Averages
Average: Definition and Calculation Methods
The Average, also known as the arithmetic mean, is a fundamental statistical measure of central tendency. It provides a single value that represents the typical or central value of a set of numbers. The average is calculated by summing all the values in a dataset and then dividing this sum by the total number of values in the dataset.
Formula for Average:
Given a set of $n$ individual values, let's denote them as $x_1, x_2, x_3, \dots, x_n$. The formula for calculating their average is:
$\boldsymbol{\text{Average} = \frac{\text{Sum of all values}}{\text{Number of values}}}$
... (i)
Using summation notation ($\sum$) which represents the sum of a series of values, the formula can be written concisely as:
$\boldsymbol{\text{Average} = \frac{\sum_{i=1}^{n} x_i}{n}}$
... (ii)
From the definition of the average, we can also derive a formula to find the sum of all values if we know the average and the number of values:
$\boldsymbol{\text{Sum of all values} = \text{Average} \times \text{Number of values}}$
... (iii)
This formula is particularly useful when dealing with averages of groups or finding the total quantity when only the average and the count are known.
Example 1. Find the average of the following numbers: 15, 25, 35, 45, 55.
Answer:
Given values are: 15, 25, 35, 45, 55.
The number of values in the dataset ($n$) is 5.
Calculate the sum of all values:
Sum $= 15 + 25 + 35 + 45 + 55 = 175$
Using the formula (i): $\text{Average} = \frac{\text{Sum of all values}}{\text{Number of values}}$
$\text{Average} = \frac{175}{5}$
Perform the division:
$\boldsymbol{\text{Average} = 35}$
The average of the given numbers is $\boldsymbol{35}$.
Example 2. The average height of 8 students in a class is 155 cm. Find the total height of all 8 students.
Answer:
Given: Average height $= 155$ cm.
Number of students ($n$) $= 8$.
We need to find the total height, which is the sum of the heights of all students.
Using the formula (iii): $\text{Sum of all values} = \text{Average} \times \text{Number of values}$
Sum of heights $= \text{Average height} \times \text{Number of students}$
Sum of heights $= 155 \text{ cm} \times 8$
Perform the multiplication:
$\begin{array}{cccc}& & 1 & 5 & 5 \\ \times & & & & 8 \\ \hline & 1 & 2 & 4 & 0 \\ \hline \end{array}$
$\boldsymbol{\text{Sum of heights} = 1240 \text{ cm}}$
The total height of all 8 students is $\boldsymbol{1240}$ cm.
Average of Consecutive Numbers / Arithmetic Progressions (AP):
A special case arises when the dataset consists of numbers that form an arithmetic progression (AP), i.e., consecutive numbers or numbers with a constant difference between successive terms (e.g., 2, 4, 6, 8 or 5, 10, 15, 20).
For an arithmetic progression, the average has a property that simplifies calculation:
- If the number of terms is odd, the average is simply the middle term of the arranged sequence.
- If the number of terms is even, the average is the arithmetic mean of the two middle terms.
More generally, for any arithmetic progression, the average is equal to the average of the first and the last term:
$\boldsymbol{\text{Average (for AP)} = \frac{\text{First Term + Last Term}}{2}}$
... (iv)
Examples:
- Find the average of the first 10 natural numbers (1, 2, 3, ..., 10). These form an AP.
- Find the average of the numbers 15, 25, 35, 45, 55 (from Example 1). These are in AP (difference is 10). Number of terms is 5 (odd).
- Find the average of the numbers 5, 10, 15, 20. These are in AP (difference is 5). Number of terms is 4 (even).
First term = 1, Last term = 10.
Average $= \frac{1 + 10}{2} = \frac{11}{2} = \boldsymbol{5.5}$
The middle term is 35.
Using formula (iv): Average $= \frac{\text{First Term + Last Term}}{2} = \frac{15 + 55}{2} = \frac{70}{2} = 35$.
Both methods give the same result, 35.
The two middle terms are 10 and 15.
Average of middle terms $= \frac{10+15}{2} = \frac{25}{2} = 12.5$.
Using formula (iv): Average $= \frac{\text{First Term + Last Term}}{2} = \frac{5 + 20}{2} = \frac{25}{2} = 12.5$.
Both methods give the same result, 12.5.
Competitive Exam Notes:
Average is a frequently tested concept, often combined with other topics. Understand the definition and formulas well.
- Basic Formula: Average = Sum / Number of values. Rearrange to get Sum = Average $\times$ Number. This is fundamental.
- Arithmetic Progressions: Use the shortcut $\frac{\text{First} + \text{Last}}{2}$ for APs (consecutive numbers, or numbers with constant difference). Recognize APs quickly.
- Weighted Average: (Discussed in a later section) Used when different values have different importance or frequencies.
- Common Mistakes: Not using Total Sum and Total Number for average, or simply averaging different averages (which is incorrect unless the number of values in each group is the same).
Calculating Average of Grouped Data
When a large amount of data is organized into a frequency distribution, either as individual values with their frequencies or as class intervals with their frequencies, calculating the average involves a slightly different approach than simply summing individual values. This is known as calculating the average of grouped data.
Formula for Average of Grouped Data (Discrete Frequency Distribution):
If you have a dataset where specific values appear multiple times, represented by their frequencies, the data is called a discrete frequency distribution. Let the distinct values be $x_1, x_2, \dots, x_k$, and their corresponding frequencies (how many times each value appears) be $f_1, f_2, \dots, f_k$.
The sum of all values is not simply $x_1 + x_2 + \dots + x_k$, but rather the sum of each value multiplied by its frequency: $(x_1 \times f_1) + (x_2 \times f_2) + \dots + (x_k \times f_k)$.
The total number of values is the sum of all frequencies: $f_1 + f_2 + \dots + f_k$.
Using the definition of average (Sum / Number of values), the formula for the average of grouped data is:
$\boldsymbol{\text{Average} = \frac{\text{Sum of (Value} \times \text{Frequency)}}{\text{Sum of Frequencies}} = \frac{\sum_{i=1}^{k} f_i x_i}{\sum_{i=1}^{k} f_i}}$
... (v)
Here, $\sum f_i x_i$ means the sum of the products of each value and its frequency, and $\sum f_i$ means the sum of all frequencies, which gives the total count of observations in the dataset.
Calculating Average for Data in Class Intervals (Continuous Frequency Distribution):
When data is grouped into class intervals (e.g., 0-10, 10-20, etc.), we assume that the values within each interval are evenly distributed around the midpoint of the interval. Therefore, we use the class mark (or midpoint) of each interval as the representative value ($x_i$) for that interval.
$\boldsymbol{\text{Class Mark} = \frac{\text{Lower Limit of Class Interval + Upper Limit of Class Interval}}{2}}$
Once the class mark for each interval is calculated, the formula for the average is the same as for a discrete frequency distribution (formula v), where $x_i$ is the class mark of the $i$-th interval and $f_i$ is the frequency of the $i$-th interval.
$\boldsymbol{\text{Average} = \frac{\sum (\text{Class Mark} \times \text{Frequency})}{\sum \text{Frequency}}}$
Example 1. Find the average of the following data representing the number of siblings per student in a survey:
| Number of Siblings ($x_i$) | Number of Students ($f_i$) |
|---|---|
| 0 | 5 |
| 1 | 12 |
| 2 | 8 |
| 3 | 3 |
| 4 | 2 |
Answer:
We need to use the formula $\text{Average} = \frac{\sum f_i x_i}{\sum f_i}$.
Create a table to calculate $f_i x_i$ and find the sums:
| Number of Siblings ($x_i$) | Number of Students ($f_i$) | $f_i x_i$ (Product of Siblings $\times$ Students) |
|---|---|---|
| 0 | 5 | $0 \times 5 = 0$ |
| 1 | 12 | $1 \times 12 = 12$ |
| 2 | 8 | $2 \times 8 = 16$ |
| 3 | 3 | $3 \times 3 = 9$ |
| 4 | 2 | $4 \times 2 = 8$ |
| Total | $\sum f_i = 5 + 12 + 8 + 3 + 2 = 30$ | $\sum f_i x_i = 0 + 12 + 16 + 9 + 8 = 45$ |
Total number of students ($\sum f_i$) = 30.
Sum of (Number of Siblings $\times$ Students) ($\sum f_i x_i$) = 45.
Using the formula (v):
$\text{Average} = \frac{\sum f_i x_i}{\sum f_i} = \frac{45}{30}$
Simplify the fraction:
$\frac{45}{30} = \frac{3}{2} = 1.5$
$\boldsymbol{\text{Average} = 1.5}$
The average number of siblings per student is $\boldsymbol{1.5}$.
Example 2. The following table shows the marks obtained by students in a test. Find the average marks.
| Marks (Class Interval) | Number of Students (Frequency, $f_i$) |
|---|---|
| 0-10 | 4 |
| 10-20 | 6 |
| 20-30 | 8 |
| 30-40 | 5 |
| 40-50 | 2 |
Answer:
The data is grouped into class intervals. We need to find the class mark ($x_i$) for each interval and then use the formula $\text{Average} = \frac{\sum f_i x_i}{\sum f_i}$.
Calculate class marks and $f_i x_i$ values:
| Marks (Class Interval) | Number of Students ($f_i$) | Class Mark ($x_i$) | $f_i x_i$ |
|---|---|---|---|
| 0-10 | 4 | $\frac{0+10}{2} = 5$ | $4 \times 5 = 20$ |
| 10-20 | 6 | $\frac{10+20}{2} = 15$ | $6 \times 15 = 90$ |
| 20-30 | 8 | $\frac{20+30}{2} = 25$ | $8 \times 25 = 200$ |
| 30-40 | 5 | $\frac{30+40}{2} = 35$ | $5 \times 35 = 175$ |
| 40-50 | 2 | $\frac{40+50}{2} = 45$ | $2 \times 45 = 90$ |
| Total | $\sum f_i = 4+6+8+5+2 = 25$ | $\sum f_i x_i = 20+90+200+175+90 = 575$ |
Total number of students ($\sum f_i$) = 25.
Sum of (Class Mark $\times$ Frequency) ($\sum f_i x_i$) = 575.
Using the formula (v):
$\text{Average} = \frac{\sum f_i x_i}{\sum f_i} = \frac{575}{25}$
Perform the division:
$\frac{575}{25} = \frac{575 \div 25}{25 \div 25} = \frac{23}{1} = 23$
$\boldsymbol{\text{Average Marks} = 23}$
The average marks obtained by students is $\boldsymbol{23}$.
Competitive Exam Notes:
Calculating the average of grouped data is common, especially in Data Interpretation problems. Be proficient in using the weighted average formula.
- Formula: Average $= \frac{\sum f_i x_i}{\sum f_i}$. This is a form of weighted average where frequencies are the weights.
- Discrete Data: $x_i$ are the given values.
- Continuous Data (Class Intervals): $x_i$ are the midpoints (class marks) of the intervals. Calculate class marks carefully.
- Calculation Steps: Create columns for $f_i$, $x_i$ (or calculate $x_i$ from intervals), and $f_i x_i$. Sum the $f_i$ column and the $f_i x_i$ column. Divide $\sum f_i x_i$ by $\sum f_i$.
Problems involving Addition, Removal, or Replacement of Items
Many problems involving averages deal with changes in the dataset, specifically when items are added, removed, or when one item is replaced by another. These operations affect the sum of the values and the number of items, consequently changing the average. Understanding how these changes relate allows us to find the value of the item that was added, removed, or replaced, or to find the new average.
Finding the Value of an Added Item:
Suppose you have a group of $n$ items with an average of $A$. The total sum of these $n$ items is $n \times A$. If a new item with value $x$ is added to this group, the total number of items becomes $n+1$, and the new sum of values becomes $(n \times A) + x$. If the new average after adding the item is $A_{\text{new}}$, then by the definition of average:
$\text{New Average} = \frac{\text{New Sum}}{\text{New Number of Items}}$
$\boldsymbol{A_{\text{new}} = \frac{nA + x}{n+1}}$
... (vi)
To find the value of the added item ($x$), rearrange equation (vi):
$(n+1) \times A_{\text{new}} = nA + x$
$\boldsymbol{x = (n+1)A_{\text{new}} - nA}$
[Value of Added Item]
This formula directly gives the value of the added item if the initial number of items, the initial average, and the new average are known.
Finding the Value of a Removed Item:
Suppose you have a group of $n$ items with an average of $A$. The total sum is $n \times A$. If an item with value $y$ is removed from this group, the total number of items becomes $n-1$, and the new sum of values becomes $(n \times A) - y$. If the new average after removing the item is $A_{\text{new}}$, then:
$\text{New Average} = \frac{\text{New Sum}}{\text{New Number of Items}}$
$\boldsymbol{A_{\text{new}} = \frac{nA - y}{n-1}}$
... (vii)
To find the value of the removed item ($y$), rearrange equation (vii):
$(n-1) \times A_{\text{new}} = nA - y$
$\boldsymbol{y = nA - (n-1)A_{\text{new}}}$
[Value of Removed Item]
Finding the Value of a Replaced Item:
Suppose you have a group of $n$ items with an average of $A$. The total sum is $n \times A$. If an item with value $y_{\text{old}}$ is replaced by a new item with value $y_{\text{new}}$, the total number of items remains $n$. The new sum of values becomes $(n \times A) - y_{\text{old}} + y_{\text{new}}$. If the new average after replacement is $A_{\text{new}}$, then:
$\text{New Average} = \frac{\text{New Sum}}{\text{Number of Items}}$
$\boldsymbol{A_{\text{new}} = \frac{nA - y_{\text{old}} + y_{\text{new}}}{n}}$
... (viii)
To find the value of the new item ($y_{\text{new}}$), rearrange equation (viii):
$\boldsymbol{n \times A_{\text{new}} = nA - y_{\text{old}} + y_{\text{new}}}}$
$\boldsymbol{y_{\text{new}} = nA_{\text{new}} - nA + y_{\text{old}}}}$
... (ix)
An alternative way to think about replacement is in terms of the change in the average. The change in the total sum is $y_{\text{new}} - y_{\text{old}}$. This change in sum is distributed among the $n$ items, causing a change in the average ($A_{\text{new}} - A$).
$\text{Change in Sum} = \text{Number of items} \times \text{Change in Average}$
$\boldsymbol{y_{\text{new}} - y_{\text{old}} = n (A_{\text{new}} - A)}$
... (x)
Rearranging equation (x) to find $y_{\text{new}}$:
$\boldsymbol{y_{\text{new}} = y_{\text{old}} + n (A_{\text{new}} - A)}$
[Value of New Item in Replacement]
This formula is often more intuitive: The new value is the old value plus the total increase/decrease in sum due to the change in average over all $n$ items.
Example 1. The average score of 10 cricketers is 30 runs. If the coach's score is included, the average score increases to 32 runs. Find the coach's score.
Answer:
Initial number of cricketers ($n$) $= 10$.
Initial average score ($A$) $= 30$ runs.
Initial total sum of scores $= n \times A = 10 \times 30 = 300$ runs.
The coach's score is included, so the new number of items ($n_{\text{new}}$) $= 10 + 1 = 11$.
The new average score ($A_{\text{new}}$) $= 32$ runs.
The new total sum of scores $= n_{\text{new}} \times A_{\text{new}} = 11 \times 32$.
Perform the multiplication:
$\begin{array}{cc}& 3 & 2 \\ \times & 1 & 1 \\ \hline & 3 & 2 \\ 3 & 2 & \times \\ \hline 3 & 5 & 2 \\ \hline \end{array}$
New sum of scores $= 352$ runs.
The coach's score is the difference between the new total sum and the initial total sum (since the coach's score was added).
Coach's score $= \text{New sum} - \text{Initial sum}$
Coach's score $= 352 \text{ runs} - 300 \text{ runs} = 52 \text{ runs}$.
The coach's score is $\boldsymbol{52}$ runs.
Using formula (vi):
$n = 10$, $A = 30$, $A_{\text{new}} = 32$. Let the coach's score be $x$.
Score of added item ($x$) $= (n+1)A_{\text{new}} - nA$
$\boldsymbol{x = (10+1) \times 32 - 10 \times 30}$
$\boldsymbol{x = 11 \times 32 - 300}$
$\boldsymbol{x = 352 - 300 = 52}$ runs.
The coach's score is $\boldsymbol{52}$ runs.
Example 2. The average age of 30 students in a class is 15 years. If a student whose age is 20 years leaves the class, what is the new average age of the remaining students?
Answer:
Initial number of students ($n$) $= 30$.
Initial average age ($A$) $= 15$ years.
Initial total sum of ages $= n \times A = 30 \times 15 = 450$ years.
A student whose age is 20 years leaves the class. Value of removed item ($y$) $= 20$ years.
The new number of students ($n_{\text{new}}$) $= 30 - 1 = 29$.
The new total sum of ages $= \text{Initial sum} - \text{Age of removed student} = 450 \text{ years} - 20 \text{ years} = 430$ years.
We need to find the new average age ($A_{\text{new}}$).
Using the definition of average: New Average $= \frac{\text{New Sum}}{\text{New Number of Items}}$
$\boldsymbol{A_{\text{new}} = \frac{430}{29}}$ years.
... (xi)
Perform the division:
$\begin{array}{r} 14.82... \\ 29{\overline{\smash{\big)}\,430.00}} \\ \underline{-~\phantom{(}29\phantom{0.00)}} \\ 140\phantom{0.00)} \\ \underline{-~\phantom{()}\,116\phantom{0.00)}} \\ 240\phantom{00)} \\ \underline{-~\phantom{()}\,232\phantom{0)}} \\ 80 \\ \underline{-~\phantom{()}\,58} \\ 22 \end{array}$
$\boldsymbol{A_{\text{new}} = \frac{430}{29} = 14\frac{24}{29}}$ years (approximately 14.83 years).
The new average age of the remaining students is $\boldsymbol{\frac{430}{29}}$ years (or $14\frac{24}{29}$ years).
Using formula (vii) to find the removed item's value (verification):
Suppose we knew the new average $A_{new}$ and wanted to find the age of the removed student ($y$).
Age of removed student ($y$) $= nA - (n-1)A_{\text{new}}$
$y = 30 \times 15 - (30-1) \times \frac{430}{29}$
$y = 450 - 29 \times \frac{430}{29} = 450 - \cancel{29} \times \frac{430}{\cancel{29}} = 450 - 430 = 20$ years.
This confirms the formula and calculation.
Example 3. The average weight of 12 persons is increased by 3 kg when one person weighing 50 kg is replaced by a new person. What is the weight of the new person?
Answer:
Initial number of persons ($n$) $= 12$.
An old person weighing 50 kg is replaced by a new person. The number of persons remains $n=12$.
Weight of the old (replaced) person ($y_{\text{old}}$) $= 50$ kg.
The average weight increases by 3 kg. This is the change in average ($A_{\text{new}} - A$) $= +3$ kg.
We need to find the weight of the new person ($y_{\text{new}}$).
Method 1: Using the change in sum.
The average increased by 3 kg for all 12 persons. The total increase in the sum of weights is the number of persons multiplied by the increase in average.
Increase in total sum $= \text{Number of persons} \times \text{Increase in average}$
Increase in total sum $= 12 \times 3 \text{ kg} = 36$ kg.
This increase in the total sum is because the new person's weight is greater than the old person's weight by this amount.
Weight of new person $=$ Weight of old person $+$ Increase in total sum
Weight of new person $= 50 \text{ kg} + 36 \text{ kg} = 86$ kg.
The weight of the new person is $\boldsymbol{86}$ kg.
Method 2: Using formula (x).
$n = 12$, $y_{\text{old}} = 50$ kg, $A_{\text{new}} - A = +3$ kg.
Weight of new person ($y_{\text{new}}$) $= y_{\text{old}} + n (A_{\text{new}} - A)$
$\boldsymbol{y_{\text{new}} = 50 + 12 \times (3)}$
$\boldsymbol{y_{\text{new}} = 50 + 36}$
$\boldsymbol{y_{\text{new}} = 86}$ kg.
The weight of the new person is $\boldsymbol{86}$ kg.
Competitive Exam Notes:
Problems involving changes to the dataset are very common in averages. The key is to relate the change in average to the change in the total sum.
- Sum is Key: Always remember Sum = Average $\times$ Number of items. The change in sum is what accounts for the change in average.
- Added Item: New Sum = Old Sum + Added Value. New Number = Old Number + 1. Added Value = (New Avg $\times$ New Number) - (Old Avg $\times$ Old Number). Formula: $x = (n+1)A_{new} - nA$. Or $x = A + n(A_{new} - A)$ if $n$ is the original number.
- Removed Item: New Sum = Old Sum - Removed Value. New Number = Old Number - 1. Removed Value = (Old Avg $\times$ Old Number) - (New Avg $\times$ New Number). Formula: $y = nA - (n-1)A_{new}$. Or $y = A - (n-1)(A - A_{new})$.
- Replaced Item: Number of items remains the same ($n$). Change in Sum = New Value - Old Value. Change in Sum = Number of items $\times$ Change in Average. Formula: $y_{new} - y_{old} = n (A_{new} - A)$, or $y_{new} = y_{old} + n(A_{new} - A)$. This last formula is very efficient.
- Change in Average: If average increases, the added/new item is heavier/older/scores more than the old average (for addition/removal) or the old item's value plus the total required increase (for replacement). If average decreases, the added/new item is less than the old average (addition/removal) or the old item's value minus the total required decrease (replacement).
Weighted Average
The Weighted Average is a type of average that is used when different values in a dataset contribute unequally to the total sum or have different levels of importance. Instead of simply dividing the sum of values by the number of values, each value is multiplied by a corresponding 'weight' that represents its relative importance or frequency. The weighted average is then calculated by dividing the sum of these weighted values by the sum of the weights.
Weighted averages are particularly useful when calculating the average of averages from groups of different sizes, or when certain data points have a higher frequency or significance than others.
Formula for Weighted Average:
If you have a set of values $x_1, x_2, \dots, x_k$, and each value $x_i$ is associated with a corresponding weight $w_i$, the weighted average is calculated as follows:
$\boldsymbol{\text{Weighted Average} = \frac{(w_1 \times x_1) + (w_2 \times x_2) + \dots + (w_k \times x_k)}{w_1 + w_2 + \dots + w_k}}$
Using summation notation:
$\boldsymbol{\text{Weighted Average} = \frac{\sum_{i=1}^{k} w_i x_i}{\sum_{i=1}^{k} w_i}}$
... (xii)
In the context of calculating the average of averages from different groups, the 'values' ($x_i$) are the averages of each group, and the 'weights' ($w_i$) are the number of items or members in each group. The product $w_i x_i$ represents the sum of values within group $i$ (since Sum = Average $\times$ Number of items).
Example 1. In a college, the average height of 50 male students is 165 cm, and the average height of 30 female students is 150 cm. Find the average height of all students in the college.
Answer:
We have two groups of students: males and females. We are given the average height (value) and the number of students (weight) for each group.
Group 1 (Male Students): Number ($w_1$) $= 50$, Average Height ($x_1$) $= 165$ cm.
Group 2 (Female Students): Number ($w_2$) $= 30$, Average Height ($x_2$) $= 150$ cm.
We need to find the average height of the entire class, which is the weighted average of the group averages.
Method 1: Calculate Total Sum and Total Number.
Sum of heights for male students $= w_1 \times x_1 = 50 \times 165$ cm.
$\begin{array}{cccc}& & 1 & 6 & 5 \\ \times & & & 5 & 0 \\ \hline &&&0&0 \\ & 8 & 2 & 5 & \times \\ \hline 8 & 2 & 5 & 0 & 0 \\ \hline \end{array}$
Sum of heights for males $= 8250$ cm.
Sum of heights for female students $= w_2 \times x_2 = 30 \times 150$ cm.
$\begin{array}{cccc}& & 1 & 5 & 0 \\ \times & & & 3 & 0 \\ \hline &&&0&0 \\ & 4 & 5 & 0 & \times \\ \hline 4 & 5 & 0 & 0 & 0 \\ \hline \end{array}$
Sum of heights for females $= 4500$ cm.
Total number of students $= w_1 + w_2 = 50 + 30 = 80$.
Total sum of heights $= (\text{Sum for males}) + (\text{Sum for females}) = 8250 \text{ cm} + 4500 \text{ cm} = 12750$ cm.
Average height of the entire class $= \frac{\text{Total Sum of Heights}}{\text{Total Number of Students}}$
Average Height $= \frac{12750 \text{ cm}}{80 \text{ students}} = \frac{1275}{8}$ cm.
Perform the division:
$\begin{array}{r} 159.375 \\ 8{\overline{\smash{\big)}\,1275.000}} \\ \underline{-~\phantom{(}8\phantom{75.000)}} \\ 47\phantom{5.000)} \\ \underline{-~\phantom{()}\,40\phantom{5.000)}} \\ 75\phantom{.000)} \\ \underline{-~\phantom{()}\,72\phantom{.000)}} \\ 3.000 \\ \underline{-~\phantom{()}\,2.400} \\ 600 \\ \underline{-~\phantom{()}\,560} \\ 40 \\ \underline{-~\phantom{()}\,40} \\ 0 \end{array}$
$\boldsymbol{\text{Average Height} = 159.375 \text{ cm}}$
Method 2: Using the Weighted Average Formula (xii).
Values ($x_i$): Average height of males ($x_1 = 165$), Average height of females ($x_2 = 150$).
Weights ($w_i$): Number of male students ($w_1 = 50$), Number of female students ($w_2 = 30$).
Weighted Average $= \frac{(w_1 \times x_1) + (w_2 \times x_2)}{w_1 + w_2}$
Average Height $= \frac{(50 \times 165) + (30 \times 150)}{50 + 30}$
Average Height $= \frac{8250 + 4500}{80} = \frac{12750}{80}$
Average Height $= \frac{1275}{8} = 159.375$ cm.
$\boldsymbol{\text{Average Height} = 159.375 \text{ cm}}$
Both methods confirm the average height of all students is $\boldsymbol{159.375}$ cm.
Competitive Exam Notes:
Weighted average is crucial for combining averages of groups of different sizes. It is a direct application of the sum formula (Sum = Average $\times$ Number of items).
- Identifying Weighted Average: Look for problems where you are given the averages of different categories or groups, and the number of items in each category/group, and asked to find the overall average.
- Formula: Weighted Average $= \frac{\sum (\text{Weight} \times \text{Value})}{\sum \text{Weight}}$. In group average problems, Value = Group Average, Weight = Number of items in the group.
- Total Sum: The numerator $\sum w_i x_i$ represents the total sum of all values across all groups. Calculate this by summing the product of (group average $\times$ number in group) for each group.
- Total Number: The denominator $\sum w_i$ represents the total number of items across all groups.
- Weighted Average is Not Simple Average of Averages: A simple average of 165 and 150 is (165+150)/2 = 157.5, which is incorrect here because there are more male students. The weighted average will be closer to the average of the larger group.
Solving Problems on Averages
This section provides more examples of problems involving averages, including scenarios that might combine different concepts or require careful interpretation of the given information. Proficiency in solving these problems requires a solid understanding of the basic definition, weighted average, and handling changes in the dataset.
Example 1. The average marks of 15 students in a test is 60. If the marks of the topper are excluded, the average marks of the remaining 14 students fall to 58. Find the marks of the topper.
Answer:
Initial number of students ($n$) $= 15$.
Initial average marks ($A$) $= 60$.
Initial total sum of marks $= n \times A = 15 \times 60 = 900$ marks.
The topper's marks are excluded, so the new number of students ($n_{\text{new}}$) $= 15 - 1 = 14$.
The new average marks ($A_{\text{new}}$) $= 58$.
The new total sum of marks for the remaining 14 students $= n_{\text{new}} \times A_{\text{new}} = 14 \times 58$.
Perform the multiplication:
$\begin{array}{cc}& 1 & 4 \\ \times & 5 & 8 \\ \hline 1 & 1 & 2 \\ 7 & 0 & \times \\ \hline 8 & 1 & 2 \\ \hline \end{array}$
New sum of marks $= 812$ marks.
The topper's marks are the difference between the initial sum (including the topper) and the new sum (excluding the topper).
Topper's marks $=$ Initial sum $-$ New sum
Topper's marks $= 900 - 812 = 88$ marks.
The marks of the topper are $\boldsymbol{88}$.
Using formula (vii) from the previous section (Value of Removed Item):
$n = 15$, $A = 60$, $A_{\text{new}} = 58$. Let the topper's marks be $y$.
Marks of removed item ($y$) $= nA - (n-1)A_{\text{new}}$
$\boldsymbol{y = 15 \times 60 - (15-1) \times 58}$
$\boldsymbol{y = 900 - 14 \times 58}$
$\boldsymbol{y = 900 - 812 = 88}$ marks.
The marks of the topper are $\boldsymbol{88}$.
Example 2. The average age of a family of 5 members is 20 years. If a new member joins the family, the average age increases by 2 years. Find the age of the new member.
Answer:
Initial number of family members ($n$) $= 5$.
Initial average age ($A$) $= 20$ years.
Initial total sum of ages $= n \times A = 5 \times 20 = 100$ years.
A new member joins, so the new number of family members ($n_{\text{new}}$) $= 5 + 1 = 6$.
The average age increases by 2 years, so the new average age ($A_{\text{new}}$) $= 20 + 2 = 22$ years.
The new total sum of ages $= n_{\text{new}} \times A_{\text{new}} = 6 \times 22 = 132$ years.
The age of the new member is the difference between the new sum and the initial sum.
Age of new member $=$ New sum $-$ Initial sum
Age $= 132 \text{ years} - 100 \text{ years} = 32 \text{ years}$.
The age of the new member is $\boldsymbol{32}$ years.
Using formula (vi) from the previous section (Value of Added Item):
$n = 5$, $A = 20$, $A_{\text{new}} = 22$. Let the new member's age be $x$.
Age of added item ($x$) $= (n+1)A_{\text{new}} - nA$
$\boldsymbol{x = (5+1) \times 22 - 5 \times 20}$
$\boldsymbol{x = 6 \times 22 - 100}$
$\boldsymbol{x = 132 - 100 = 32}$ years.
The age of the new member is $\boldsymbol{32}$ years.
Example 3. The average weight of 20 students in a class is 45 kg. If a student whose weight is 50 kg leaves the class, what is the new average weight of the remaining students?
Answer:
Initial number of students ($n$) $= 20$.
Initial average weight ($A$) $= 45$ kg.
Initial total sum of weights $= n \times A = 20 \times 45 = 900$ kg.
A student whose weight is 50 kg leaves. Value of removed item ($y$) $= 50$ kg.
The new number of students ($n_{\text{new}}$) $= 20 - 1 = 19$.
The new total sum of weights $= \text{Initial sum} - \text{Weight of removed student} = 900 \text{ kg} - 50 \text{ kg} = 850$ kg.
We need to find the new average weight ($A_{\text{new}}$) of the remaining 19 students.
Using the definition of average: New Average $= \frac{\text{New Sum}}{\text{New Number of Items}}$
$\boldsymbol{A_{\text{new}} = \frac{850}{19}}$ kg.
... (xiii)
Perform the division:
$\begin{array}{r} 44.73... \\ 19{\overline{\smash{\big)}\,850.00}} \\ \underline{-~\phantom{(}76\phantom{0.00)}} \\ 90\phantom{0.00)} \\ \underline{-~\phantom{()}\,76\phantom{0.00)}} \\ 140\phantom{00)} \\ \underline{-~\phantom{()}\,133\phantom{0)}} \\ 70 \\ \underline{-~\phantom{()}\,57} \\ 13 \end{array}$
$\boldsymbol{A_{\text{new}} = \frac{850}{19} = 44\frac{14}{19}}$ kg (approximately 44.74 kg).
The new average weight of the remaining students is $\boldsymbol{\frac{850}{19}}$ kg (or $44\frac{14}{19}$ kg).
Competitive Exam Notes:
Solving average problems requires correctly identifying the total sum and the total number of items. Problems involving changes are very common.
- Sum = Average $\times$ Number: This is the most important formula. Use it to find the total sum when the average and count are known.
- Changes in Dataset:
- Addition: New Sum = Old Sum + Added Value. New Number = Old Number + 1.
- Removal: New Sum = Old Sum - Removed Value. New Number = Old Number - 1.
- Replacement: New Sum = Old Sum - Old Value + New Value. Number remains the same.
- Finding Value of Changed Item/New Average: Use the Sum = Average $\times$ Number relationship for both the initial and the final state, and relate them. The formulas derived in the previous section (I3) provide quick ways to find the value of added/removed/replaced items.
- Combining Concepts: Some problems might involve calculating average speed (Total Distance / Total Time) or weighted average (for combining group averages).