1. **Problem Statement:** We are given a frequency distribution of single digits from 0 to 9, with frequencies for digits 1 and 5 as 8 and 15 respectively, and others presumably zero or not given. 2. **Given Data:** Digits $x$: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Frequencies $f$: 0, 8, 0, 0, 0, 15, 0, 0, 0, 0 3. **Find:** Mean, median, and mode of the distribution. 4. **Formulas:** - Mean: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ - Median: The middle value when data is ordered. - Mode: The value with the highest frequency. 5. **Calculate total frequency:** $$\sum f_i = 8 + 15 = 23$$ 6. **Calculate mean:** $$\sum f_i x_i = (1 \times 8) + (5 \times 15) = 8 + 75 = 83$$ $$\bar{x} = \frac{83}{23}$$ Intermediate step showing cancellation: $$\bar{x} = \frac{\cancel{83}}{\cancel{23}}$$ (no common factors to cancel) So, $$\bar{x} \approx 3.61$$ 7. **Calculate median:** Total frequency is 23, median position is $$\frac{23+1}{2} = 12^{th}$$ value. Frequencies cumulative: - Up to digit 1: 8 - Up to digit 5: 8 + 15 = 23 Since 12th value lies between 9 and 23, median digit is 5. 8. **Calculate mode:** Mode is the digit with highest frequency, which is 5 (frequency 15). **Final answers:** - Mean $\approx 3.61$ - Median $= 5$ - Mode $= 5$