User should create a table with columns: Hours, Number of Students (fᵢ), Class Mark (xᵢ), and fᵢxᵢ. For class 0-5: Class mark xᵢ = (0+5)/2 = 2.5, fᵢ = 5, fᵢxᵢ = 5 × 2.5 = 12.5. For class 5-10: Class mark xᵢ = (5+10)/2 = 7.5, fᵢ = 12, fᵢxᵢ = 12 × 7.5 = 90. For class 10-15: Class mark xᵢ = (10+15)/2 = 12.5, fᵢ = 15, fᵢxᵢ = 15 × 12.5 = 187.5. For class 15-20: Class mark xᵢ = (15+20)/2 = 17.5, fᵢ = 6, fᵢxᵢ = 6 × 17.5 = 105. For class 20-25: Class mark xᵢ = (20+25)/2 = 22.5, fᵢ = 2, fᵢxᵢ = 2 × 22.5 = 45. User should calculate Σfᵢ = 5 + 12 + 15 + 6 + 2 = 40. User should calculate Σfᵢxᵢ = 12.5 + 90 + 187.5 + 105 + 45 = 440. User should apply the formula: Mean (x̄) = Σfᵢxᵢ / Σfᵢ = 440 / 40 = 11. Final answer: Mean = 11 hours.User should identify the modal class as the class with the highest frequency. The highest frequency is 18, corresponding to the class 145-150. Modal class = 145-150. User should identify: l = 145 (lower limit of modal class). User should identify: f₁ = 18 (frequency of modal class). User should identify: f₀ = 8 (frequency of class preceding modal class, which is 140-145). User should identify: f₂ = 12 (frequency of class succeeding modal class, which is 150-155). User should identify: h = 5 (class width). User should apply the formula: Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h. Mode = 145 + [(18 - 8)/(2×18 - 8 - 12)] × 5. Mode = 145 + [10/(36 - 20)] × 5 = 145 + (10/16) × 5 = 145 + 3.125 = 148.125. Mode = 148.13 cm.User should create a cumulative frequency table. For class 50-100: frequency = 10, cumulative frequency (cf) = 10. For class 100-150: frequency = 15, cumulative frequency = 10 + 15 = 25. For class 150-200: frequency = 18, cumulative frequency = 25 + 18 = 43. For class 200-250: frequency = 12, cumulative frequency = 43 + 12 = 55. For class 250-300: frequency = 5, cumulative frequency = 55 + 5 = 60. User should calculate n = 60, so n/2 = 30. User should identify the median class as the class where cumulative frequency is just greater than n/2 = 30. The median class is 150-200 (cf = 43, which is greater than 30). User should identify: l = 150 (lower limit of median class). User should identify: n = 60 (total frequency). User should identify: cf = 25 (cumulative frequency of class preceding median class). User should identify: f = 18 (frequency of median class). User should identify: h = 50 (class width). User should apply the formula: Median = l + [(n/2 - cf)/f] × h. Median = 150 + [(30 - 25)/18] × 50 = 150 + (5/18) × 50 = 150 + 13.89 = 163.89. Median pocket money = ₹163.89.User should create a table with columns: Class, fᵢ, xᵢ, dᵢ = xᵢ - a, and fᵢdᵢ. Given assumed mean a = 35. For class 10-20: xᵢ = 15, fᵢ = 4, dᵢ = 15 - 35 = -20, fᵢdᵢ = 4 × (-20) = -80. For class 20-30: xᵢ = 25, fᵢ = 8, dᵢ = 25 - 35 = -10, fᵢdᵢ = 8 × (-10) = -80. For class 30-40: xᵢ = 35, fᵢ = 14, dᵢ = 35 - 35 = 0, fᵢdᵢ = 14 × 0 = 0. For class 40-50: xᵢ = 45, fᵢ = 10, dᵢ = 45 - 35 = 10, fᵢdᵢ = 10 × 10 = 100. For class 50-60: xᵢ = 55, fᵢ = 4, dᵢ = 55 - 35 = 20, fᵢdᵢ = 4 × 20 = 80. User should calculate Σfᵢ = 4 + 8 + 14 + 10 + 4 = 40. User should calculate Σfᵢdᵢ = -80 + (-80) + 0 + 100 + 80 = 20. User should apply the formula: Mean (x̄) = a + (Σfᵢdᵢ / Σfᵢ) = 35 + (20/40) = 35 + 0.5 = 35.5. Final answer: Mean = 35.5.User should take assumed mean a = 325 (middle class mark) and h = 50(Since dᵢ is 50). User should create a table with columns: Wages, fᵢ, xᵢ, dᵢ = xᵢ - a, uᵢ = dᵢ/h, and fᵢuᵢ. For class 200-250: xᵢ = 225, fᵢ = 5, dᵢ = 225 - 325 = -100, uᵢ = -100/50 = -2, fᵢuᵢ = 5 × (-2) = -10. For class 250-300: xᵢ = 275, fᵢ = 10, dᵢ = 275 - 325 = -50, uᵢ = -50/50 = -1, fᵢuᵢ = 10 × (-1) = -10. For class 300-350: xᵢ = 325, fᵢ = 20, dᵢ = 325 - 325 = 0, uᵢ = 0/50 = 0, fᵢuᵢ = 20 × 0 = 0. For class 350-400: xᵢ = 375, fᵢ = 10, dᵢ = 375 - 325 = 50, uᵢ = 50/50 = 1, fᵢuᵢ = 10 × 1 = 10. For class 400-450: xᵢ = 425, fᵢ = 5, dᵢ = 425 - 325 = 100, uᵢ = 100/50 = 2, fᵢuᵢ = 5 × 2 = 10. User should calculate Σfᵢ = 5 + 10 + 20 + 10 + 5 = 50. User should calculate Σfᵢuᵢ = -10 + (-10) + 0 + 10 + 10 = 0. User should apply the formula: Mean (x̄) = a + h(Σfᵢuᵢ / Σfᵢ) = 325 + 50(0/50) = 325 + 0 = 325. Final answer: Mean wage = ₹325.Given: Total frequency = 230, Median = 46. User should write: 12 + 30 + x + 65 + y + 25 + 18 = 230. Simplifying: 150 + x + y = 230, so x + y = 80. ... (equation 1) User should create cumulative frequency table. For class 10-20: f = 12, cf = 12. For class 20-30: f = 30, cf = 12 + 30 = 42. For class 30-40: f = x, cf = 42 + x. For class 40-50: f = 65, cf = 42 + x + 65 = 107 + x. For class 50-60: f = y, cf = 107 + x + y. n/2 = 230/2 = 115. Since median is 46, the median class is 40-50. This means cf of class 30-40 must be less than 115, and cf of class 40-50 must be greater than or equal to 115. So 42 + x < 115 and 107 + x ≥ 115. From 107 + x ≥ 115, we get x ≥ 8. User should apply median formula: l = 40, cf = 42 + x, f = 65, h = 10. Median = 40 + [(115 - (42 + x))/65] × 10 = 46. Solving: [(115 - 42 - x)/65] × 10 = 6. [(73 - x)/65] × 10 = 6. (73 - x)/65 = 0.6. 73 - x = 39. x = 34. From equation 1: x + y = 80, so 34 + y = 80, therefore y = 46. Final answer: x = 34, y = 46.(a) Convert to Simple Frequency Distribution: User should convert cumulative frequency to simple frequency distribution. For class 0-10: Frequency = 3 - 0 = 3. For class 10-20: Frequency = 12 - 3 = 9. For class 20-30: Frequency = 27 - 12 = 15. For class 30-40: Frequency = 57 - 27 = 30. For class 40-50: Frequency = 75 - 57 = 18. For class 50-60: Frequency = 80 - 75 = 5. (b) Create "More than or equal to" Cumulative Frequency: For marks ≥ 0: All 80 students, cf = 80. For marks ≥ 10: 80 - 3 = 77 students, cf = 77. For marks ≥ 20: 77 - 9 = 68 students, cf = 68. For marks ≥ 30: 68 - 15 = 53 students, cf = 53. For marks ≥ 40: 53 - 30 = 23 students, cf = 23. For marks ≥ 50: 23 - 18 = 5 students, cf = 5. For marks ≥ 60: 5 - 5 = 0 students, cf = 0. User should present this in table format. (c) Finding Median: Using simple frequency distribution with "less than" cumulative frequencies: 3, 12, 27, 57, 75, 80. n = 80, n/2 = 40. Median class is 30-40 (cf = 57 is just greater than 40). l = 30, cf = 27, f = 30, h = 10. Median = 30 + [(40 - 27)/30] × 10 = 30 + (13/30) × 10 = 30 + 4.33 = 34.33. Final answer: (a) Simple frequency: 3, 9, 15, 30, 18, 5. (b) More than type cf: 80, 77, 68, 53, 23, 5, 0. (c) Median = 34.33 marks.User should create table with: Class, fᵢ, xᵢ, fᵢxᵢ. For class 0-20: xᵢ = 10, fᵢ = 17, fᵢxᵢ = 170. For class 20-40: xᵢ = 30, fᵢ = 28, fᵢxᵢ = 840. For class 40-60: xᵢ = 50, fᵢ = 32, fᵢxᵢ = 1600. For class 60-80: xᵢ = 70, fᵢ = p, fᵢxᵢ = 70p. For class 80-100: xᵢ = 90, fᵢ = 19, fᵢxᵢ = 1710. Σfᵢ = 17 + 28 + 32 + p + 19 = 96 + p. Σfᵢxᵢ = 170 + 840 + 1600 + 70p + 1710 = 4320 + 70p. Given Mean = 50. Using formula: Mean = Σfᵢxᵢ / Σfᵢ. 50 = (4320 + 70p) / (96 + p). 50(96 + p) = 4320 + 70p. 4800 + 50p = 4320 + 70p. 4800 - 4320 = 70p - 50p. 480 = 20p. p = 24. Final answer: p = 24.User should identify ERROR 1: The modal class is WRONG. The highest frequency is 22, not 15. Therefore, the modal class should be 40-50, not 30-40. User should identify ERROR 2: The values of f₀, f₁, and f₂ are therefore all wrong. User should identify ERROR 3: Division by zero (7/0) is mathematically undefined, which indicates wrong values were used. User should identify ERROR 4: The final answer 35.5 is impossible when division by zero occurred. User should note: The student incorrectly identified the modal class and then tried to force a calculation. Above errors can be listed in any order.Error numbers might vary accordingly. CORRECT SOLUTION: Modal class = 40-50 (frequency 22 is highest). l = 40 (lower limit of modal class). f₁ = 22 (frequency of modal class 40-50). f₀ = 15 (frequency of preceding class 30-40). f₂ = 12 (frequency of succeeding class 50-60). h = 10 (class width). Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h. Mode = 40 + [(22 - 15)/(2×22 - 15 - 12)] × 10. Mode = 40 + [7/(44 - 27)] × 10 = 40 + (7/17) × 10 = 40 + 4.12 = 44.12. Final answer: Errors - Wrong modal class identified (should be 40-50, not 30-40), wrong f values, division by zero. Correct mode = 44.12.(a) Finding Mean: User should create table with: Class, fᵢ, xᵢ, fᵢxᵢ. For 0-15: xᵢ = 7.5, fᵢ = 12, fᵢxᵢ = 90. For 15-30: xᵢ = 22.5, fᵢ = 28, fᵢxᵢ = 630. For 30-45: xᵢ = 37.5, fᵢ = 35, fᵢxᵢ = 1312.5. For 45-60: xᵢ = 52.5, fᵢ = 15, fᵢxᵢ = 787.5. For 60-75: xᵢ = 67.5, fᵢ = 7, fᵢxᵢ = 472.5. For 75-90: xᵢ = 82.5, fᵢ = 3, fᵢxᵢ = 247.5. Σfᵢ = 100, Σfᵢxᵢ = 3540. Mean = 3540/100 = 35.4 minutes. (b) Finding Median: User should create cumulative frequency table. cf: 12, 40, 75, 90, 97, 100. n/2 = 50, median class is 30-45 (cf = 75 is greater than 50). l = 30, cf = 40, f = 35, h = 15. Median = 30 + [(50 - 40)/35] × 15 = 30 + (10/35) × 15 = 30 + 4.29 = 34.29 minutes. (c) Claim Verification: The median is 34.29 minutes, which means exactly 50% of patients wait less than 34.29 minutes. Since the claim states "less than 35 minutes" and the median is 34.29, approximately 50% wait less than 35 minutes. The claim "most patients" typically implies more than 50% (a clear majority). Therefore, the claim is NOT fully justified, though it's very close (median 34.29 is just under 35). More accurately, about half the patients wait less than 35 minutes, not "most." Final answers: (a) Mean = 35.4 minutes, (b) Median = 34.29 minutes, (c) Claim not fully justified - median shows only ~50% wait less than 35 min, not "most" (>50%).(a) Finding Mean (with unequal class widths): User should note: Class widths are 1, 2, 3, 5, 8 (UNEQUAL). User should still use direct method with class marks. For class 1-2: xᵢ = 1.5, fᵢ = 45, fᵢxᵢ = 67.5. For class 2-4: xᵢ = 3, fᵢ = 52, fᵢxᵢ = 156. For class 4-7: xᵢ = 5.5, fᵢ = 30, fᵢxᵢ = 165. For class 7-12: xᵢ = 9.5, fᵢ = 15, fᵢxᵢ = 142.5. For class 12-20: xᵢ = 16, fᵢ = 8, fᵢxᵢ = 128. Σfᵢ = 150, Σfᵢxᵢ = 659. Mean = 659/150 = 4.39 days (approximately). (b) Finding Modal Class: Highest frequency is 52, so modal class is 2-4 days. INTERPRETATION: Most orders (52 out of 150, which is about 35%) are delivered within 2-4 days. This tells the store that their delivery performance is quite good and efficient. The majority of orders arrive quickly within 2-4 days, which indicates good logistics. This is positive information showing customer satisfaction with delivery speed. User may give 2 or 3 reasons for this. (c) Why Mean ≠ Mode: The mean (4.39 days) is higher than the modal class midpoint (3 days) because: There are some orders with much longer delivery times (7-12 days, 12-20 days). Even though these long deliveries are fewer in number, they pull the mean upward significantly. The distribution is positively skewed (right-skewed) with a long tail toward higher delivery times. The mode represents the most common experience, while mean includes effect of all deliveries including delayed ones.(a) Calculating Mode: Modal class = 0-10 (highest frequency is 25). l = 0, f₁ = 25, f₀ = 0 (no preceding class), f₂ = 18, h = 10. Since there's no preceding class, we use f₀ = 0. Mode = 0 + [(25 - 0)/(2×25 - 0 - 18)] × 10. Mode = 0 + [25/(50 - 18)] × 10 = 0 + (25/32) × 10 = 0 + 7.81 = 7.81 hours. This verifies that the modal class is indeed 0-10 hours with mode value of 7.81 hours. (b) Explanation of Discrepancy: The modal class (0-10 hours) and mode (7.81 hours) show that most households (25 out of 80) watch TV for less than 10 hours per week. However, the mean (21.5 hours) is much higher - almost triple the mode value. This discrepancy occurs because: Looking at the distribution, while 25 households watch 0-10 hours, there are significant numbers watching much more. Specifically: 12 households watch 30-40 hours/week and 10 households watch 40-50 hours/week. These 22 heavy TV watchers (watching 30-50 hours) significantly pull the mean upward. INTERPRETATION about TV habits: This tells us the town has two distinct viewing patterns: 1. A majority (25 households) are light/moderate viewers (0-10 hrs/week). 2. A significant minority (22 households) are very heavy viewers (30-50 hrs/week). The mean (21.5) is not representative of "typical" behavior because it's inflated by extreme values. The median would be a better measure here to understand typical viewing habits. This suggests potential concern about excessive TV watching among some households. Final answers: (a) Mode = 7.81 hours (verifying modal class 0-10), (b) Mean much higher due to heavy TV watchers (30-50 hrs) pulling average up significantly; indicates two distinct groups: moderate viewers (most common) and heavy viewers (pulling mean up); mean not representative of typical behavior.(a) Prediction WITHOUT Calculation: User should carefully observe and compare the frequency distributions. Distribution A: Highest frequency (20) is in class 20-30 (middle-upper class). Distribution A frequencies: 5, 10, 20, 10, 5 - concentrated in middle (20-30). Distribution B: Frequencies more spread toward lower classes. Distribution B frequencies: 10, 15, 10, 10, 5 - more weight in lower classes (0-20). REASONING: Distribution A has more observations concentrated in middle-upper range (class 20-30). Distribution B has more observations in lower range (classes 0-10 and 10-20 have 10+15=25). Since Distribution A's data is concentrated toward higher class values, it should have higher mean. PREDICTION: Distribution A will have a higher mean than Distribution B. (b) Calculating Both Means: Using class marks: 5, 15, 25, 35, 45 for classes 0-10, 10-20, 20-30, 30-40, 40-50. For Distribution A: fᵢxᵢ values: 5×5=25, 10×15=150, 20×25=500, 10×35=350, 5×45=225. Σfᵢxᵢ = 25 + 150 + 500 + 350 + 225 = 1250. Σfᵢ = 5 + 10 + 20 + 10 + 5 = 50. Mean A = 1250/50 = 25. For Distribution B: fᵢxᵢ values: 10×5=50, 15×15=225, 10×25=250, 10×35=350, 5×45=225. Σfᵢxᵢ = 50 + 225 + 250 + 350 + 225 = 1100. Σfᵢ = 10 + 15 + 10 + 10 + 5 = 50. Mean B = 1100/50 = 22. VERIFICATION: Mean A (25) > Mean B (22). Prediction confirmed! The difference is 3 units (25 - 22 = 3).(a) Finding Combined Mean: User should understand that combined mean requires total marks and total students. For Group A: Mean = 72, Number of students (n₁) = 40. Total marks of Group A = Mean × Number of students = 72 × 40 = 2880 marks. For Group B: Mean = 68, Number of students (n₂) = 60. Total marks of Group B = Mean × Number of students = 68 × 60 = 4080 marks. When groups are combined: Combined total marks = Total of Group A + Total of Group B = 2880 + 4080 = 6960 marks. Combined number of students = n₁ + n₂ = 40 + 60 = 100 students. Combined mean = Combined total marks / Combined number of students. Combined mean = 6960 / 100 = 69.6 marks. (b) Explanation of Why It's Different from Simple Average: Simple average of the two means = (72 + 68)/2 = 140/2 = 70. But we calculated combined mean = 69.6 (which is different from 70).User should convert cumulative frequency to simple frequency distribution. For class 0-10: Frequency = 4 - 0 = 4. For class 10-20: Frequency = 10 - 4 = 6. For class 20-30: Frequency = 22 - 10 = 12. For class 30-40: Frequency = 35 - 22 = 13. For class 40-50: Frequency = 50 - 35 = 15. User should identify the modal class as the class with the highest frequency. The highest frequency is 15, corresponding to the class 40-50. Modal class = 40-50. User should identify: l = 40 (lower limit of modal class). User should identify: f₁ = 15 (frequency of modal class). User should identify: f₀ = 13 (frequency of class preceding modal class, which is 30-40). User should identify: f₂ = 0 (frequency of class succeeding modal class - none exists, so 0). User should identify: h = 10 (class width). User should apply the formula: Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h. Mode = 40 + [(15 - 13)/(2×15 - 13 - 0)] × 10. Mode = 40 + [2/(30 - 13)] × 10 = 40 + (2/17) × 10 = 40 + 1.18 = 41.18. Final answer: Mode = 41.18 wickets.User should convert "more than" cumulative frequency to simple frequency distribution. For class 140-145: Frequency = 60 - 52 = 8. For class 145-150: Frequency = 52 - 38 = 14. For class 150-155: Frequency = 38 - 20 = 18. For class 155-160: Frequency = 20 - 8 = 12. For class 160-165: Frequency = 8 - 0 = 8. Total frequency = 8 + 14 + 18 + 12 + 8 = 60. User should create "less than" cumulative frequency table. For class 140-145: frequency = 8, cumulative frequency (cf) = 8. For class 145-150: frequency = 14, cumulative frequency = 8 + 14 = 22. For class 150-155: frequency = 18, cumulative frequency = 22 + 18 = 40. For class 155-160: frequency = 12, cumulative frequency = 40 + 12 = 52. For class 160-165: frequency = 8, cumulative frequency = 52 + 8 = 60. User should calculate n = 60, so n/2 = 30. User should identify the median class as the class where cumulative frequency is just greater than n/2 = 30. The median class is 150-155 (cf = 40, which is greater than 30). User should identify: l = 150 (lower limit of median class). User should identify: n = 60 (total frequency). User should identify: cf = 22 (cumulative frequency of class preceding median class). User should identify: f = 18 (frequency of median class). User should identify: h = 5 (class width). User should apply the formula: Median = l + [(n/2 - cf)/f] × h. Median = 150 + [(30 - 22)/18] × 5. Median = 150 + (8/18) × 5 = 150 + 2.22 = 152.22. Final answer: Median height = 152.22 cm.
