1. **Problem:** Find which batsman, A or B, is more consistent in scoring based on their scores in ten innings. 2. **Step 1:** Calculate the mean (average) score for each batsman. For batsman A: Scores = 32, 28, 47, 63, 71, 39, 10, 60, 96, 14 Mean $\bar{x}_A = \frac{32 + 28 + 47 + 63 + 71 + 39 + 10 + 60 + 96 + 14}{10} = \frac{460}{10} = 46$. For batsman B: Scores = 19, 31, 48, 53, 67, 90, 10, 62, 40, 80 Mean $\bar{x}_B = \frac{19 + 31 + 48 + 53 + 67 + 90 + 10 + 62 + 40 + 80}{10} = \frac{500}{10} = 50$. 3. **Step 2:** Calculate the standard deviation (S.D.) for each batsman to measure consistency. Standard deviation formula: $$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2}$$ Calculate for A: $\sum (x_i - \bar{x}_A)^2 = (32-46)^2 + (28-46)^2 + (47-46)^2 + (63-46)^2 + (71-46)^2 + (39-46)^2 + (10-46)^2 + (60-46)^2 + (96-46)^2 + (14-46)^2$ $= 196 + 324 + 1 + 289 + 625 + 49 + 1296 + 196 + 2500 + 1024 = 7499$ $\sigma_A = \sqrt{\frac{7499}{10}} = \sqrt{749.9} \approx 27.38$ Calculate for B: $\sum (x_i - \bar{x}_B)^2 = (19-50)^2 + (31-50)^2 + (48-50)^2 + (53-50)^2 + (67-50)^2 + (90-50)^2 + (10-50)^2 + (62-50)^2 + (40-50)^2 + (80-50)^2$ $= 961 + 361 + 4 + 9 + 289 + 1600 + 1600 + 144 + 100 + 900 = 5968$ $\sigma_B = \sqrt{\frac{5968}{10}} = \sqrt{596.8} \approx 24.43$ 4. **Step 3:** Compare the standard deviations. Batsman B has a lower standard deviation ($24.43$) compared to A ($27.38$), indicating B's scores are less spread out and more consistent. 5. **Answer:** Batsman B is more consistent in scoring because B has a lower standard deviation. --- 1. **Problem:** Calculate mean and standard deviation of the frequency distribution: | Class Interval | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | | Frequency | 14 | 23 | 27 | 21 | 15 | 2. **Step 1:** Find midpoints ($x_i$) of each class interval: $5, 15, 25, 35, 45$ 3. **Step 2:** Calculate mean: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate total frequency: $N = 14 + 23 + 27 + 21 + 15 = 100$ Calculate $\sum f_i x_i$: $14 \times 5 + 23 \times 15 + 27 \times 25 + 21 \times 35 + 15 \times 45 = 70 + 345 + 675 + 735 + 675 = 2500$ Mean: $$\bar{x} = \frac{2500}{100} = 25$$ 4. **Step 3:** Calculate variance and standard deviation: Variance formula: $$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$$ Calculate each squared deviation: $(5-25)^2 = 400$, $(15-25)^2 = 100$, $(25-25)^2 = 0$, $(35-25)^2 = 100$, $(45-25)^2 = 400$ Calculate $\sum f_i (x_i - \bar{x})^2$: $14 \times 400 + 23 \times 100 + 27 \times 0 + 21 \times 100 + 15 \times 400 = 5600 + 2300 + 0 + 2100 + 6000 = 16000$ Variance: $$\sigma^2 = \frac{16000}{100} = 160$$ Standard deviation: $$\sigma = \sqrt{160} = 12.65$$ 5. **Answer:** Mean = 25, Standard Deviation $\approx 12.65$. --- 1. **Problem:** Find variance and coefficient of variation for the distribution: | Class Interval | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | | Frequency | 4 | 6 | 10 | 18 | 9 | 3 | 2. **Step 1:** Find midpoints ($x_i$): $15, 25, 35, 45, 55, 65$ 3. **Step 2:** Calculate mean: Total frequency: $N = 4 + 6 + 10 + 18 + 9 + 3 = 50$ Calculate $\sum f_i x_i$: $4 \times 15 + 6 \times 25 + 10 \times 35 + 18 \times 45 + 9 \times 55 + 3 \times 65 = 60 + 150 + 350 + 810 + 495 + 195 = 2060$ Mean: $$\bar{x} = \frac{2060}{50} = 41.2$$ 4. **Step 3:** Calculate variance: Calculate squared deviations: $(15 - 41.2)^2 = 26.2^2 = 686.44$ $(25 - 41.2)^2 = 16.2^2 = 262.44$ $(35 - 41.2)^2 = 6.2^2 = 38.44$ $(45 - 41.2)^2 = 3.8^2 = 14.44$ $(55 - 41.2)^2 = 13.8^2 = 190.44$ $(65 - 41.2)^2 = 23.8^2 = 566.44$ Calculate $\sum f_i (x_i - \bar{x})^2$: $4 \times 686.44 + 6 \times 262.44 + 10 \times 38.44 + 18 \times 14.44 + 9 \times 190.44 + 3 \times 566.44$ $= 2745.76 + 1574.64 + 384.4 + 259.92 + 1713.96 + 1699.32 = 8377.99$ Variance: $$\sigma^2 = \frac{8377.99}{50} = 167.56$$ Standard deviation: $$\sigma = \sqrt{167.56} \approx 12.95$$ 5. **Step 4:** Calculate coefficient of variation (C.V.): $$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 = \frac{12.95}{41.2} \times 100 \approx 31.44\%$$ 6. **Answer:** Variance $\approx 167.56$, Coefficient of Variation $\approx 31.44\%$.