1. **State the problem:** We are given the frequency distribution of the number of chairs per household and need to find the mean, median, and mode. 2. **Given data:** Number of Chairs ($x$): 1, 2, 3, 4, 5, 6 Frequency ($f$): 10, 11, 9, 6, 4, 15 3. **Calculate the mean ($\bar{x}$):** The formula for mean is: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate $\sum f_i x_i$: $$10 \times 1 + 11 \times 2 + 9 \times 3 + 6 \times 4 + 4 \times 5 + 15 \times 6 = 10 + 22 + 27 + 24 + 20 + 90 = 193$$ Calculate total frequency $\sum f_i$: $$10 + 11 + 9 + 6 + 4 + 15 = 55$$ Calculate mean: $$\bar{x} = \frac{193}{55}$$ Show cancellation: $$\bar{x} = \frac{\cancel{193}}{\cancel{55}}$$ Since 193 and 55 have no common factors, simplify by division: $$\bar{x} \approx 3.5$$ 4. **Calculate the median:** Total frequency $N = 55$, median position is at $\frac{N+1}{2} = \frac{56}{2} = 28^{th}$ value. Cumulative frequencies: - Up to 1 chair: 10 - Up to 2 chairs: 10 + 11 = 21 - Up to 3 chairs: 21 + 9 = 30 The 28th value lies in the 3 chairs group because 21 < 28 \leq 30. So, median = 3 chairs. 5. **Calculate the mode:** Mode is the value(s) with the highest frequency. Frequencies: 10, 11, 9, 6, 4, 15 Highest frequency is 15 corresponding to 6 chairs. So, mode = 6 chairs. **Final answers:** - Mean: 3.5 chairs - Median: 3 chairs - Mode: 6 chairs