1. **Problem statement:** We have a frequency distribution of 120 potatoes by mass intervals and frequencies. We need to estimate the number of potatoes with mass greater than 270 grams. 2. **Given data:** Mass intervals and frequencies: - $0 \leq m < 100$: frequency = 14 - $100 \leq m < 150$: frequency = 28 - $150 \leq m < 200$: frequency = 37 - $200 \leq m < 250$: frequency = 21 - $250 \leq m < 350$: frequency = 20 3. **Step for (ii): Estimate number of potatoes with mass > 270 grams.** - The interval $250 \leq m < 350$ has frequency 20 over a width of $100$ grams. - Frequency density for this interval is $\frac{20}{100} = 0.2$. - The sub-interval from 270 to 350 grams has width $350 - 270 = 80$ grams. - Estimated frequency for $270 \leq m < 350$ is frequency density $\times$ width: $$0.2 \times 80 = 16$$ So, approximately 16 potatoes have mass greater than 270 grams. 4. **Step for (iii): Find the interval containing the upper quartile (75th percentile).** - Total potatoes = 120 - Upper quartile position = $\frac{3}{4} \times 120 = 90$th potato - Cumulative frequencies: - $0 \leq m < 100$: 14 - $100 \leq m < 150$: 14 + 28 = 42 - $150 \leq m < 200$: 42 + 37 = 79 - $200 \leq m < 250$: 79 + 21 = 100 - The 90th potato lies between 79 and 100, so in the interval $200 \leq m < 250$. 5. **Step for (iv): Probability that a potato has mass less than 150 grams.** - Frequencies for $m < 150$ are 14 and 28, total $14 + 28 = 42$. - Probability = $\frac{42}{120}$. - Simplify fraction: $$\frac{42}{120} = \frac{\cancel{6}7}{\cancel{6}20} = \frac{7}{20}$$ **Final answers:** (ii) Estimated number of potatoes with mass > 270 grams: $16$ (iii) Upper quartile interval: $200 \leq m < 250$ (iv) Probability mass < 150 grams: $\frac{7}{20}$