Question

Find the median of the following data:
\[ \begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 40-45 & 2 \\ 45-50 & 3 \\ 50-55 & 8 \\ 55-60 & 6 \\ 60-65 & 6 \\ 65-70 & 3 \\ 70-75 & 2 \\ \hline \end{array} \]

Hint:

To find the median of grouped data, use the formula \( \text{Median} = L + \frac{\frac{N}{2} - CF}{f} \times h \), where \( N \) is the total frequency, \( CF \) is the cumulative frequency of the class before the median class, \( f \) is the frequency of the median class, and \( h \) is the class width.

Answer 1

To find the median, we first calculate the cumulative frequency (CF). \[ \begin{array}{|c|c|c|} \hline \text{Class Interval} & \text{Frequency (f)} & \text{Cumulative Frequency (CF)} \\ \hline 40-45 & 2 & 2 \\ 45-50 & 3 & 5 \\ 50-55 & 8 & 13 \\ 55-60 & 6 & 19 \\ 60-65 & 6 & 25 \\ 65-70 & 3 & 28 \\ 70-75 & 2 & 30 \\ \hline \end{array} \] The total frequency, \( N = 30 \). The median class is the class where the cumulative frequency exceeds \( \frac{N}{2} = \frac{30}{2} = 15 \). From the cumulative frequency table, the median class is \( 50-55 \) because the cumulative frequency just exceeds 15. Now, we use the following formula for the median: \[ \text{Median} = L + \frac{\frac{N}{2} - CF}{f} \times h \] Where: - \( L \) is the lower limit of the median class = 50, - \( N \) is the total frequency = 30, - \( CF \) is the cumulative frequency of the class before the median class = 5, - \( f \) is the frequency of the median class = 8, - \( h \) is the class width = 5 (since \( 50-45 = 5 \)). Substituting the values: \[ \text{Median} = 50 + \frac{15 - 5}{8} \times 5 \] \[ = 50 + \frac{10}{8} \times 5 \] \[ = 50 + \frac{50}{8} \] \[ = 50 + 6.25 \] \[ \text{Median} = 56.25 \] Thus, the median of the data is \( \mathbf{56.25} \). \hrule