1. **Problem Statement:** Organize the given marks of 40 candidates into a grouped frequency distribution with class intervals 0-9, 10-19, etc. Then, from this distribution, calculate the standard deviation and the 90th percentile. 2. **Step 1: Group the data into class intervals** Class intervals: 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99 Count the frequency of marks in each interval: - 0-9: 2 (8,5) - 10-19: 2 (11,19) - 20-29: 4 (26,28,24) - 30-39: 7 (34,35,38,32,39,33,32) - 40-49: 11 (40,40,44,46,49,43,47,46,45,42,44,45,46) - 50-59: 6 (55,54,55,57,59) - 60-69: 2 (67,64) - 70-79: 3 (77,76,74) - 80-89: 1 (86) - 90-99: 1 (97) 3. **Step 2: Calculate midpoints ($x_i$) for each class:** - 0-9: 4.5 - 10-19: 14.5 - 20-29: 24.5 - 30-39: 34.5 - 40-49: 44.5 - 50-59: 54.5 - 60-69: 64.5 - 70-79: 74.5 - 80-89: 84.5 - 90-99: 94.5 4. **Step 3: Calculate mean ($\bar{x}$):** $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate $\sum f_i x_i$: $$2\times4.5 + 2\times14.5 + 4\times24.5 + 7\times34.5 + 11\times44.5 + 6\times54.5 + 2\times64.5 + 3\times74.5 + 1\times84.5 + 1\times94.5 = 9 + 29 + 98 + 241.5 + 489.5 + 327 + 129 + 223.5 + 84.5 + 94.5 = 1725.5$$ Total frequency $\sum f_i = 40$ $$\bar{x} = \frac{1725.5}{40} = 43.1375$$ 5. **Step 4: Calculate variance ($\sigma^2$) and standard deviation ($\sigma$):** Formula: $$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$$ Calculate each term $f_i (x_i - \bar{x})^2$: - For 0-9: $2(4.5 - 43.1375)^2 = 2 \times 1493.5 = 2987$ - 10-19: $2(14.5 - 43.1375)^2 = 2 \times 812.9 = 1625.8$ - 20-29: $4(24.5 - 43.1375)^2 = 4 \times 345.3 = 1381.2$ - 30-39: $7(34.5 - 43.1375)^2 = 7 \times 74.7 = 522.9$ - 40-49: $11(44.5 - 43.1375)^2 = 11 \times 1.86 = 20.46$ - 50-59: $6(54.5 - 43.1375)^2 = 6 \times 129.3 = 775.8$ - 60-69: $2(64.5 - 43.1375)^2 = 2 \times 458.7 = 917.4$ - 70-79: $3(74.5 - 43.1375)^2 = 3 \times 992.3 = 2976.9$ - 80-89: $1(84.5 - 43.1375)^2 = 1 \times 1711.3 = 1711.3$ - 90-99: $1(94.5 - 43.1375)^2 = 1 \times 2630.3 = 2630.3$ Sum: $$2987 + 1625.8 + 1381.2 + 522.9 + 20.46 + 775.8 + 917.4 + 2976.9 + 1711.3 + 2630.3 = 14548.16$$ Variance: $$\sigma^2 = \frac{14548.16}{40} = 363.7$$ Standard deviation: $$\sigma = \sqrt{363.7} \approx 19.07$$ 6. **Step 5: Calculate the 90th percentile** Position of 90th percentile: $$P_{90} = 0.9 \times 40 = 36^{th} \text{ value}$$ Cumulative frequencies: - 0-9: 2 - 10-19: 4 - 20-29: 8 - 30-39: 15 - 40-49: 26 - 50-59: 32 - 60-69: 34 - 70-79: 37 - 80-89: 38 - 90-99: 39 The 36th value lies in the 70-79 class interval. Use formula for percentile: $$P_k = L + \left(\frac{kN - F}{f}\right) \times c$$ Where: - $L = 69.5$ (lower boundary of 70-79) - $kN = 36$ - $F = 34$ (cumulative frequency before class) - $f = 3$ (frequency of class) - $c = 10$ (class width) Calculate: $$P_{90} = 69.5 + \left(\frac{36 - 34}{3}\right) \times 10 = 69.5 + \frac{2}{3} \times 10 = 69.5 + 6.67 = 76.17$$ **Final answers:** - Standard deviation $\approx 19.07$ - 90th percentile $\approx 76.17$