Mean median mode calculator for grouped data

Mean, median and mode

Mean, median, mode are the measures of central tendency. They are also known as averages. Averages are the measures which condense a huge set of numerical data into a single numerical value which is representative of the entire data.

They give us an idea about the concentration of the values in the central part of data. In brief, average of a statistical data is the value of the variable which is representative of the entire population.

Two data sets are not comparable because of the unsystematic variation generally present in the data sets, but constants make it possible to compare the data sets easily.

Averages are very much useful for :

  • describing the distribution in concise manner.
  • comparative study of different distributions.
  • computing various other statistical measures such as dispersion (variation), skewness (lack of symmetry) and so on.

Mean, Median and Mode for grouped data

Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution.

Sample mean

The mean of $X$ is denoted by $\overline{x}$ and is given by

$$\overline{x} =\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i$$

In case of continuous frequency distribution, $x_i$'s are the mid-values of the respective classes.

Steps for the calculation of Mean

Step 1: Calculate the total frequency $N$.

Step 2: Calculate mid-value for each class as $x=\dfrac{\text{Lower Boundry}+\text{Upper Boundry}}{2}$.

Step 3: Calculate $f*x$.

Step 4: Find the mean using above formula.

Sample median

The median of a frequency distribution is given by

$$\text{Median } = l + \bigg(\dfrac{\frac{N}{2} - F_<}{f}\bigg)\times h$$

where

  • $N$, total number of observations
  • $l$, the lower limit of the median class
  • $f$, frequency of the median class
  • $F_<$, cumulative frequency of the pre median class
  • $h$, the class width

Steps for the calculation of Median

Step 1: Calculate the total frequency $N$.

Step 2: Calculate $\dfrac{N}{2}$.

Step 3: Calculate cumulative frequency. Select Cumulative frequency just greater than or equal to $\dfrac{N}{2}$.

Step 4: The class corresponding to the cumulative frequency selected in Step 3 is the median class.

Step 5: Find the median using above formula.

Sample mode

The mode of the distribution is given by

$$\text{Mode } = l + \bigg(\dfrac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h$$

where

  • $l$, the lower limit of the modal class
  • $f_m$, frequency of the modal class
  • $f_1$, frequency of the class pre-modal class
  • $f_2$, frequency of the class post-modal class
  • $h$, the class width

Steps for the calculation of Mode

Step 1: Select the maximum frequency.

Step 2: The class corresponding to the maximum frequency is the modal class.

Step 3: Find the mode using above formula.

Mean, median and mode calculator for grouped data

Use this calculator to find the mean, median and mode for grouped (frequency distribution) data.

Mean, median and mode Calculator (Grouped Data)
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Results
Number of Obs. (N):
Mean :
Median :
Mode :
freq dist :

Mean median and mode for grouped data Example 1

Following is the data about the daily number of car accidents during a month

No. of car accidents 1 2 3 4 5 6 7
No. of days 2 4 4 10 7 2 1

Calculate mean, median and mode.

Solution

x f f*x cf
1 2 2 2
2 4 8 6
3 4 12 10
4 10 40 20
5 7 35 27
6 2 12 29
7 1 7 30
Total 30 116

Mean

The mean number of accidents is

$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{116}{30}\\ &=3.8667 \end{aligned} $$

Median

Median number of accidents is

$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{30}{2}\bigg)^{th}\text{ value}\\ &=\big(15\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $15$ is $20$. The corresponding value of $x$ is median. That is, $M =4$.

Thus, median number of accidents $M$ = $4$.

Mode

Mode is the value of $x$ with maximum frequency.

The maximum frequency is $10$. The value of $x$ corresponding to the maximum frequency $10$ is $\text{Mode }=4$.

Mean median and mode for grouped data Example 2

The number of defective items in successive groups of fifteen items were counted as they came off a production line. The results can be summarized as follows:

No. of defective 0 1 2 3 4 $>$ 4
Frequency 57 57 18 5 3 0

Calculate mean, median and mode for the above frequenct distribution.

Solution

x f f*x cf
0 57 0 57
1 57 57 114
2 18 36 132
3 5 15 137
4 3 12 140
Total 140 120

Mean

The mean number of defectives is

$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{120}{140}\\ &=0.8571 \end{aligned} $$

Median

Median number of defectives is

$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{140}{2}\bigg)^{th}\text{ value}\\ &=\big(70\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $70$ is $114$. The corresponding value of $x$ is median. That is, $M =1$.

Thus, median number of defectives $M$ = $1$.

Mode

Mode is the value of $x$ with maximum frequency.

The maximum frequency is $57$. The value of $x$ corresponding to the maximum frequency $57$ is the mode. Since the maximum frequency occurs for two values of $x$ (i.e., 0 and 1). Hence the frequency distribution is bimodal.

Mean median and mode for grouped data Example 3

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute mean, median and mode for the following frequency distribution.

Time spent on Internet ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 24 2

Solution

The classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries mid-value (x) Freq (f) f*x cf
10-12 9.5-12.5 11 3 33 3
13-15 12.5-15.5 14 12 168 15
16-18 15.5-18.5 17 15 255 30
19-21 18.5-21.5 20 24 480 54
22-24 21.5-24.5 23 2 46 56
Total 56 982

Mean

The mean time spent on internet is

$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \text{ minutes} \end{aligned} $$

Median

Median time spent on internet by the students is

$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{56}{2}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $28$ is $30$. The corresponding class $15.5-18.5$ is the median class.

Thus

  • $N=56$, total number of observations
  • $l = 15.5$, the lower limit of the median class
  • $f =15$, frequency of the median class
  • $F_< = 15$, cumulative frequency of the pre median class
  • $h =3$, the class width

The median can be computed as follows:

$$ \begin{aligned} \text{Median } &= l + \bigg(\frac{\frac{N}{2} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{56}{2} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned} $$

Mode

The maximum frequency is $24$, the corresponding class $18.5-21.5$ is the modal class.

Mode of the given frequency distribution is:

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$

where

  • $l = 18.5$, the lower limit of the modal class
  • $f_m =24$, frequency of the modal class
  • $f_1 = 15$, frequency of the pre-modal class
  • $f_2 = 2$, frequency of the post-modal class
  • $h =3$, the class width

Thus mode of a frequency distribution is

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 18.5 + \bigg(\frac{24 - 15}{2\times24 - 15 - 2}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{9}{31}\bigg)\times 3\\ &= 18.5 + \big(0.2903\big)\times 3\\ &= 18.5 + \big(0.871\big)\\ &= 19.371 \text{ minutes} \end{aligned} $$

Mean median and mode for grouped data Example 4

The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:

Maximum load No. of Cables
9.25-9.75 2
9.75-10.25 5
10.25-10.75 12
10.75-11.25 17
11.25-11.75 14
11.75-12.25 6
12.25-12.75 3
12.75-13.25 1

Compute mean, median and mode for the above frequency distribution.

Solution

Class Interval Class Boundries mid-value (x) Freq (f) f*x cf
9.25-9.75 8.75-10.25 9.5 2 19 2
9.75-10.25 9.25-10.75 10 5 50 7
10.25-10.75 9.75-11.25 10.5 12 126 19
10.75-11.25 10.25-11.75 11 17 187 36
11.25-11.75 10.75-12.25 11.5 14 161 50
11.75-12.25 11.25-12.75 12 6 72 56
12.25-12.75 11.75-13.25 12.5 3 37.5 59
12.75-13.25 12.25-13.75 13 1 13 60
Total 60 665.5

Mean

The mean maximum load is

$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{665.5}{60}\\ &=11.0917 \text{ tons} \end{aligned} $$

Median

Median maximum load is

$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{60}{2}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $30$ is $36$. The corresponding class $10.25-11.75$ is the median class.

Thus

  • $N=60$, total number of observations
  • $l = 10.25$, the lower limit of the median class
  • $f =17$, frequency of the median class
  • $F_< = 19$, cumulative frequency of the pre median class
  • $h =0.5$, the class width

The median can be computed as follows:

$$ \begin{aligned} \text{Median } &= l + \bigg(\frac{\frac{N}{2} - F_<}{f}\bigg)\times h\\ &= 10.25 + \bigg(\frac{\frac{60}{2} - 19}{17}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{30 - 19}{17}\bigg)\times 0.5\\ &= 10.25 + \big(0.6471\big)\times 0.5\\ &= 10.25 + 0.3235\\ &= 10.5735 \text{ tons} \end{aligned} $$

Mode

The maximum frequency is $17$, the corresponding class $10.25-11.75$ is the modal class.

Mode of the given frequency distribution is:

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$

where

  • $l = 10.25$, the lower limit of the modal class
  • $f_m =17$, frequency of the modal class
  • $f_1 = 12$, frequency of the pre-modal class
  • $f_2 = 14$, frequency of the post-modal class
  • $h =0.5$, the class width

Thus mode of a frequency distribution is

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 10.25 + \bigg(\frac{17 - 12}{2\times17 - 12 - 14}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{5}{8}\bigg)\times 0.5\\ &= 10.25 + \big(0.625\big)\times 0.5\\ &= 10.25 + \big(0.3125\big)\\ &= 10.5625 \text{ tons} \end{aligned} $$

Mean median and mode for grouped data Example 5

Following table shows the weight of 100 pumpkin produced from a farm :

Weight ('00 grams) Frequency
$4 \leq x < 6$ 4
$6 \leq x < 8$ 14
$8 \leq x < 10$ 34
$10 \leq x < 12$ 28
$12 \leq x < 14$ 20

Compute mean, median and mode for the above frequency distribution.

Solution

Class Interval Class Boundries mid-value (x) Freq (f) f*x cf
4-6 3.5-6.5 5 4 20 4
6-8 5.5-8.5 7 14 98 18
8-10 7.5-10.5 9 34 306 52
10-12 9.5-12.5 11 28 308 80
12-14 11.5-14.5 13 20 260 100
Total 100 992

Mean

The mean weight of pumpkin is

$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{992}{100}\\ &=9.92 \text{ ('00 grams)} \end{aligned} $$

Median

Median weight of pumpkin is

$$ \begin{aligned} \text{Median} &=\bigg(\dfrac{N}{2}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{100}{2}\bigg)^{th}\text{ value}\\ &=\big(50\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $50$ is $52$. The corresponding class $7.5-10.5$ is the median class.

Thus

  • $N=100$, total number of observations
  • $l = 7.5$, the lower limit of the median class
  • $f =34$, frequency of the median class
  • $F_< = 18$, cumulative frequency of the pre median class
  • $h =2$, the class width

The median can be computed as follows:

$$ \begin{aligned} \text{Median } &= l + \bigg(\frac{\frac{N}{2} - F_<}{f}\bigg)\times h\\ &= 7.5 + \bigg(\frac{\frac{100}{2} - 18}{34}\bigg)\times 2\\ &= 7.5 + \bigg(\frac{50 - 18}{34}\bigg)\times 2\\ &= 7.5 + \big(0.9412\big)\times 2\\ &= 7.5 + 1.8824\\ &= 9.3824 \text{ ('00 grams)} \end{aligned} $$

Mode

The maximum frequency is $34$, the corresponding class $7.5-10.5$ is the modal class.

Mode of the given frequency distribution is:

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ \end{aligned} $$

where

  • $l = 7.5$, the lower limit of the modal class
  • $f_m =34$, frequency of the modal class
  • $f_1 = 14$, frequency of the pre-modal class
  • $f_2 = 28$, frequency of the post-modal class
  • $h =2$, the class width

Thus mode of a frequency distribution is

$$ \begin{aligned} \text{Mode } &= l + \bigg(\frac{f_m - f_1}{2f_m-f_1-f_2}\bigg)\times h\\ &= 7.5 + \bigg(\frac{34 - 14}{2\times34 - 14 - 28}\bigg)\times 2\\ &= 7.5 + \bigg(\frac{20}{26}\bigg)\times 2\\ &= 7.5 + \big(0.7692\big)\times 2\\ &= 7.5 + \big(1.5385\big)\\ &= 9.0385 \text{ ('00 grams)} \end{aligned} $$

Conclusion

In this tutorial, you learned about formula for mean, median and mode for grouped data and how to calculate mean, median and mode for grouped data. You also learned about how to solve numerical problems based on mean, median and mode for grouped data.

To learn more about other descriptive statistics measures, please refer to the following tutorials:

Descriptive Statistics

Let me know in the comments if you have any questions on Mean, median and mode calculator for grouped data with examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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