Q.If the median of the following data is 32, find the values of \(x\) and \(y\) when the total frequency is 100:
Class interval0–1010–2020–3030–4040–5050–60
Frequency10\(x\)2530\(y\)10

Frequency distribution table:
Class IntervalFrequencyCumulative Frequency (less than type)
0–101010
10–20\(x\)10 + \(x\)
20–302535 + \(x\)
30–403065 + \(x\)
40–50\(y\)65 + \(x + y\)
50–601075 + \(x + y = n\)
Given: \(n = 100\) From the condition: \(75 + x + y = 100\) ⇒ \(x + y = 25\) ----(i) Again, since the **median = 32**, the median class is (30–40) To find median: \[ \text{Median} = l + \left[\frac{n}{2} - cf}{f} \right] \times h \] Where, \(l = 30\), \(n = 100\), \(cf = 35 + x\), \(f = 30\), \(h = 10\) So, \[ \text{Median} = 30 + \left[\frac{50 - (35 + x)}{30} \right] \times 10 = 30 + \frac{15 - x}{3} \] Given median = 32 So, \[ 30 + \frac{15 - x}{3} = 32 \Rightarrow \frac{15 - x}{3} = 2 \Rightarrow 15 - x = 6 \Rightarrow x = 9 \] Substitute into (i): \(9 + y = 25\) ⇒ \(y = 16\) ∴ Required values are \(x = 9\), \(y = 16\) (Answer)
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