
If${L_1} = 44.5$,$C = 10$,${F_2} = 10$, ${F_1} = 28$, $f = 30$, then find mode.
A. $45.41$
B. $47.31$
C. $49.61$
D. $53$
Answer
493.8k+ views
Hint : First, we shall analyze the given data so that we are able to solve the given problem. Here, we are asked to calculate the mode. We all know that the mode of the data is the most frequently occurring value. In this question, we are directly given the values that are usually found using the frequency distribution. We need to substitute the given values in the formula directly.
Formula used:
The formula to calculate the mode for the grouped data is as follows.
$ Mode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$
Where $l$ is the lower limit, $h$ is the size of the class interval,${f_1}$ denotes the frequency of the modal class, ${f_0}$ is the frequency of class preceding (before) the modal class, and${f_2}$ denotes the frequency of the class succeeding (after) the modal class.
Complete step-by-step solution:
We are given${L_1} = 44.5$,$C = 10$,${F_2} = 10$, ${F_1} = 28$, $f = 30$
Here, we can note that${L_1}$denotes the lower limit of the modal class, $C$ is the size of the class interval l, $f$ denotes the frequency of the modal class,${F_1}$ is the frequency of class preceding (before) the modal class and${F_2}$ is the frequency of class succeeding (after) the modal class
Now, we shall apply the given values in the formula.
Thus, we get
\[Mode = {L_1} + \dfrac{{f - {F_1}}}{{2f - {F_1} - {F_2}}} \times C\]
\[ \Rightarrow Mode = 44.5 + \dfrac{{30 - 28}}{{2 \times 30 - 28 - 10}} \times 10\]
\[ \Rightarrow Mode = 44.5 + \dfrac{2}{{60 - 38}} \times 10\]
\[ \Rightarrow Mode = 44.5 + \dfrac{{20}}{{22}}\]
\[ \Rightarrow Mode = 44.5 + 0.9090\]
\[ \Rightarrow Mode = 45.409090\]
\[ \Rightarrow Mode = 45.41\] (approximately)
Hence, option A is correct.
Note: Here in this problem, we are directly given the values that are needed for the formula. The modal class is the class interval that contains the maximum frequency. Suppose if we are given a frequency distribution table, we need to first find the modal class. The given formula is applicable for finding the mode for grouped data. For ungrouped data, we need to just pick the most frequently occurred value for mode.
Formula used:
The formula to calculate the mode for the grouped data is as follows.
$ Mode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$
Where $l$ is the lower limit, $h$ is the size of the class interval,${f_1}$ denotes the frequency of the modal class, ${f_0}$ is the frequency of class preceding (before) the modal class, and${f_2}$ denotes the frequency of the class succeeding (after) the modal class.
Complete step-by-step solution:
We are given${L_1} = 44.5$,$C = 10$,${F_2} = 10$, ${F_1} = 28$, $f = 30$
Here, we can note that${L_1}$denotes the lower limit of the modal class, $C$ is the size of the class interval l, $f$ denotes the frequency of the modal class,${F_1}$ is the frequency of class preceding (before) the modal class and${F_2}$ is the frequency of class succeeding (after) the modal class
Now, we shall apply the given values in the formula.
Thus, we get
\[Mode = {L_1} + \dfrac{{f - {F_1}}}{{2f - {F_1} - {F_2}}} \times C\]
\[ \Rightarrow Mode = 44.5 + \dfrac{{30 - 28}}{{2 \times 30 - 28 - 10}} \times 10\]
\[ \Rightarrow Mode = 44.5 + \dfrac{2}{{60 - 38}} \times 10\]
\[ \Rightarrow Mode = 44.5 + \dfrac{{20}}{{22}}\]
\[ \Rightarrow Mode = 44.5 + 0.9090\]
\[ \Rightarrow Mode = 45.409090\]
\[ \Rightarrow Mode = 45.41\] (approximately)
Hence, option A is correct.
Note: Here in this problem, we are directly given the values that are needed for the formula. The modal class is the class interval that contains the maximum frequency. Suppose if we are given a frequency distribution table, we need to first find the modal class. The given formula is applicable for finding the mode for grouped data. For ungrouped data, we need to just pick the most frequently occurred value for mode.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

