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1309 LearnersLast updated on November 27, 2025

The mean is the average of arithmetical data. The mean of grouped data is the mean data of grouped sets, categories, objects, etc. The mean of a set of values is found by adding all the values together and dividing by the total number of values.
The mean of grouped data is the average value of the data organized into classes or groups. Such data, arranged in groups or categories, is called grouped data, in which each class interval shows a range of values, and the frequency indicates how many observations fall within that range. The mean, also known as the central value, helps us understand the overall trend of the data. For grouped data, we use the midpoint of each class to calculate the mean.
In statistics, the mean is the sum of all observations divided by the total number of observations. The mean is denoted as \(\bar x\) (x bar). That is:
\( x = x_1 + x_2 + x_3 + …. + x_n.\)
The mean of grouped data is calculated using the midpoint (class mark) of each class interval along with its frequency. The central value of its interval is known as the midpoint. By multiplying these midpoints by their respective frequencies and dividing the total by the overall frequency. The formula for the mean of grouped data is:
\({\bar x} = {{∑{f_i} {x_i} \over N }}\)
Where,
\(\bar x\) is the mean value of the set of given data
\(f_i\) is the frequency of the individual data
\(x_i\) is the midpoint of each class interval
N is the sum of the frequencies
The mean of grouped data is a measure of central tendency that represents the average value of a dataset organized into classes. The different methods to calculate the mean of grouped data are:
Direct Method
The direct method is the simplest method used to calculate the mean of grouped data. If the values of the observations are \(x_1, x_2,..., x_n\), with their corresponding frequencies being \(f_1, f_2,..., f_n\). Then the mean of the data is given by,
\(x = \frac{x_1f_1 + x_2f_2 + \cdots + x_n f_n}{f_1 + f_2 + \cdots + f_n} \)
\(x = \frac{\Sigma (f \times x)}{\Sigma f} \), where i = 1, 2, 3, 4,...n
To find the mean of grouped data, follow these steps:
For example, a class of students scored the following marks in a test. Find the mean marks.
| Mark | Number of Students |
| 0-10 | 2 |
| 10-20 | 3 |
| 20-30 | 5 |
| 30-40 | 4 |
| 40-50 | 6 |
To find the midpoints (xi) of each class using the formula: \(x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \)
| Mark | Number of Students | Midpoint (\(x_i\)) | \({{f_i} {x_i}}\) |
| 0-10 | 2 | \({{0 + 10}\over 2 } = 5\) | \(2 \times 5 = 10 \) |
| 10-20 | 3 | \({{10 + 20}\over 2 } = 15\) | \(3 \times 15 = 45\) |
| 20-30 | 5 | \({{20 + 30}\over 2 } = 25\) | \(5 \times 25 = 125 \) |
| 30-40 | 4 | \({{30 + 40}\over 2 } = 140\) | \(4 \times 35 = 140\) |
| 40-50 | 6 | \({{40 + 50}\over 2 } = 270\) | \( 6 \times 45 = 270\) |
Sum the frequencies and sum the products:
\(Σf_i = 2 + 3 + 5 + 4 + 6 = 20\)
\(Σf_ix_i = 10 + 45 + 125 + 140 + 270 = 590\)
Finding the mean of grouped data using the formula: \(\bar{x} = \frac{\sum f_i x_i}{N} \)
\({\bar x} ={{590 \over 20 }} \)
= 29.5
Assumed Mean Method
The assumed mean method is used to estimate the mean of a large dataset. In this method, a value from the dataset is chosen as an assumed mean (a), and the deviations of all midpoints from this assumed mean are calculated.
Let the deviation of each midpoint from the assumed mean be: \(d_i = x_i - a\)
Then the mean is calculated as:
\(\bar{x} = \frac{\sum (a + d_i) f_i}{\sum f_i} \)
\(\bar{x} = a + \frac{\sum ( f_i d_i)}{\sum f_i}\)
Find the mean of the following grouped data using the assumed mean method.
| Class Interval | Frequency (f) |
| 10-20 | 5 |
| 20-30 | 8 |
| 30-40 | 12 |
| 40-50 | 7 |
| 50-60 | 3 |
To find the midpoint, xi, we use the formula:
\(x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \)
Here, the midpoints are 15, 25, 35, 45, and 55
Let a = 35
Then deviations, di is calculated using the formula \(d_i = x_i - a\)
| Class Interval | Frequency (f) | \(x_i\) | \({d_i} = {x_i} - a\) | \({f_i} \cdot {d_i}\) |
| 10-20 | 5 | 15 | -20 | -100 |
| 20-30 | 8 | 25 | -10 | -80 |
| 30-40 | 12 | 35 | 0 | 0 |
| 40-50 | 7 | 45 | 10 | 70 |
| 50-60 | 3 | 55 | 20 | 60 |
Applying the assumed mean formula: \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \)
\({d_if_i} = (-100) + (-80) + 0 + 70 + 60 = -50\)
\(Σf_i = 5 + 8 + 12 + 7 + 3 = 35 \)
\( = {{35 + -50} \over { 35 }}\)
\(= 35 - 1.43\)
= 33.57
Step Deviation Method
The step deviation method is used when the values in a data set are large and far from the assumed mean. Here, deviations are divided by the class size h to simplify calculations:
\(\bar{x} = a + h \cdot \frac{\sum f_i u_i}{\sum f_i} \)
Where,
a is the assumed mean
h is the class size
ui = \(\frac{d_i}{h} \)
For example, find the mean of the following data using the step deviation method.
| Class Interval | Frequency |
| 100-110 | 5 |
| 110-120 | 8 |
| 120-130 | 12 |
| 130-140 | 7 |
| 140-150 | 3 |
Finding the midpoints xi using:
\(x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \)
Here, the midpoints are 105, 115, 125, 135, 145
Let the assumed mean: a = 125
Calculating deviation and step deviation:
\(d_i = x_i - a \)
\(u_i = \frac{d_i}{h} \), where h = 10
| Class Interval | Frequency | \(x_i\) | \(d_i = x_i - a\) | \(u_i = {d_i \over h}\) | \({f_i u_i}\) |
| 100-110 | 5 | 105 | -20 | -2 | -10 |
| 110-120 | 8 | 115 | -10 | -1 | -8 |
| 120-130 | 12 | 125 | 0 | 0 | 0 |
| 130-140 | 7 | 135 | 10 | 1 | 7 |
| 140-150 | 3 | 145 | 20 | 2 | 6 |
\(f_iu_i = -10 + -8 + 0 + 7 + 6 = -5 \)
\(f_i = 5 + 8 + 12 + 7 + 3 \)
Substituting the values in the equation: \(\bar{x} = a + h \cdot \frac{\sum f_i u_i}{\sum f_i} \)
\(\bar{x} = 125 + 10 \left( -\frac{5}{35} \right) \)
\(= {125 - 1.43}\)
\(\bar {x} = 123.57\)


Mean of Grouped data is a complex mathematical topic and therefore some tips and tricks can be helpful. In this section we discuss such tips and tricks.
The possibility of kids making mistakes while doing mean grouped data is high. This is not similar to finding the ordinary mean value. Here are some of the few common mistakes that kids might make and how to avoid them.
The mean of grouped data is used in many real-life situations where data is collected in ranges. Here are some important applications:
A teacher records the test scores of students in a class as follows: Score Range 40 - 50, 50-60, 60-70, 70-80 Frequency 3, 5, 8, 4. Find the mean score.
The mean score is 63.75.
Find class midpoints
\(\begin{align*} \frac{40 + 50}{2} &= 45 \\ \ \\ \frac{50 + 60}{2} &= 55 \\ \ \\ \frac{60 + 70}{2} &= 65 \\ \ \\ \frac{70 + 80}{2} &= 75 \end{align*} \)
Multiply each midpoint by its frequency:
\((45 \times 3) + (55 \times 5) + (65 \times 8) + (75 \times 4)\)
\(= 135 + 275 + 520 + 300\)
= 1230
Sum of frequencies: \(3 + 5 + 8 + 4 = 20\)
Mean = \(\frac{1230}{20} = 61.5 \)
The following table shows the number of hours students spend studying in a week. Hours studied : 0-5 , 5-10 , 10-15, 15-20 , frequency : 6, 10, 8, 6. Find the mean number of hours studied.
The mean study time is 9.58 hours.
Find the class midpoints
\(\begin{align*} \frac{0 + 5}{2} &= 2.5 \\ \ \\ \frac{5 + 10}{2} &= 7.5 \\ \ \\ \frac{10 + 15}{2} &= 12.5 \\ \ \\ \frac{15 + 20}{2} &= 17.5 \end{align*} \)
Multiplying each midpoint by its frequency
\((2.5 × 6) + (7.5 × 10) + (12.5 × 8) + (17.5 × 6)\)
\(= 15 + 75 + 100 + 105\)
\(= 295\)
Sum of frequencies: \(6 + 10 + 8 + 6 = 30\)
Mean = \(\frac{295}{30} = 9.83\overline{3} \)
= 9.58
The speed (in km/h) of vehicles on a highway is recorded for 50 vehicles. Speed (km/h) : 40 - 50 , 50 - 60 , 60 - 70 , 70 - 80 , 80 - 90 Frequency (Vehicles) : 5. 10, 15, 12 , 8. Find the mean speed of vehicles.
The mean speed is 66 km/h.
Find class midpoints:
\(\begin{align*} \frac{40 + 50}{2} &= 45 \\ \ \\ \frac{50 + 60}{2} &= 55 \\ \ \\ \frac{60 + 70}{2} &= 65 \\ \ \\ \frac{70 + 80}{2} &= 75 \\ \ \\ \frac{80 + 90}{2} &= 85 \end{align*} \)
Multiply each midpoint by its frequency
\((45 × 5) + (55 × 10) + (65 × 15) + (75 × 12) + (85 × 8)\)
\(= 225 + 550 + 975 + 900 + 680 \)
\(= 3330\)
Sum of frequencies: 5 + 10 + 15 + 12 + 8 = 50
Mean = \(\frac{3330}{50} = 66.6 \)
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!





