1. **State the problem:** We have grouped data for monthly expenses with class intervals and frequencies. We need to calculate the median and mode expenses for 24 months. 2. **Given data:** Class intervals (c.i): 140-149, 150-159, 160-169, 170-179 Frequencies (f): 3, 7, 5, 9 Total frequency $n = 3 + 7 + 5 + 9 = 24$ 3. **Calculate the median:** - Median class is the class where the cumulative frequency reaches or exceeds $\frac{n}{2} = 12$ - Cumulative frequencies: - 140-149: 3 - 150-159: 3 + 7 = 10 - 160-169: 10 + 5 = 15 - 170-179: 15 + 9 = 24 - Median class is 160-169 because cumulative frequency 15 $\geq 12$ - Median class lower boundary $l = 159.5$ - Frequency of median class $f_m = 5$ - Cumulative frequency before median class $F = 10$ - Class width $h = 10$ Median formula: $$\text{Median} = l + \left(\frac{\frac{n}{2} - F}{f_m}\right) \times h$$ Substitute values: $$\text{Median} = 159.5 + \left(\frac{12 - 10}{5}\right) \times 10 = 159.5 + \frac{2}{5} \times 10$$ Simplify: $$\text{Median} = 159.5 + 4 = 163.5$$ 4. **Calculate the mode:** - Mode class is the class with highest frequency, which is 170-179 with frequency 9 - Mode class lower boundary $l = 169.5$ - Frequency of mode class $f_1 = 9$ - Frequency of class before mode class $f_0 = 5$ - Frequency of class after mode class $f_2 = 0$ (no class after 170-179) - Class width $h = 10$ Mode formula: $$\text{Mode} = l + \left(\frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)}\right) \times h$$ Substitute values: $$\text{Mode} = 169.5 + \left(\frac{9 - 5}{(9 - 5) + (9 - 0)}\right) \times 10 = 169.5 + \left(\frac{4}{4 + 9}\right) \times 10$$ Simplify: $$\text{Mode} = 169.5 + \frac{4}{13} \times 10 = 169.5 + 3.08 = 172.58$$ **Final answers:** - Median expense = $163.5$ million rupees - Mode expense = $172.58$ million rupees