1. **State the problem:** Find the interquartile range (IQR) from the given frequency distribution with intervals and frequencies: | Interval | Frequency (f) | |----------|--------------| | 5 - 9 | 4 | | 10 - 14 | 8 | | 15 - 19 | 3 | | 20 - 24 | 2 | Total frequency $n = 4 + 8 + 3 + 2 = 17$. 2. **Formula and explanation:** The interquartile range is defined as: $$\text{IQR} = Q_3 - Q_1$$ where $Q_1$ is the first quartile (25th percentile) and $Q_3$ is the third quartile (75th percentile). To find $Q_1$ and $Q_3$ in grouped data, use the formula: $$Q_k = L + \left(\frac{\frac{k}{4}n - F}{f_m}\right) \times w$$ where: - $L$ = lower boundary of the quartile class - $n$ = total frequency - $F$ = cumulative frequency before the quartile class - $f_m$ = frequency of the quartile class - $w$ = class width - $k$ = 1 for $Q_1$, 3 for $Q_3$ 3. **Calculate cumulative frequencies:** | Interval | f | Cumulative Frequency (CF) | |----------|---|---------------------------| | 5 - 9 | 4 | 4 | | 10 - 14 | 8 | 12 | | 15 - 19 | 3 | 15 | | 20 - 24 | 2 | 17 | 4. **Find $Q_1$ position:** $$\frac{1}{4}n = \frac{1}{4} \times 17 = 4.25$$ The 4.25th value lies in the first class (5-9) because CF up to first class is 4, and next class CF is 12. - $L = 5$ (lower boundary of 5-9) - $F = 0$ (no cumulative frequency before first class) - $f_m = 4$ - $w = 5$ (class width: 9 - 5 + 1 = 5) Calculate $Q_1$: $$Q_1 = 5 + \left(\frac{4.25 - 0}{4}\right) \times 5 = 5 + (1.0625) \times 5 = 5 + 5.3125 = 10.3125$$ 5. **Find $Q_3$ position:** $$\frac{3}{4}n = \frac{3}{4} \times 17 = 12.75$$ The 12.75th value lies in the second class (10-14) because CF up to second class is 12, and next class CF is 15. - $L = 10$ - $F = 4$ (CF before second class) - $f_m = 8$ - $w = 5$ Calculate $Q_3$: $$Q_3 = 10 + \left(\frac{12.75 - 4}{8}\right) \times 5 = 10 + (1.09375) \times 5 = 10 + 5.46875 = 15.46875$$ 6. **Calculate IQR:** $$\text{IQR} = Q_3 - Q_1 = 15.46875 - 10.3125 = 5.15625$$ **Final answer:** The interquartile range is approximately $5.16$.