1. **Problem Statement:** Find the mean, median, and mode of the grouped data given by the class intervals and frequencies. 2. **Given Data:** Class Intervals and Frequencies (f): 11-20: 5, 21-30: 10, 31-40: 11, 41-50: 7, 51-60: 23, 61-70: 56, 71-80: 6, 81-90: 8, 91-100: 4 3. **Step 1: Calculate Class Marks (X):** Class mark $X = \frac{\text{Lower limit} + \text{Upper limit}}{2}$ | Class Interval | Frequency (f) | Class Mark (X) | |---------------|---------------|---------------| | 11 - 20 | 5 | $\frac{11+20}{2} = 15.5$ | | 21 - 30 | 10 | $25.5$ | | 31 - 40 | 11 | $35.5$ | | 41 - 50 | 7 | $45.5$ | | 51 - 60 | 23 | $55.5$ | | 61 - 70 | 56 | $65.5$ | | 71 - 80 | 6 | $75.5$ | | 81 - 90 | 8 | $85.5$ | | 91 - 100 | 4 | $95.5$ | 4. **Step 2: Calculate $fX$ (frequency times class mark):** Multiply each frequency by its class mark: | Class Interval | f | X | $fX$ | |----------------|----|------|------------| | 11 - 20 | 5 | 15.5 | $5 \times 15.5 = 77.5$ | | 21 - 30 | 10 | 25.5 | $255$ | | 31 - 40 | 11 | 35.5 | $390.5$ | | 41 - 50 | 7 | 45.5 | $318.5$ | | 51 - 60 | 23 | 55.5 | $1276.5$ | | 61 - 70 | 56 | 65.5 | $3668$ | | 71 - 80 | 6 | 75.5 | $453$ | | 81 - 90 | 8 | 85.5 | $684$ | | 91 - 100 | 4 | 95.5 | $382$ | 5. **Step 3: Calculate Total Frequency (N) and Sum of $fX$:** $N = 5 + 10 + 11 + 7 + 23 + 56 + 6 + 8 + 4 = 130$ $\sum fX = 77.5 + 255 + 390.5 + 318.5 + 1276.5 + 3668 + 453 + 684 + 382 = 8555$ 6. **Step 4: Calculate Mean ($\bar{x}$):** $$\bar{x} = \frac{\sum fX}{N} = \frac{8555}{130} = 65.81$$ 7. **Step 5: Calculate Median:** - Find cumulative frequencies (CF): | Class Interval | f | CF | |----------------|----|-----| | 11 - 20 | 5 | 5 | | 21 - 30 | 10 | 15 | | 31 - 40 | 11 | 26 | | 41 - 50 | 7 | 33 | | 51 - 60 | 23 | 56 | | 61 - 70 | 56 | 112 | | 71 - 80 | 6 | 118 | | 81 - 90 | 8 | 126 | | 91 - 100 | 4 | 130 | - Median class is where $\frac{N}{2} = 65$ lies, which is in 61-70 class. - Median formula: $$\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f_m} \right) \times CI$$ Where: - $L = 60.5$ (lower boundary of median class) - $F = 56$ (cumulative frequency before median class) - $f_m = 56$ (frequency of median class) - $CI = 10$ (class width) Calculate: $$\text{Median} = 60.5 + \left( \frac{65 - 56}{56} \right) \times 10 = 60.5 + \frac{9}{56} \times 10 = 60.5 + 1.61 = 62.11$$ 8. **Step 6: Calculate Mode:** - Mode class is the class with highest frequency, which is 61-70 with $f_m = 56$. - Let $d_1 = f_m - f_{previous} = 56 - 23 = 33$ - Let $d_2 = f_m - f_{next} = 56 - 6 = 50$ - Mode formula: $$\text{Mode} = L + \left( \frac{d_1}{d_1 + d_2} \right) \times CI$$ Where $L = 60.5$, $CI = 10$ Calculate: $$\text{Mode} = 60.5 + \left( \frac{33}{33 + 50} \right) \times 10 = 60.5 + \frac{33}{83} \times 10 = 60.5 + 3.98 = 64.48$$ **Final answers:** - Mean = 65.81 - Median = 62.11 - Mode = 64.48