The median of a frequency distribution is given by the formula:
Median Formula:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h, \]
where:
The cumulative frequency is calculated by adding the frequencies sequentially:
Class Interval | Frequency | Cumulative Frequency |
---|---|---|
10 - 25 | 3 | 3 |
25 - 40 | 10 | 13 |
40 - 55 | 20 | 33 |
55 - 70 | 13 | 46 |
70 - 85 | 4 | 50 |
The total frequency \( N = 50 \). To find the median class, we calculate \( \frac{N}{2} = \frac{50}{2} = 25 \).
The cumulative frequency just greater than or equal to 25 is 33, which corresponds to the class interval \( 40 - 55 \). This is the median class.
The median class is \( 40 - 55 \), so:
Substitute these values into the median formula:
\[ \text{Median} = 40 + \left( \frac{25 - 13}{20} \right) \times 15 = 40 + \left( \frac{12}{20} \right) \times 15 = 40 + \left( 0.6 \right) \times 15 = 40 + 9 = 49. \]
The median is 49.
The coefficient of correlation of the above two data series will be equal to \(\underline{\hspace{1cm}}\)
\[\begin{array}{|c|c|} \hline X & Y \\ \hline -3 & 9 \\ -2 & 4 \\ -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ \hline \end{array}\]
Identify the median class for the following grouped data:
\[\begin{array}{|c|c|} \hline \textbf{Class interval} & \textbf{Frequency} \\ \hline 5-10 & 5 \\ 10-15 & 15 \\ 15-20 & 22 \\ 20-25 & 25 \\ 25-30 & 10 \\ 30-35 & 3 \\ \hline \end{array}\]
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
Two concentric circles are of radii $8\ \text{cm}$ and $5\ \text{cm}$. Find the length of the chord of the larger circle which touches (is tangent to) the smaller circle.