Correct Answer: Option D

Explanation

Given the frequency distribution:

\(\begin{array}{|c|c|}
\hline
\text{Class Interval} & \text{Frequency} \\
\hline
0-9 & 1 \\
10-19 & 5 \\
20-29 & 6 \\
30-39 & 12 \\
40-49 & 8 \\
50-59 & 3 \\
\hline\end{array}\)

The modal class is \(30-39\) with frequency \(f_1 = 12\), \(f_0 = 6\) (previous class), \(f_2 = 8\) (next class), and the lower boundary \(L = 29.5\) with class width \(h = 10\).

Using the mode formula:

\(\text{Mode} = L + \left( \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \right) \times h\)

Substituting the values:

\(\text{Mode} = 29.5 + \left( \frac{12 - 6}{(12 - 6) + (12 - 8)} \right) \times 10\)

\(\text{Mode} = 29.5 + \left( \frac{6}{6 + 4} \right) \times 10 = 29.5 + \left( \frac{6}{10} \right) \times 10 = 29.5 + 6 = 35.5\)

There is an explanation video available below.

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