1. **State the problem:** We need to estimate the fraction of videos that are between 2 minutes and 3 minutes long, i.e., between 120 seconds and 180 seconds. 2. **Understand the data:** The histogram shows frequency density over intervals of video lengths in seconds. The intervals given are 0-160, 160-200, 200-320, and 320-400 seconds. 3. **Identify the target interval:** The target interval 2 to 3 minutes corresponds to 120 to 180 seconds. This interval overlaps with the first two histogram intervals: 0-160 and 160-200. 4. **Calculate the fraction for the overlapping intervals:** - For 0-160 seconds interval, frequency density is $d_1$ (unknown exact value, but we will denote it as $d_1$). - For 160-200 seconds interval, frequency density is $d_2$. 5. **Calculate the area (fraction) for the part of the intervals that fall within 120-180 seconds:** - From 120 to 160 seconds (within 0-160 interval): length = $160 - 120 = 40$ seconds. - From 160 to 180 seconds (within 160-200 interval): length = $180 - 160 = 20$ seconds. 6. **Fraction estimate:** $$\text{Fraction} = d_1 \times 40 + d_2 \times 20$$ 7. **Simplify the fraction:** Since exact frequency densities $d_1$ and $d_2$ are not provided, the fraction is expressed in terms of these densities. If the histogram data is available, substitute the values to get the numerical fraction. **Final answer:** The fraction of videos between 2 and 3 minutes is estimated by $$\boxed{40 d_1 + 20 d_2}$$ where $d_1$ and $d_2$ are the frequency densities for intervals 0-160 and 160-200 seconds respectively.