For the following frequency distribution

Class:	3-5	5-7	7-9	9-11
Frequency:	1	4	2	1

the value of mode is:

The correct answer is 6.20

Key Points

The formula of the mode used in a frequency distribution is:

\(\mathbf{M_{0}=\text{l }+(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}})*h }\)

where ,\(M_{0}\)=Mode, l=lower limit of the modal class, h=size of the class interval,\(f_{1}\)=frequency of modal class.\(f_{1}\) =frequency of the class preceding, the modal class and \(f_{2}\)= The frequency of the class succeeding the modal class.

According to the problem, class 5-7 is the modal class, since it has the highest frequency.

Substituting the respective values in the mode equation gives us:

\(M_{0}=5+(\frac{4-1}{(2*4)-1-2})*2\\ \implies M_{0}=5+1.20\\ \implies M_{0}=6.20 \)

Additional Information

•  The value of h here is symmetrical which is \( 7-5=2\).
• The real upper limit is produced by adding 0.5 to the highest number, if it is expressed as a whole number, or by adding 0.05 to the highest number if it is represented as a decimal. The upper limit is the highest value of the class interval.
• Similarly to the upper limit, the lower limit is the smallest value in the class interval. To determine the actual lower limit, take the smallest number and either deduct 0.5 from it if it's a whole number or 0.05 from it if it's a decimal.

Hence for the given frequency distribution the value of mode is 6.20