1. The problem asks us to estimate the number of children for whom $50 < m \leq 80$ minutes based on the histogram data. 2. The histogram shows frequency density for intervals of time $m$ in minutes. The frequency density multiplied by the width of the interval gives the frequency (number of children) in that interval. 3. We know from the problem that for $m \leq 20$, there are 10 children. The first bar covers $0$ to $20$ minutes with frequency density approximately $1$. The width of this interval is $20$ minutes. 4. Using the formula for frequency: $$\text{frequency} = \text{frequency density} \times \text{width}$$ For the first bar: $$10 = 1 \times 20$$ This confirms the frequency density scale is correct. 5. We want to find the number of children for $50 < m \leq 80$. This covers two intervals: $40$ to $60$ and $60$ to $80$. 6. From the histogram: - For $40$ to $60$, frequency density is about $2$, width is $20$. - For $60$ to $80$, frequency density is about $2.5$, width is $20$. 7. Calculate frequency for each interval: $$\text{frequency}_{40-60} = 2 \times 20 = 40$$ $$\text{frequency}_{60-80} = 2.5 \times 20 = 50$$ 8. But we only want $50 < m \leq 80$, so for the $40$ to $60$ interval, we only count from $50$ to $60$, which is half the interval width ($10$ minutes). 9. Calculate frequency for $50$ to $60$: $$\text{frequency}_{50-60} = 2 \times 10 = 20$$ 10. Add frequencies for $50$ to $60$ and $60$ to $80$: $$20 + 50 = 70$$ 11. Therefore, the estimate for the number of children for whom $50 < m \leq 80$ is $70$. This method uses simple multiplication of frequency density by interval width and partial interval consideration for $50$ to $60$ minutes.