
Compute the mode for the following frequency distribution.
Classes 100-120 120-140 140-160 160-180 180-200 200-220 220-240 Frequency 2 5 10 25 20 15 12
Classes | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
Frequency | 2 | 5 | 10 | 25 | 20 | 15 | 12 |
Answer
509.4k+ views
Hint: We start solving this problem by first considering the given frequency distribution. Then we find the modal class, that is, the class which has the highest frequency. Then we find the mode of the given frequency distribution by using the formula $l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h$ .
Complete step by step answer:
Let us consider the given data,
Now, we find the modal class. The class with the highest frequency is the modal class.
By observing the given data, we can say that the class 160-180 has the highest frequency.
So, the modal class is 160-180 class.
Now, let us consider the formula for calculating mode,
$\text{Mode}=l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h$.
Where, $l$ is the lower boundary of the modal class.
${{f}_{1}}$ is the frequency of the modal class.
${{f}_{0}}$ is the frequency of the class preceding the modal class.
${{f}_{2}}$ is the frequency of the class succeeding the modal class.
$h$ is the height of the modal class.
So, we get,
So, we get, ${{f}_{0}}=10,{{f}_{1}}=25,{{f}_{2}}=20,l=160$ and $h=180-160=20$ .
By using the above mentioned formula,
We get the mode of the given frequency distribution as
$\begin{align}
& l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h \\
& =160+\left( \dfrac{25-10}{2\left( 25 \right)-10-20} \right)20 \\
& =160+\left( \dfrac{15}{50-30} \right)20 \\
& =160+\left( \dfrac{15}{20} \right)20 \\
& =160+15 \\
& =175 \\
\end{align}$
Therefore, the mode of the given frequency distribution is 175.
Note:
The possibilities for making mistakes in this type of problems are, one may make a mistake by considering ${{f}_{0}}$ and ${{f}_{2}}$ as below.
${{f}_{0}}$ is the frequency of the class succeeding the modal class.
${{f}_{2}}$ is the frequency of the class preceding the modal class.
Complete step by step answer:
Let us consider the given data,
Classes | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
Frequency | 2 | 5 | 10 | 25 | 20 | 15 | 12 |
Now, we find the modal class. The class with the highest frequency is the modal class.
By observing the given data, we can say that the class 160-180 has the highest frequency.
So, the modal class is 160-180 class.
Now, let us consider the formula for calculating mode,
$\text{Mode}=l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h$.
Where, $l$ is the lower boundary of the modal class.
${{f}_{1}}$ is the frequency of the modal class.
${{f}_{0}}$ is the frequency of the class preceding the modal class.
${{f}_{2}}$ is the frequency of the class succeeding the modal class.
$h$ is the height of the modal class.
So, we get,
Classes | Frequency |
100-120 | 2 |
120-140 | 5 |
140-160 | 10(${{f}_{0}}$) |
160-180 | 25(${{f}_{1}}$) |
180-200 | 20(${{f}_{2}}$) |
200-220 | 15 |
220-240 | 12 |
So, we get, ${{f}_{0}}=10,{{f}_{1}}=25,{{f}_{2}}=20,l=160$ and $h=180-160=20$ .
By using the above mentioned formula,
We get the mode of the given frequency distribution as
$\begin{align}
& l+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h \\
& =160+\left( \dfrac{25-10}{2\left( 25 \right)-10-20} \right)20 \\
& =160+\left( \dfrac{15}{50-30} \right)20 \\
& =160+\left( \dfrac{15}{20} \right)20 \\
& =160+15 \\
& =175 \\
\end{align}$
Therefore, the mode of the given frequency distribution is 175.
Note:
The possibilities for making mistakes in this type of problems are, one may make a mistake by considering ${{f}_{0}}$ and ${{f}_{2}}$ as below.
${{f}_{0}}$ is the frequency of the class succeeding the modal class.
${{f}_{2}}$ is the frequency of the class preceding the modal class.
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