Question: The histogram below shows information about the masses of the gold rings in a jewellery shop. a) Work out an estimate for the number of gold rings with a mass between 1.4 g and 3 g. b) Write a sentence to explain why the value you calculated in part a) is only an estimate. Histogram shape in the bottom-right: frequency density vs mass (g), with bars approximately 0–1 at height 10, 1–1.8 at 25, 1.8–2.2 at 50, 2.2–3.2 at 30, 3.2–3.4 at 75, and 3.4–5.4 at 5.

1. **State the problem:** We need to estimate the number of gold rings with mass between $1.4$ g and $3$ g using the histogram data. 2. **Recall the formula for frequency from a histogram:** $$\text{Frequency} = \text{Frequency density} \times \text{Class width}$$ 3. **Identify relevant bars and their intervals:** - The interval $1.4$ g to $3$ g spans parts of the bars: - From $1.4$ to $1.8$ in the $1$–$1.8$ bar (height $25$) - The full $1.8$–$2.2$ bar (height $50$) - Part of the $2.2$–$3.2$ bar (height $30$) from $2.2$ to $3$ 4. **Calculate frequencies for each part:** - For $1.4$ to $1.8$ (part of $1$–$1.8$ bar): - Class width of full bar = $1.8 - 1 = 0.8$ - Partial width = $1.8 - 1.4 = 0.4$ - Frequency density = $25$ - Frequency = $25 \times 0.4 = 10$ - For $1.8$ to $2.2$ (full bar): - Class width = $2.2 - 1.8 = 0.4$ - Frequency density = $50$ - Frequency = $50 \times 0.4 = 20$ - For $2.2$ to $3$ (part of $2.2$–$3.2$ bar): - Class width of full bar = $3.2 - 2.2 = 1.0$ - Partial width = $3 - 2.2 = 0.8$ - Frequency density = $30$ - Frequency = $30 \times 0.8 = 24$ 5. **Add frequencies to estimate total number of rings:** $$10 + 20 + 24 = 54$$ 6. **Answer for part a):** The estimated number of gold rings with mass between $1.4$ g and $3$ g is $54$. 7. **Answer for part b):** This is only an estimate because the histogram groups data into intervals and assumes uniform distribution of rings within each interval, which may not be exactly true. Also, partial intervals require approximation of frequency density over smaller ranges.