Consider the following frequency distribution table: \[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class Interval} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \textbf{Frequency} & 180 & F_1 & 34 & 180 & 136 & F_2 & 50 \\ \hline \end{array} \] If the total frequency is 686 and the median is 42.6, then the value of \( F_1 \) and \( F_2 \) are.
Since the total frequency is 686, we have:
\[
F_1 + F_2 = 686 - \text{sum of all other frequencies}
\]
\[
F_1 + F_2 = 106
\]
Given that the median is 42.6, the median class is 40-50. The formula for the median is:
\[
\ell_1 + \frac{\frac{n}{2} - C}{f} (\ell_2 - \ell_1)
\]
Where:
- \( \ell_1, \ell_2 \) are the lower and upper limits of the class interval,
- \( n \) is the total frequency,
- \( C \) is the cumulative frequency of the pre-median class,
- \( f \) is the frequency of the median class.
Thus, we have:
\[
42.6 = 40 + \frac{343 - C}{180} \times 10
\]
Solving for \( C \):
\[
C = 296
\]
Now, from the equation:
\[
180 + 34 + F_1 = 296
\]
We find:
\[
F_1 = 82
\]
Hence:
\[
F_2 = 24
\]
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