1. **Problem Statement:** Find the Mean (Direct Method), Median, and Mode of the delivery time for the given frequency distribution. 2. **Given Data:** | Time (mins) | Frequency (f) | |-------------|---------------| | 20–29 | 2 | | 30–39 | 4 | | 40–49 | 7 | | 50–59 | 10 | | 60–69 | 15 | | 70–79 | 9 | | 80–89 | 6 | | 90–99 | 4 | | 100–109 | 3 | 3. **Step 1: Calculate class midpoints (x):** Midpoint $x = \frac{\text{lower limit} + \text{upper limit}}{2}$ - 20–29: $\frac{20+29}{2} = 24.5$ - 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: $94.5$ - 100–109: $104.5$ 4. **Step 2: Calculate $f \times x$ for each class:** - $2 \times 24.5 = 49$ - $4 \times 34.5 = 138$ - $7 \times 44.5 = 311.5$ - $10 \times 54.5 = 545$ - $15 \times 64.5 = 967.5$ - $9 \times 74.5 = 670.5$ - $6 \times 84.5 = 507$ - $4 \times 94.5 = 378$ - $3 \times 104.5 = 313.5$ 5. **Step 3: Sum frequencies and $f \times x$ values:** - $\sum f = 2+4+7+10+15+9+6+4+3 = 60$ - $\sum f x = 49 + 138 + 311.5 + 545 + 967.5 + 670.5 + 507 + 378 + 313.5 = 3880$ 6. **Step 4: Calculate Mean using direct method formula:** $$\text{Mean} = \frac{\sum f x}{\sum f} = \frac{3880}{60} = 64.67$$ 7. **Step 5: Find Median class:** - Median position = $\frac{N}{2} = \frac{60}{2} = 30$ - Cumulative frequencies: - 20–29: 2 - 30–39: 6 - 40–49: 13 - 50–59: 23 - 60–69: 38 (first cumulative freq $\geq 30$) - Median class is 60–69 8. **Step 6: Apply Median formula:** $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times h$$ Where: - $L = 60$ (lower boundary of median class) - $N = 60$ - $F = 23$ (cumulative frequency before median class) - $f_m = 15$ (frequency of median class) - $h = 10$ (class width) Calculate: $$\text{Median} = 60 + \left(\frac{30 - 23}{15}\right) \times 10 = 60 + \left(\frac{7}{15}\right) \times 10 = 60 + 4.67 = 64.67$$ 9. **Step 7: Find Mode class:** - Highest frequency is 15 for class 60–69 - Mode class is 60–69 10. **Step 8: Apply Mode formula:** $$\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ Where: - $L = 60$ - $f_1 = 15$ (frequency of modal class) - $f_0 = 10$ (frequency before modal class) - $f_2 = 9$ (frequency after modal class) - $h = 10$ Calculate: $$\text{Mode} = 60 + \left(\frac{15 - 10}{2 \times 15 - 10 - 9}\right) \times 10 = 60 + \left(\frac{5}{30 - 19}\right) \times 10 = 60 + \left(\frac{5}{11}\right) \times 10 = 60 + 4.55 = 64.55$$ **Final answers:** - Mean = 64.67 minutes - Median = 64.67 minutes - Mode = 64.55 minutes