To determine the value of [tex]\( x \)[/tex] given the median of the data, we need to follow a step-by-step algorithm, using the concept of median in a frequency distribution.
### Step-by-Step Solution
Step 1: Organize the given data
Given data:
- Marks obtained: [tex]\( 0-5, \ 5-10, \ 10-15, \ 15-20, \ 20-25 \)[/tex]
- Number of students: [tex]\( 14, \ 25, \ 27, \ x, \ 15 \)[/tex]
- Median (M) = 15
Step 2: Calculate the cumulative frequency
First, calculate the cumulative frequency (CF) for each class interval.
[tex]\[
\begin{array}{c|c|c}
\text{Class Interval} & \text{Frequency (f)} & \text{Cumulative Frequency (CF)} \\
\hline
0-5 & 14 & 14 \\
5-10 & 25 & 14 + 25 = 39 \\
10-15 & 27 & 39 + 27 = 66 \\
15-20 & x & 66 + x \\
20-25 & 15 & 66 + x + 15 = 81 + x \\
\end{array}
\][/tex]
Step 3: Determine the median class
The median class is the interval where the cumulative frequency exceeds half the total number of observations. To find this, we need the total number of students:
[tex]\[
N = 14 + 25 + 27 + x + 15 = 81 + x
\][/tex]
Since the median is given as 15, the median class must be [tex]\( 10-15 \leq M < 20-25 \)[/tex].
[tex]\( M \)[/tex] falls in the [tex]\( 15-20 \)[/tex] class interval.
Step 4: Apply the median formula
The formula for the median in a frequency distribution is:
[tex]\[
M = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h
\][/tex]
where:
- [tex]\( M \)[/tex] is the median.
- [tex]\( L \)[/tex] is the lower boundary of the median class.
- [tex]\( N \)[/tex] is the total number of students.
- [tex]\( CF \)[/tex] is the cumulative frequency of the class preceding the median class.
- [tex]\( f \)[/tex] is the frequency of the median class.
- [tex]\( h \)[/tex] is the class width.
Given:
- [tex]\( M = 15 \)[/tex]
- [tex]\( L = 15 \)[/tex]
- [tex]\( CF_{\text{before median class}} = 66 \)[/tex]
- [tex]\( f_{\text{median class}} = x \)[/tex]
- [tex]\( h = 5 \)[/tex]
We set up the equation:
[tex]\[
15 = 15 + \left( \frac{\frac{81 + x}{2} - 66}{x} \right) \times 5
\][/tex]
Subtracting 15 from both sides:
[tex]\[
0 = \left( \frac{\frac{81 + x}{2} - 66}{x} \right) \times 5
\][/tex]
Simplifying inside the parentheses:
[tex]\[
0 = \left( \frac{40.5 + 0.5x - 66}{x} \right) \times 5
\][/tex]
[tex]\[
0 = \left( \frac{0.5x - 25.5}{x} \right) \times 5
\][/tex]
Further simplifying:
[tex]\[
0 = \left( \frac{0.5x}{x} - \frac{25.5}{x} \right) \times 5
\][/tex]
[tex]\[
0 = \left(0.5 - \frac{25.5}{x}\right) \times 5
\][/tex]
[tex]\[
0 = 2.5 - \frac{127.5}{x}
\][/tex]
Equating:
[tex]\[
2.5 = \frac{127.5}{x}
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{127.5}{2.5} = 51
\][/tex]
### Conclusion
The number of students, [tex]\( x \)[/tex], in the [tex]\( 15-20 \)[/tex] class interval is 51.
