Answer :
Answer:
To find the median from this grouped data, we first need to determine the cumulative frequency and then find the class interval that contains the median. The formula to find the median of grouped data is:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times w \]
Where:
- \( L \) = Lower boundary of the median class
- \( N \) = Total frequency
- \( F \) = Cumulative frequency of the class before the median class
- \( f \) = Frequency of the median class
- \( w \) = Width of the median class
Given data:
\[
\begin{array}{|c|c|c|}
\hline
\text{Age (years)} & \text{No. of Students} \\
\hline
20-25 & 50 \\
25-30 & 70 \\
30-35 & 100 \\
35-40 & 180 \\
40-45 & 150 \\
45-50 & 120 \\
50-55 & 70 \\
55-60 & 60 \\
\hline
\end{array}
\]
1. Calculate the cumulative frequencies:
\[
\begin{array}{|c|c|}
\hline
\text{Age (years)} & \text{Cumulative Frequency} \\
\hline
20-25 & 50 \\
25-30 & 120 \\
30-35 & 220 \\
35-40 & 400 \\
40-45 & 550 \\
45-50 & 670 \\
50-55 & 740 \\
55-60 & 800 \\
\hline
\end{array}
\]
2. Identify the median class, which is the class that contains the middle value. Since the total frequency is 800, the median will be the \( \frac{800}{2} = 400 \)th value.
3. From the cumulative frequency table, the median class is 35-40, which has a cumulative frequency of 400.
4. Calculate the median:
\[ L = 35 \text{ (lower boundary of the median class)} \]
\[ N = 800 \text{ (total frequency)} \]
\[ F = 220 \text{ (cumulative frequency before the median class)} \]
\[ f = 180 \text{ (frequency of the median class)} \]
\[ w = 5 \text{ (width of the median class)} \]
\[ \text{Median} = 35 + \left( \frac{\frac{800}{2} - 220}{180} \right) \times 5 \]
\[ \text{Median} = 35 + \left( \frac{400 - 220}{180} \right) \times 5 \]
\[ \text{Median} = 35 + \left( \frac{180}{180} \right) \times 5 \]
\[ \text{Median} = 35 + 1 \times 5 \]
\[ \text{Median} = 35 + 5 \]
\[ \text{Median} = 40 \]
So, the median age is 40 years.