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Question: Calculate the mode.![](https://www.vedantu.com/question-sets/231fad97-eeb9-4d0d-82c0-6c9f99b3a61c748...

Calculate the mode.

Explanation

Solution

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=fixiNmean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}. By this we will get the mean of the class. Here, xi{{x}_{i}} is mid-value of the class interval and fi{{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 mode=l+ff12ff1f2×h\bmod e=l+\dfrac{f-{{f}_{1}}}{2f-{{f}_{1}}-{{f}_{2}}}\times h. Here, ll= lower limit of modal class, hh = class width, f=f= frequency of the modal class, f1={{f}_{1}}= frequency of the class before modal class and f2={{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 fixi{{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=fixiNmean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{N}
Here, xi{{x}_{i}} is mid-value of the class interval and fi{{f}_{i}} is the frequency and N is the total frequency.
fixi=25+120+175+630+1260+1100+650+750=4440\sum{{{f}_{i}}{{x}_{i}}}=25+120+175+630+1260+1100+650+750=4440
and N=5+8+7+18+28+20+10+10=106N=5+8+7+18+28+20+10+10=106
So, mean=fixiN=4440106=41.88mean=\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,
mode=l+ff12ff1f2×h\bmod e=l+\dfrac{f-{{f}_{1}}}{2f-{{f}_{1}}-{{f}_{2}}}\times h
Here, ll = lower limit of modal class, hh = class width, f=f= frequency of the modal class, f1={{f}_{1}}= frequency of the class before modal class and f2={{f}_{2}}= frequency of the class after modal class.
So, we have
ll = 40
hh = 10
f=f= 28
f1={{f}_{1}}= 18
f2={{f}_{2}}= 20
So, the equation becomes,
mode=40+28182(28)1820×10\bmod e=40+\dfrac{28-18}{2(28)-18-20}\times 10
mode=40+10018\bmod e=40+\dfrac{100}{18}
On solving we get,
mode=45.55\bmod e=45.55
So, the mode is 45.55.

Note: We must be careful while taking the value of f1{{f}_{1}} and f2{{f}_{2}}, sometimes we get confused while taking these values. Also be careful while finding fixi\sum{{{f}_{i}}{{x}_{i}}} because sometimes we might miss out 1 or 2 values.