To calculate the median for the given distribution, follow these steps:
### Step 1: Determine the Cumulative Frequencies
1. Class intervals and frequencies:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Class Interval} & f \\
\hline
50-60 & 3 \\
60-70 & 7 \\
70-80 & 11 \\
80-90 & 16 \\
90-100 & 8 \\
100-110 & 5 \\
\hline
\end{array}
\][/tex]
2. Cumulative frequencies (cf):
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Class Interval} & f & cf \\
\hline
50-60 & 3 & 3 \\
60-70 & 7 & 10 \\
70-80 & 11 & 21 \\
80-90 & 16 & 37 \\
90-100 & 8 & 45 \\
100-110 & 5 & 50 \\
\hline
\end{array}
\][/tex]
### Step 2: Determine the Median Class
1. Total number of observations (N):
[tex]\[
N = \sum f = 50
\][/tex]
2. Calculate [tex]\( \frac{N}{2} \)[/tex]:
[tex]\[
\frac{N}{2} = \frac{50}{2} = 25
\][/tex]
3. Finding the median class (The class whose cumulative frequency is just greater than or equal to [tex]\( \frac{N}{2} \)[/tex]):
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Class Interval} & f & cf \\
\hline
50-60 & 3 & 3 \\
60-70 & 7 & 10 \\
70-80 & 11 & 21 \\
80-90 & 16 & 37 \quad \leftarrow \text{(Median Class, as 37 > 25)} \\
90-100 & 8 & 45 \\
100-110 & 5 & 50 \\
\hline
\end{array}
\][/tex]
### Step 3: Extracting Necessary Values for the Median Calculation
1. The lower boundary (L) of the median class:
[tex]\[
L = 80
\][/tex]
2. The frequency (f) of the median class:
[tex]\[
f = 16
\][/tex]
3. The cumulative frequency (cf) of the class before the median class:
[tex]\[
cf = 21
\][/tex]
4. The class width (h):
[tex]\[
h = 90 - 80 = 10
\][/tex]
### Step 4: Calculate the Median
Using the formula for the median in grouped data:
[tex]\[
\text{Median} = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h
\][/tex]
Substitute the values:
[tex]\[
\text{Median} = 80 + \left(\frac{25 - 21}{16}\right) \times 10
\][/tex]
[tex]\[
\text{Median} = 80 + \left(\frac{4}{16}\right) \times 10
\][/tex]
[tex]\[
\text{Median} = 80 + \frac{1}{4} \times 10
\][/tex]
[tex]\[
\text{Median} = 80 + 2.5
\][/tex]
[tex]\[
\text{Median} = 82.5
\][/tex]
Therefore, the median of the given distribution is [tex]\(82.5\)[/tex].
