Problem: The masses of all of the Roman coins in a museum are recorded in the histogram below. 108 of the coins each weigh between 8 g and 17 g. Work out how many Roman coins are in the museum's collection in total. 1. State the method: For a histogram with class width 5 the number of coins in any class equals the frequency density multiplied by the class width. 2. Label the frequency densities for the five classes 0–5, 5–10, 10–15, 15–20 and 20–25 as $f_1,f_2,f_3,f_4,f_5$ respectively. 3. Express the total number of coins $T$ in terms of the densities. $$T=5(f_1+f_2+f_3+f_4+f_5)$$ 4. Translate the information that 108 coins lie between 8 g and 17 g into an equation for the densities. The interval 8–17 covers the final 2 units of the 5–10 class, all 5 units of the 10–15 class and the first 2 units of the 15–20 class. So the count between 8 and 17 is $2f_2+5f_3+2f_4$ and we are told this equals 108. $$2f_2+5f_3+2f_4=108$$ 5. Solve this for $f_3$ in terms of $f_2$ and $f_4$. $$5f_3=108-2f_2-2f_4$$ $$f_3=\frac{108-2f_2-2f_4}{5}$$ 6. Substitute this expression for $f_3$ into the formula for $T$ and simplify. Start with $T=5f_1+5f_2+5f_3+5f_4+5f_5$. Replace $5f_3$ by $108-2f_2-2f_4$ to get $T=5f_1+5f_2+(108-2f_2-2f_4)+5f_4+5f_5$. Collect like terms to obtain the simplified formula. $$T=108+5f_1+3f_2+3f_4+5f_5$$ 7. Conclusion: The total number of coins cannot be uniquely determined from the information given alone because $f_1,f_2,f_4,f_5$ are not specified. 8. Final instruction: If you can supply the frequency densities (the heights of the histogram bars) for the five classes then substitute them into $$T=108+5f_1+3f_2+3f_4+5f_5$$ to get the numeric total. Final answer: The total number of coins is given by the formula above and additional information (the histogram bar heights) is required to compute a numeric value.