Question

Find the mode of the following classified data. 
\[ \begin{array}{|c|c|c|c|c|} \hline \text{Class interval} & 0\text{-}10 & 10\text{-}20 & 20\text{-}30 & 30\text{-}40 \\ \hline \text{Frequency} & 3 & 4 & 6 & 5 \\ \hline \end{array} \]

• A. \(25\dfrac{2}{3}\)
• B. \(26\dfrac{2}{3}\)
• C. \(27\dfrac{2}{3}\)
• D. \(28\dfrac{2}{3}\)

Hint:

For grouped data, \[ \boxed{ \text{Mode} = l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h } \] where
• \(l\) = lower limit of modal class,
• \(h\) = class width,
• \(f_1\) = frequency of modal class,
• \(f_0\) = frequency preceding the modal class,
• \(f_2\) = frequency succeeding the modal class.

Answer 1

The Correct Option is B

Step 1: Identify the modal class. The highest frequency is \[ 6, \] which corresponds to the class interval \[ 20-30. \] Hence, \[ l=20,\quad h=10,\quad f_1=6,\quad f_0=4,\quad f_2=5. \]

Step 2: Apply the mode formula. \[ \text{Mode} = l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h. \] Substituting the values, \[ = 20+\frac{6-4}{2(6)-4-5}\times10 \] \[ = 20+\frac{2}{3}\times10 \] \[ = 20+\frac{20}{3} = \frac{80}{3} = 26\dfrac{2}{3}. \]

Step 3: Final conclusion. \[ \boxed{26\dfrac{2}{3}} \] Hence, the correct option is \(\boxed{(B)}\).