Median class of the following frequency distribution will be:
\[ \begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 7 \\ \hline 10-20 & 12 \\ \hline 20-30 & 18 \\ \hline 30-40 & 15 \\ \hline 40-50 & 10 \\ \hline 50-60 & 3 \\ \hline \end{array} \]
Step 1: Find total frequency ($N$)
\[ N = 7 + 12 + 18 + 15 + 10 + 3 = 65 \]
Step 2: Find $\tfrac{N{2}$}
\[ \frac{N}{2} = \frac{65}{2} = 32.5 \]
Step 3: Construct cumulative frequency (CF)
\[ \begin{array}{|c|c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 0-10 & 7 & 7 \\ \hline 10-20 & 12 & 19 \\ \hline 20-30 & 18 & 37 \\ \hline 30-40 & 15 & 52 \\ \hline 40-50 & 10 & 62 \\ \hline 50-60 & 3 & 65 \\ \hline \end{array} \]
Step 4: Identify median class
- The median class is the class interval where the cumulative frequency first becomes $\geq 32.5$.
- Here, CF = 37 for class $20$-$30$.
Thus, the median class is $20$-$30$.
Step 5: Conclusion
Therefore, the median class is $20$-$30$.
The correct answer is option (A).
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be:
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
The median class of the following frequency distribution will be:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-Interval} & \text{$0$--$10$} & \text{$10$--$20$} & \text{$20$--$30$} & \text{$30$--$40$} & \text{$40$--$50$} \\ \hline \text{Frequency} & \text{$7$} & \text{$8$} & \text{$15$} & \text{$10$} & \text{$5$} \\ \hline \end{array}\]