Mean of grouped and Ungrouped Data: Formula, Examples, Questions, Calculator

What are Ungrouped data?

When a data set is large, a frequency distribution table is often used to display the data in an organized way. A frequency distribution table lists the data values, as well as the number of times each value appears in the data set. A frequency distribution table is easy to both read and interpret and in this concept is used for ungrouped data, or data that is listed.

What is a frequency distribution table and tally column?

The numbers in a frequency distribution table do not have to be put in order. To make it easier to enter the values in the table, a tally column is often inserted. Inserting a tally column allows you to account for every value in the data set, without having to continually scan the numbers to find them in the list. A slash (/) is used to represent the presence of a value in the list, and the total number of slashes will be the frequency. If a tally column is inserted, the table will consist of 3 columns, and if no tally column is inserted, the table will consist of 2 columns.

How to find the mean of Ungrouped data?

The mean of ungrouped data often means the arithmetic mean or normal average of a given set of values.

The formula is:

Mean = $\frac{\text{Sum of all the Observations}} {\text{Total number of Observations}}$

Determining mean from a given list

Determine the mean of the given data.

5, 7, 9, 7, 3, 7

Sum of all observations = 5 + 7 + 9 + 8 + 3 + 10 = 42

Total number of observations = 6

Hence, the mean = $\frac{42}{7}$ = 6

Determining mean using frequency distribution table

Consider the following frequency distribution table.

To calculate the mean we have to use the following formula.

Mean = $\frac{\sum\text{(Mid Point × Frequencies)}}{\sum \text{Frequencies}}$

Required mean = $\frac{200+1800+4500+6000}{40+90+100+80}=\frac{12500}{310}$ = 40.32

What is grouped data?

Data that has been organized into groups like classes or intervals is called Grouped data. This type of data presentation is useful for large datasets, making it easier to analyze and interpret the data. Grouping data helps in identifying patterns and trends more effectively than working with raw, unorganized data.

Representation of Grouped Data

Grouped data can be represented in many different ways like

• Bar graph

• Histogram

• Frequency polygon

Bar graph:

A histogram is a type of bar graph that represents the frequency distribution of continuous data. Unlike bar graphs, histograms display data in intervals (bins) with no gaps between the bars. Each bar represents the frequency of data within that interval.

Histogram:

A histogram is a type of bar graph that represents the frequency distribution of continuous data. Unlike bar graphs, histograms display data in intervals (bins) with no gaps between the bars. Each bar represents the frequency of data within that interval.

Frequency polygon:

A frequency polygon is a graphical representation of the distribution of a dataset. It is created by plotting points corresponding to the frequency of each class interval and then connecting these points with straight lines. Frequency polygons are useful for comparing multiple distributions on the same graph.

Frequency distribution table for Grouped data

A frequency distribution table organizes data into intervals (classes) and shows the number of observations (frequency) in each interval. It provides a summary of the dataset and helps in understanding the distribution of data.

Terms used in or derived from the frequency table

There are some key terms used in or can be derived from the frequency table.

Class size:

Class size is the difference between the upper and lower boundaries of a class interval.

For example, in the class interval 0-10, the class size is 10 - 0 = 10

Classmark(Midpoint):

Classmark is the average of the upper and lower boundaries of a class interval.

It is calculated by dividing the sum of upper and lower intervals by 2.

For example, in the class interval 0-10, the classmark is $\frac{0+10}{2}$ = 5

Class intervals:

Class intervals are the ranges into which data is grouped. There are two types of class intervals:

• Continuous Class Interval: There are no gaps between intervals, and each interval includes the upper boundary of the previous interval. For example, 0-10, 10-20, 20-30, etc.

• Discontinuous Class Interval: There are gaps between intervals, and each interval does not include the upper boundary of the previous interval. For example, 0-9, 10-19, 20-29, etc.

Frequency:

Frequency is the number of observations that fall within a particular class interval. It indicates how often a particular value or range of values occurs in the dataset.

Calculation of Mean for Grouped data using frequency distribution table

There are three basic methods to find the mean of grouped data

• Direct Method

• Assumed mean method

• Step deviation method

Direct Method

In the Direct method to calculate the mean, we will use the midpoint of the class intervals and frequencies.

The formula is:

Mean = $\frac{\sum\text{(Mid Point × Frequencies)}}{\sum \text{Frequencies}}$

In notation, it can be written as follows:

Mean, $\bar{x}= \frac{ \sum f_i x_i}{\sum f_i}$,

Where

fi is the frequency of the ith class

xi is the midpoint of the ith class

Assumed mean method

In the Assumed mean method, we have to assume a mean and proceed with it to calculate the original mean.

The formula is:

Mean, $\bar{x}=a + \frac{ \sum f_i d_i}{\sum f_i}$,

Where

di = xi – A, is a derivation of the ith class

A is the assumed mean

fi is the frequency of the ith class

xi is the midpoint of the ith class

Assumed Mean Method = $\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}$

Step deviation method

The Step deviation method is used when all the class intervals are equal. It simplifies the calculation further.

The formula is:

Mean = $A + \frac{h \sum u_if_i}{\sum f_i}$,

Where

$u_i=\frac{x_i-A}{h}$, is a derivation of the ith class

A is the assumed mean

h is the class size

fi is the frequency of the ith class

xi is the midpoint of the ith class

Step Deviation Method $

=A+\frac{h\left(\sum u_{i}f_i\right)}{\sum f_i}

$

More Detailed Information

To learn more about Average or Arithmetic mean, read the below article.

Average

To learn more about the mode of grouped and ungrouped data, read the below article.

Mode

To learn more about the median of grouped and ungrouped data, read the below article.

Median

Important points

• Mean = $\frac{\text{Sum of all the Observations}} {\text{Total number of Observations}}$

• Using the Direct method to calculate, Mean = $\frac{\sum\text{(Mid Point × Frequencies)}}{\sum \text{Frequencies}}$

• Using the Assumed mean method to calculate,
  Mean, $\bar{x}=a + \frac{ \sum f_i d_i}{\sum f_i}$,

Where

di = xi – A, is a derivation of the ith class

A is the assumed mean

fi is the frequency of the ith class

xi is the midpoint of the ith class

• Using the step deviation method to calculate,

Mean = $A + \frac{h \sum u_if_i}{\sum f_i}$,

Where

$u_i=\frac{x_i-A}{h}$, is a derivation of the ith class

A is the assumed mean

h is the class size

fi is the frequency of the ith class

xi is the midpoint of the ith class

Practice Questions

Q1. Calculate the mean from the following table.

1. 34.2

2. 33.4

3. 32.6

4. 35.8

Hint: Mean = $\frac{\sum\text{(Mid Point × Frequencies)}}{\sum \text{Frequencies}}$

Use this formula to calculate the mean.

Answer:

Mean = $\frac{\sum \text{ (Mid Point × Frequencies)}}{\sum \text{Frequencies}}$

= $\frac{10+60+300+735+270+165+130}{2+4+12+21+6+3+2}$

= $\frac{1670}{50}$

= 33.4

Hence, the correct answer is 33.4.

Q2. The mean of 20 observations was 42. It was found later that observation 54 was wrongly taken as 45. The corrected new mean is:

1. 45.42

2. 43.5

3. 44.25

4. 42.45

Hint: First, find the sum of initial observations. Then subtract the wrong value, add the correct value and find the final sum.

Answer:

The mean of 20 initial observations = 42

Sum of initial observations = 42 × 20 = 840

Wrong entry = 45

Correct entry = 54

Sum after correction = 840 – 45 + 54 = 849

$\therefore$ Final mean = $\frac{849}{20}$ = 42.45

Hence, the correct answer is 42.45.

Q3. Find the standard deviation of the following data (rounded off to two decimal places).

5, 3, 4, 7

1. 1.48

2. 3.21

3. 4.12

4. 3.45

Hint: Standard deviation = $\sqrt{\frac{\sum {x_i}^2}{n}-(\frac{\sum x_i}{n})^2}$

where $x_i$ are the numbers and $n$ is the count of numbers.

Answer:

Standard deviation of 5, 3, 4, 7

= $\sqrt{\frac{\sum {x_i}^2}{n}-(\frac{\sum x_i}{n})^2}$

where $x_i$ are the numbers and $n$ is the count of numbers.

= $\sqrt{\frac{5^2+3^2+4^2+7^2}{4}-(\frac{5+3+4+7)}{4})^2}$

= $\sqrt{\frac{99}{4}-\frac{361}{16}}$

= $\sqrt{\frac{35}{16}}$

= 1.48

Hence, the correct answer is 1.48.

Q4. In data analysis, which of the following statements about the mean (average) is true?

1. Outliers are not sensitive to the mean.

2. Categorical data can be analyzed using the mean.

3. The mean is the best way to describe the variability of data.

4. The mean reflects the dataset's central value.

Answer:

The core value of a dataset is represented by the mean.
The mean is a metric of central tendency that offers a single value that serves as the "typical" value to sum up a dataset.
D) The core value of a dataset is represented by the mean: This claim is true. The mean is a metric of central tendency that offers a single value that serves as the "typical" value to sum up a dataset.

A) The statement "The mean is not sensitive to outliers" is false. Extreme values have a considerable impact on the mean and are very sensitive to outliers.

B) The mean shouldn't be utilized for categorical data because it is normally used for numerical data.

C) Although the mean helps determine central tendency, it is not a good way to represent data variability.

Hence, the correct answer is "The mean reflects the dataset's central value."

Q5. The mean of the following numbers is 12, 15, 18, 21, 24, and 27.

1. 15.789

2. 17.83

3. 18.26

4. 17.65

Answer:

The Mean of ungrouped data is given by $\sum$ total values/Total no of values

Here,

The sum of $2,15,18,21,24$, and $27$ is $107$.

Total no of values $=6$

Therefore, the mean $=\frac{107 }{ 6}=17.83$

Hence, the correct answer is 17.83.

Q6. The table below shows the number of people belonging to specific age groups. Find the mean of the given data.

The class marks $x_i$ can be found by adding the upper and lower bounds and dividing them by $2 \cdot(20+0 / 2=10,20+40 / 2=30$ etc $\ldots)$

1. 34.78

2. 29.46

3. 36.59

4. 40.01

Answer:

$\begin{aligned} \text { Mean } & =\Sigma \mathrm{X}_{\mathrm{i}} \mathrm{f}_{\mathrm{i}} / \underline{\boldsymbol{\Sigma}} \mathrm{f}_{\mathrm{i}} \\ & =\frac{22550 }{615} \\ & =36.59\end{aligned}$

Hence, the correct answer is 36.59.

Q7. The following table contains the number of patients admitted for cholera in St. Johns Hospital according to their age group. Find the mean of the following data.

1. 28.90

2. 31.45

3. 26.81

4. 25.73

Answer:

The class marks can be found by adding the upper and lower bounds and dividing them by 2. (10 + 0 / 2 = 5, 20 + 10 / 2 = 15 etc...)

$\therefore$ The Mean of a grouped data is given by $\Sigma \mathrm{X}_{\mathrm{i}} \mathrm{f}_{\mathrm{i}} / \underline{\underline{\Sigma}} \mathrm{f}_{\mathrm{i}}$

Applying this, we get,
$
\begin{aligned}
& =\frac{1415 }{ 55} \\
& =25.73
\end{aligned}
$

Hence, the correct answer is 25.73.