Contents
- 1 Mean, median and mode
- 2 Mean, Median and Mode for grouped data
- 3 Sample mean
- 4 Sample median
- 5 Steps for the calculation of Median
- 6 Sample mode
- 7 Mean, median and mode calculator for grouped data
- 8 Mean median and mode for grouped data Example 1
- 9 Mean median and mode for grouped data Example 2
- 10 Mean median and mode for grouped data Example 3
- 11 Mean median and mode for grouped data Example 4
- 12 Mean median and mode for grouped data Example 5
- 13 Conclusion
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:
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.