)×wi=a+i=1∑nwii=1∑ndi×wi
Example 9
A group of students were asked by their teacher to count the number of leaves in a tree. 20%,15%,25%,30% and 10% of the students responded 3345,3346,3348,3349 and 3350 respectively. What was the average number reported by the students?
Solution
Let the assumed average, a=3348
| Leaves (xi) |
wi |
a |
di=(xi−a) |
di×wi |
| 3345 |
0.20 |
3348 |
−3 |
−0.60 |
| 3346 |
0.15 |
3348 |
−2 |
−0.30 |
| 3348 |
0.25 |
3348 |
0 |
0 |
| 3349 |
0.30 |
3348 |
1 |
0.30 |
| 3350 |
0.10 |
3348 |
2 |
0.20 |
| Total |
|
|
|
−0.40 |
Average
=a+i=1∑nwii=1∑ndi×wi=3348+1−0.4=3347.6
Answer: 3347.6