
Calculate the mode.

Answer
507.6k+ views
Hint: To find the mode in this question, first of all we have to find the maximum frequency. So, to find the maximum frequency we have to use the equation of mean. That is, $mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}$. By this we will get the mean of the class. Here, ${{x}_{i}}$ is mid-value of the class interval and ${{f}_{i}}$ is the frequency and N is the total frequency. After finding the mean of the class we can know the maximum frequency, that is the frequency of the modal class. Then, we have to find the mode using the equation $\bmod e=l+\dfrac{f-{{f}_{1}}}{2f-{{f}_{1}}-{{f}_{2}}}\times h$. Here, $l$= lower limit of modal class, $h$ = class width, $f=$ frequency of the modal class, ${{f}_{1}}=$ frequency of the class before modal class and ${{f}_{2}}=$ frequency of the class after modal class.
Complete step-by-step answer:
First of all we have to redraw the table that contains the mid-value and ${{f}_{i}}{{x}_{i}}$ columns. So, we will have a table as such,
So, now we can find the modal class by using the equation,
$mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}$
Here, ${{x}_{i}}$ is mid-value of the class interval and ${{f}_{i}}$ is the frequency and N is the total frequency.
$\sum{{{f}_{i}}{{x}_{i}}}=25+120+175+630+1260+1100+650+750=4440$
and $N=5+8+7+18+28+20+10+10=106$
So, $mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}=\dfrac{4440}{106}=41.88$
So, the modal class is 40 – 50.
Therefore, the maximum frequency = 28.
Now, we have to find the mode of the class by using the equation,
$\bmod e=l+\dfrac{f-{{f}_{1}}}{2f-{{f}_{1}}-{{f}_{2}}}\times h$
Here, $l$ = lower limit of modal class, $h$ = class width, $f=$ frequency of the modal class, ${{f}_{1}}=$ frequency of the class before modal class and ${{f}_{2}}=$ frequency of the class after modal class.
So, we have
$l$ = 40
$h$ = 10
$f=$ 28
${{f}_{1}}=$ 18
${{f}_{2}}=$ 20
So, the equation becomes,
$\bmod e=40+\dfrac{28-18}{2(28)-18-20}\times 10$
$\bmod e=40+\dfrac{100}{18}$
On solving we get,
$\bmod e=45.55$
So, the mode is 45.55.
Note: We must be careful while taking the value of ${{f}_{1}}$ and ${{f}_{2}}$, sometimes we get confused while taking these values. Also be careful while finding $\sum{{{f}_{i}}{{x}_{i}}}$ because sometimes we might miss out 1 or 2 values.
Complete step-by-step answer:
First of all we have to redraw the table that contains the mid-value and ${{f}_{i}}{{x}_{i}}$ columns. So, we will have a table as such,

So, now we can find the modal class by using the equation,
$mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}$
Here, ${{x}_{i}}$ is mid-value of the class interval and ${{f}_{i}}$ is the frequency and N is the total frequency.
$\sum{{{f}_{i}}{{x}_{i}}}=25+120+175+630+1260+1100+650+750=4440$
and $N=5+8+7+18+28+20+10+10=106$
So, $mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}=\dfrac{4440}{106}=41.88$
So, the modal class is 40 – 50.
Therefore, the maximum frequency = 28.
Now, we have to find the mode of the class by using the equation,
$\bmod e=l+\dfrac{f-{{f}_{1}}}{2f-{{f}_{1}}-{{f}_{2}}}\times h$
Here, $l$ = lower limit of modal class, $h$ = class width, $f=$ frequency of the modal class, ${{f}_{1}}=$ frequency of the class before modal class and ${{f}_{2}}=$ frequency of the class after modal class.
So, we have
$l$ = 40
$h$ = 10
$f=$ 28
${{f}_{1}}=$ 18
${{f}_{2}}=$ 20
So, the equation becomes,
$\bmod e=40+\dfrac{28-18}{2(28)-18-20}\times 10$
$\bmod e=40+\dfrac{100}{18}$
On solving we get,
$\bmod e=45.55$
So, the mode is 45.55.
Note: We must be careful while taking the value of ${{f}_{1}}$ and ${{f}_{2}}$, sometimes we get confused while taking these values. Also be careful while finding $\sum{{{f}_{i}}{{x}_{i}}}$ because sometimes we might miss out 1 or 2 values.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

The non protein part of an enzyme is a A Prosthetic class 11 biology CBSE

Which of the following blood vessels in the circulatory class 11 biology CBSE

What is a zygomorphic flower Give example class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

The deoxygenated blood from the hind limbs of the frog class 11 biology CBSE
