1. **Problem Statement:** Organize the marks of 40 candidates into grouped frequency distribution with class intervals 0-9, 10-19, ..., 90-99. Calculate the standard deviation and the 90th percentile from this grouped data. 2. **Step 1: Group the data into class intervals and count frequencies** Class intervals and frequencies ($f_i$): - 0-9: 2 - 10-19: 2 - 20-29: 3 - 30-39: 7 - 40-49: 11 - 50-59: 6 - 60-69: 2 - 70-79: 3 - 80-89: 1 - 90-99: 3 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 $\sum f_i x_i$: $$2\times4.5 + 2\times14.5 + 3\times24.5 + 7\times34.5 + 11\times44.5 + 6\times54.5 + 2\times64.5 + 3\times74.5 + 1\times84.5 + 3\times94.5$$ $$= 9 + 29 + 73.5 + 241.5 + 489.5 + 327 + 129 + 223.5 + 84.5 + 283.5 = 1889$$ Total frequency $\sum f_i = 40$ $$\bar{x} = \frac{1889}{40} = 47.225$$ 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 $f_i (x_i - \bar{x})^2$: - 0-9: $2(4.5 - 47.225)^2 = 2 \times 1817.5 = 3635$ - 10-19: $2(14.5 - 47.225)^2 = 2 \times 1076.7 = 2153.4$ - 20-29: $3(24.5 - 47.225)^2 = 3 \times 515.3 = 1545.9$ - 30-39: $7(34.5 - 47.225)^2 = 7 \times 161.9 = 1133.3$ - 40-49: $11(44.5 - 47.225)^2 = 11 \times 7.4 = 81.4$ - 50-59: $6(54.5 - 47.225)^2 = 6 \times 52.7 = 316.2$ - 60-69: $2(64.5 - 47.225)^2 = 2 \times 299.3 = 598.6$ - 70-79: $3(74.5 - 47.225)^2 = 3 \times 743.5 = 2230.5$ - 80-89: $1(84.5 - 47.225)^2 = 1 \times 1394.3 = 1394.3$ - 90-99: $3(94.5 - 47.225)^2 = 3 \times 2230.3 = 6690.9$ Sum: $$3635 + 2153.4 + 1545.9 + 1133.3 + 81.4 + 316.2 + 598.6 + 2230.5 + 1394.3 + 6690.9 = 19379.5$$ Variance: $$\sigma^2 = \frac{19379.5}{40} = 484.49$$ Standard deviation: $$\sigma = \sqrt{484.49} \approx 22.01$$ 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: 7 - 30-39: 14 - 40-49: 25 - 50-59: 31 - 60-69: 33 - 70-79: 36 - 80-89: 37 - 90-99: 40 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) - $kN = 36$ - $F = 33$ (cumulative frequency before class) - $f = 3$ (frequency of class) - $c = 10$ (class width) Calculate: $$P_{90} = 69.5 + \left(\frac{36 - 33}{3}\right) \times 10 = 69.5 + 1 \times 10 = 79.5$$ **Final answers:** - Standard deviation $\approx 22.01$ - 90th percentile $\approx 79.5$