Find the median of the following frequency table: 
Step 1: Calculate cumulative frequency. 
Step 2: Identify median class.
Total frequency \( N = 400 \). \[ \frac{N}{2} = 200 \] The cumulative frequency just greater than 200 is 216, so the median class is \( 30–35 \).
Step 3: Use the median formula.
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf_{\text{before}}}{f}\right) \times h \] Here, \( l = 30, \; cf_{\text{before}} = 130, \; f = 86, \; h = 5 \).
Step 4: Substitute the values.
\[ \text{Median} = 30 + \left(\frac{200 - 130}{86}\right) \times 5 = 30 + \frac{70}{86} \times 5 \] \[ = 30 + 4.07 = 34.07 \] Step 5: Conclusion.
Hence, the median of the data is \( \boxed{34.07} \).
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:
The mean of the following table will be:
| Class-interval | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 |
|---|---|---|---|---|---|
| Frequency (f) | 3 | 1 | 5 | 4 | 7 |
From the following data, the modal class of the table will be:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency (f)} & 11 & 21 & 23 & 14 & 5 \\ \hline \end{array} \]
Find the Mean from the Following Table
Given data:
\[ \begin{array}{|c|c|} \hline \text{Class-interval} & \text{Frequency (f)} \\ \hline 0-10 & 3 \\ 10-20 & 10 \\ 20-30 & 11 \\ 30-40 & 9 \\ 40-50 & 7 \\ \hline \end{array} \]