Frequency Distribution Table: | Class Interval | Frequency | Cumulative Frequency (Less than type) | |----------------|-----------|----------------------------------------| | 0–10 | 10 | 10 | | 10–20 | \(x\) | \(10 + x\) | | 20–30 | 25 | \(35 + x\) | | 30–40 | 30 | \(65 + x\) | | 40–50 | \(y\) | \(65 + x + y\) | | 50–60 | 10 | \(75 + x + y = n\) | Given: Total frequency \(n = 100\) So, \[ 75 + x + y = 100 \Rightarrow x + y = 25 \quad \text{—— (i)} \] Also, since the **median is 32**, the **median class** is (30–40) Using the median formula: \[ \text{Median} = l + \left[\frac{\frac{n}{2} - cf}{f}\right] \times h \] Where: - \(l = 30\) (lower boundary of median class) - \(n = 100\) - \(cf = 35 + x\) (cumulative frequency before median class) - \(f = 30\) (frequency of median class) - \(h = 10\) (class width) Substitute into the formula: \[ = 30 + \left[\frac{50 - (35 + x)}{30}\right] \times 10 = 30 + \frac{15 - x}{3} \] Given median = 32: \[ 30 + \frac{15 - x}{3} = 32 \Rightarrow \frac{15 - x}{3} = 2 \Rightarrow 15 - x = 6 \Rightarrow x = 9 \] Now substitute \(x = 9\) into equation (i): \[ 9 + y = 25 \Rightarrow y = 16 \] Therefore, the required values are: \(x = 9\), \(y = 16\).