
Calculate the value of mode for the following frequency distribution:
Class 1-4 5-8 9-12 13-16 17-20 21-24 25-28 29-32 33-36 37-40 Frequency 2 5 8 9 12 14 14 15 11 13
| Class | 1-4 | 5-8 | 9-12 | 13-16 | 17-20 | 21-24 | 25-28 | 29-32 | 33-36 | 37-40 |
| Frequency | 2 | 5 | 8 | 9 | 12 | 14 | 14 | 15 | 11 | 13 |
Answer
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Hint: In this question, we have to calculate mode for the given inclusive series. For this, we will first change the inclusive series to exclusive series by subtracting $\dfrac{n}{2}$ from the lower limit and adding $\dfrac{n}{2}$ to the upper limit where n is the difference between the upper limit of a class and the lower limit of next class. As we can see, the frequency of three intervals becomes almost the same and we have changed to an exclusive series, so, we will have to use a grouping method for solving this sum. By grouping method, we will find the modal class and then use the formula of mode for continuous series to get the required answer. We will use the following formula:
\[\text{Mode}={{l}_{1}}+\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times i\]
Where, ${{l}_{1}}$ is the lower limit of modal class, ${{f}_{1}}$ is the frequency of modal class, ${{f}_{0}}$ is the frequency of pre-modal class and ${{f}_{2}}$ is the frequency of next highest class and i is the size of the modal group.
Complete step-by-step solution:
Let us solve the summing step by step:
Step 1: We will convert the series into exclusive series by adding $\dfrac{n}{2}$ to the upper limit of classes and subtracting $\dfrac{n}{2}$ from the lower limit of classes. Here, n is 1 for the given series (5-4=1, 9-8=1)
Hence, we will subtract and add 0.5 from the upper and lower class respectively. Series becomes
Step 2: To find a modal class, let us make a grouping table. In this, we will add frequencies in the table by specific method given in the table and then mark the highest frequency for each column.
Here, 1+2 represents adding the first two rows and then the next two and similarly, till last or only one row is left.
2.+3+4 represents adding three rows at a time starting from the second row only.
Similarly, the table is constructed by taking different columns and making the highest frequency in each column.
Step 3: Now, let's make an analysis table by marking bars for classes with the highest frequencies marked earlier.
Step 4: From the analysis table, we can see that class 24.5-28.5 has maximum bars. So, it is our modal class. Now, as we can see from frequency distribution table made earlier, ${{f}_{1}}=14$ (frequency of modal class), ${{f}_{0}}=14$ (frequency of class preceding modal class) and ${{f}_{2}}=15$ (frequency of class succeeding modal class).
Also, ${{l}_{1}}=24.5$ (lower limit of modal class)
Let us calculate i which is the class interval of the modal class. As we know, class interval = upper limit - lower limit. Therefore, i = 28.5-24.5 = 4
Step 5: Using formula for calculating mode,
\[\text{Mode}={{l}_{1}}+\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times i\]
Putting all values obtained earlier, we get:
\[\begin{align}
& \text{Mode}=24.5+\dfrac{14-14}{28-14-15}\times 4 \\
& \Rightarrow 24.5+0\times 4 \\
& \Rightarrow 24.5 \\
\end{align}\]
Hence, the mode of the given series is 24.5.
Note: Students should note that, while changing the inclusive series to exclusive series, make sure that class interval remains the same for all classes. Also, while making a grouping table, if there are only terms left at the end but you are taking a sum of two or three rows then it should be left and not just take that frequency. While making the analysis table, make sure that you put bars for all classes from which the highest frequency is obtained. Students should do these sums carefully and step by step to avoid mistakes.
\[\text{Mode}={{l}_{1}}+\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times i\]
Where, ${{l}_{1}}$ is the lower limit of modal class, ${{f}_{1}}$ is the frequency of modal class, ${{f}_{0}}$ is the frequency of pre-modal class and ${{f}_{2}}$ is the frequency of next highest class and i is the size of the modal group.
Complete step-by-step solution:
Let us solve the summing step by step:
Step 1: We will convert the series into exclusive series by adding $\dfrac{n}{2}$ to the upper limit of classes and subtracting $\dfrac{n}{2}$ from the lower limit of classes. Here, n is 1 for the given series (5-4=1, 9-8=1)
Hence, we will subtract and add 0.5 from the upper and lower class respectively. Series becomes
| Class | Frequency |
| 0.5-4.5 | 2 |
| 4.5-8.5 | 5 |
| 8.5-12.5 | 8 |
| 12.5-16.5 | 9 |
| 16.5-20.5 | 12 |
| 20.5-24.5 | 14 |
| 24.5-28.5 | 14 |
| 28.5-32.5 | 15 |
| 32.5-36.5 | 11 |
| 36.5-40.5 | 13 |
Step 2: To find a modal class, let us make a grouping table. In this, we will add frequencies in the table by specific method given in the table and then mark the highest frequency for each column.
| Class Interval | I | II (1+2) | III (2+3) | IV (1+2+3) | V (2+3+4) | VI (3+4+5) |
| 0.5-4.5 | 2 | 7 | 15 | |||
| 4.5-8.5 | 5 | 13 | 22 | |||
| 8.5-12.5 | 8 | 17 | 29 | |||
| 12.5-16.5 | 9 | 21 | 35 | |||
| 16.5-20.5 | 12 | 26 | 40 | |||
| 20.5-24.5 | 14 | 28 | 43 | |||
| 24.5-28.5 | 14 | 29 | 40 | |||
| 28.5-32.5 | 15 | 26 | 39 | |||
| 32.5-36.5 | 11 | 24 | ||||
| 36.5-40.5 | 13 |
Here, 1+2 represents adding the first two rows and then the next two and similarly, till last or only one row is left.
2.+3+4 represents adding three rows at a time starting from the second row only.
Similarly, the table is constructed by taking different columns and making the highest frequency in each column.
Step 3: Now, let's make an analysis table by marking bars for classes with the highest frequencies marked earlier.
| Column No | 0.5-4.5 | 4.5-8.5 | 8.5-12.5 | 12.5-16.5 | 16.5-20.5 | 20.5-24.5 | 24.5-28.5 | 28.5-32.5 | 32.5-36.5 | 36.5-40.5 |
| I | 1 | |||||||||
| II | 1 | 1 | ||||||||
| III | 1 | 1 | ||||||||
| IV | 1 | 1 | 1 | |||||||
| V | 1 | 1 | 1 | |||||||
| VI | 1 | 1 | 1 | |||||||
| Total | 1 | 3 | 5 | 4 | 1 |
Step 4: From the analysis table, we can see that class 24.5-28.5 has maximum bars. So, it is our modal class. Now, as we can see from frequency distribution table made earlier, ${{f}_{1}}=14$ (frequency of modal class), ${{f}_{0}}=14$ (frequency of class preceding modal class) and ${{f}_{2}}=15$ (frequency of class succeeding modal class).
Also, ${{l}_{1}}=24.5$ (lower limit of modal class)
Let us calculate i which is the class interval of the modal class. As we know, class interval = upper limit - lower limit. Therefore, i = 28.5-24.5 = 4
Step 5: Using formula for calculating mode,
\[\text{Mode}={{l}_{1}}+\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times i\]
Putting all values obtained earlier, we get:
\[\begin{align}
& \text{Mode}=24.5+\dfrac{14-14}{28-14-15}\times 4 \\
& \Rightarrow 24.5+0\times 4 \\
& \Rightarrow 24.5 \\
\end{align}\]
Hence, the mode of the given series is 24.5.
Note: Students should note that, while changing the inclusive series to exclusive series, make sure that class interval remains the same for all classes. Also, while making a grouping table, if there are only terms left at the end but you are taking a sum of two or three rows then it should be left and not just take that frequency. While making the analysis table, make sure that you put bars for all classes from which the highest frequency is obtained. Students should do these sums carefully and step by step to avoid mistakes.
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