1. **Stating the problem:** Find the first quartile $Q_1$ from a grouped frequency distribution with a total number of observations $N=50$. 2. **Given data:** Class intervals, Class frequencies $f$, and Cumulative frequencies $F$ are approximately: - $9-11$: $f=1$, $F=1$ - $12-14$: $f=2$, $F=3$ - $15-17$: $f=0$, $F=3$ - $18-20$: $f=3$, $F=6$ - $21-23$: $f=3$, $F=9$ - $24-26$: $f=8$, $F=17$ - $27-29$: $f=13$, $F=30$ - ... rest don't affect $Q_1$ calculation since $Q_1$ is below median 3. **Calculate the position of $Q_1$:** $$Q_1 = \frac{1}{4} N = \frac{1}{4} imes 50 = 12.5$$ This means the first quartile is at the 12.5-th observation. 4. **Locate the class containing the 12.5-th observation:** Check cumulative frequencies: - Up to $12-14$: $F=3$ (less than 12.5) - Up to $15-17$: $F=3$ (still less than 12.5) - Up to $18-20$: $F=6$ (less than 12.5) - Up to $21-23$: $F=9$ (less than 12.5) - Up to $24-26$: $F=17$ (greater than 12.5) So, $Q_1$ lies in the class interval $24-26$. 5. **Apply the formula for quartile in grouped data:** $$Q_1 = L + \left(\frac{\frac{N}{4} - F_{prev}}{f_{class}}\right) \times h$$ Where: - $L$ = lower class boundary of quartile class = 23.5 (because class is 24-26, lower boundary is 24 - 0.5) - $F_{prev}$ = cumulative frequency before quartile class = 9 - $f_{class}$ = class frequency for quartile class = 8 - $h$ = class width = 3 6. **Calculate $Q_1$ value:** $$Q_1 = 23.5 + \left(\frac{12.5 - 9}{8}\right) \times 3 = 23.5 + \left(\frac{3.5}{8}\right) \times 3 = 23.5 + 1.3125 = 24.8125$$ **Answer:** $$Q_1 = 24.81$$ (rounded to two decimals)