1. **State the problem:** We need to estimate the probability that a randomly chosen blood sample has a glucose concentration less than 5.6 mmol/l based on the histogram. 2. **Recall the formula:** Probability for a range in a histogram is the area under the bars for that range. Area = frequency density $ \times $ width of the bar. 3. **Identify relevant bars:** Concentration less than 5.6 mmol/l includes bars from 4.5 to 5, 5 to 5.5, and part of 5.5 to 6. 4. **Calculate areas:** - From 4.5 to 5: frequency density = 1, width = 0.5, area = $1 \times 0.5 = 0.5$ - From 5 to 5.5: frequency density = 3, width = 0.5, area = $3 \times 0.5 = 1.5$ - From 5.5 to 6: frequency density = 5, width = 0.5, but we only want up to 5.6, so width = $5.6 - 5.5 = 0.1$, area = $5 \times 0.1 = 0.5$ 5. **Sum the areas:** $$0.5 + 1.5 + 0.5 = 2.5$$ 6. **Interpretation:** The total area under the histogram is 1 (since it represents the whole sample space). The sum of areas for concentration less than 5.6 mmol/l is 2.5, which is not possible for a probability. This suggests the histogram's frequency densities are scaled such that the total area is not 1. We must find the total area to normalize. 7. **Calculate total area:** - 4.5 to 5: $1 \times 0.5 = 0.5$ - 5 to 5.5: $3 \times 0.5 = 1.5$ - 5.5 to 6: $5 \times 0.5 = 2.5$ - 6 to 6.5: $2.5 \times 0.5 = 1.25$ - 6.5 to 7.5: $3.5 \times 1 = 3.5$ - 7.5 to 8: $0.3 \times 0.5 = 0.15$ Total area = $0.5 + 1.5 + 2.5 + 1.25 + 3.5 + 0.15 = 9.4$ 8. **Calculate probability:** $$\text{Probability} = \frac{\text{area for } < 5.6}{\text{total area}} = \frac{2.5}{9.4} \approx 0.2660$$ 9. **Round to 2 decimal places:** $$0.27$$ **Final answer:** The estimated probability that the glucose concentration is less than 5.6 mmol/l is **0.27**.