If the median of the distribution given below is 28.5, find the values of x and y. Class interval Frequency 0 - 10 5 10 - 20 x 20 - 30 20 30 - 40 15 40 - 50 y 50 - 60 5 Total 60
Step-by-Step Solution
Key Concept: For grouped data, the median is found using the formula \(M = L + \frac{\frac{N}{2} - C_f}{f}\,h\), where \(L\) is the lower limit of the median class, \(h\) is the class width, \(f\) is the frequency of the median class and \(C_f\) is the cumulative frequency before the median class. The total frequency gives a relation between the unknown frequencies.
1. Total frequency
\[5 + x + 20 + 15 + y + 5 = 60 \]
\[45 + x + y = 60 \]
\[x + y = 15 \] (1)
2. Position of the median
\[N = 60 \Rightarrow \frac{N}{2}=30\]
The median is the 30th observation.
3. Cumulative frequencies
\[\begin{aligned}
\text{0-10}: &\; C_f = 5\\
\text{10-20}: &\; C_f = 5 + x\\
\text{20-30}: &\; C_f = 5 + x + 20 = 25 + x\\
\text{30-40}: &\; C_f = 40 + x\
\end{aligned}\]
Since the 30th observation must lie in the class where the cumulative frequency just exceeds 30, we need \(25 + x \ge 30\) i.e. \(x \ge 5\). Hence the median class is 20‑30.
4. Apply the median formula
\[L = 20,\; h = 10,\; f = 20,\; C_f = 5 + x\]
\[\begin{aligned}
28.5 &= 20 + \frac{30 - (5 + x)}{20}\times 10\\
28.5 - 20 &= \frac{25 - x}{20}\times 10\\
8.5 &= \frac{25 - x}{2}\\
25 - x &= 17\\
x &= 8\
\end{aligned}\]
5. Find y using (1)
\[y = 15 - x = 15 - 8 = 7\]
6. Verification
Cumulative frequency before the median class = 5 + 8 = 13 < 30,
Cumulative frequency after the median class = 13 + 20 = 33 ≥ 30, confirming that the median indeed lies in the 20‑30 class.
Correct Answer: x = 8, \; y = 7