1. **Problem Statement:** Given a grouped frequency distribution table, find the minimum, mode, median, variance, standard deviation, 30th percentile (P30), 80th percentile (PP80), Z-score, and T-score. 2. **Given Data:** Classes and frequencies: - 20-29: 3 - 30-39: 6 - 40-49: 11 - 50-59: 14 - 60-69: 9 - 70-79: 5 - 80-89: 2 3. **Step 1: Minimum** The minimum value is the lower boundary of the first class: $\boxed{20}$. 4. **Step 2: Mode** Mode class is the class with highest frequency: 50-59 with frequency 14. Mode formula for grouped data: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ Where: - $L=50$ (lower class boundary of modal class) - $f_1=14$ (frequency of modal class) - $f_0=11$ (frequency before modal class) - $f_2=9$ (frequency after modal class) - $h=10$ (class width) Calculate: $$\text{Mode} = 50 + \frac{(14 - 11)}{(2 \times 14 - 11 - 9)} \times 10 = 50 + \frac{3}{(28 - 20)} \times 10 = 50 + \frac{3}{8} \times 10 = 50 + 3.75 = 53.75$$ 5. **Step 3: Median** Total frequency $N = 3+6+11+14+9+5+2=50$. Median class is where cumulative frequency $rac{N}{2} = 25$ lies. Cumulative frequencies: 3, 9, 20, 34, ... so median class is 50-59. Median formula: $$\text{Median} = L + \frac{\frac{N}{2} - F}{f} \times h$$ Where: - $L=50$ (lower boundary of median class) - $F=20$ (cumulative frequency before median class) - $f=14$ (frequency of median class) - $h=10$ Calculate: $$\text{Median} = 50 + \frac{25 - 20}{14} \times 10 = 50 + \frac{5}{14} \times 10 = 50 + 3.57 = 53.57$$ 6. **Step 4: Variance and Standard Deviation** Mean $\bar{x} = \frac{\sum kX}{N} = \frac{73.5 + 207 + 489.5 + 763 + 580.5 + 372.5 + 169}{50} = \frac{2655}{50} = 53.1$ Variance formula: $$s^2 = \frac{\sum kX^2}{N} - \bar{x}^2$$ Sum $kX^2 = 1800.75 + 7141.5 + 21782.75 + 41583.5 + 37442.25 + 27751.25 + 14280.5 = 151782.5$ Calculate variance: $$s^2 = \frac{151782.5}{50} - (53.1)^2 = 3035.65 - 2819.61 = 216.04$$ Standard deviation: $$s = \sqrt{216.04} = 14.7$$ 7. **Step 5: Percentiles P30 and PP80** - P30: Position $= 0.30 \times 50 = 15$. Cumulative frequencies: 3, 9, 20,... so P30 lies in 40-49 class. $$P_{30} = L + \frac{(15 - F)}{f} \times h = 40 + \frac{(15 - 9)}{11} \times 10 = 40 + \frac{6}{11} \times 10 = 40 + 5.45 = 45.45$$ - PP80: Position $= 0.80 \times 50 = 40$. Cumulative frequencies: 3, 9, 20, 34, 43,... so PP80 lies in 60-69 class. $$PP_{80} = 60 + \frac{(40 - 34)}{9} \times 10 = 60 + \frac{6}{9} \times 10 = 60 + 6.67 = 66.67$$ 8. **Step 6: Z-score and T-score** For a value $X=60$ (example): $$Z = \frac{X - \bar{x}}{s} = \frac{60 - 53.1}{14.7} = \frac{6.9}{14.7} = 0.47$$ T-score formula: $$T = 50 + 10Z = 50 + 10 \times 0.47 = 54.7$$ **Final answers:** - Minimum = 20 - Mode = 53.75 - Median = 53.57 - Variance = 216.04 - Standard Deviation = 14.7 - P30 = 45.45 - PP80 = 66.67 - Z-score (for X=60) = 0.47 - T-score (for X=60) = 54.7