Solution

(i) The median is the 110th value (since \( \dfrac{220}{2} = 110 \)). From the cumulative frequency table, the median lies in the interval \(45\text{–}50\ \text{g}\).

(ii) The lower quartile (LQ) is the 55th value, which lies in the interval \(40\text{–}45\ \text{g}\). The upper quartile (UQ) is the 165th value, which lies in the interval \(50\text{–}60\ \text{g}\). Therefore, the smallest interquartile range (IQR) could be \(5\ \text{g}\) (if \( \text{LQ} = 45 \) and \( \text{UQ} = 50 \)), and the largest could be \(20\ \text{g}\) (if \( \text{LQ} = 40 \) and \( \text{UQ} = 60 \)).

(iii) The number of sausages weighing between \(50\ \text{g}\) and \(60\ \text{g}\) is the difference in cumulative frequencies: \( 210 - 160 = 50 \).

(iv) To draw the histogram, first find class frequencies from cumulative frequencies, then compute frequency density \( \big( \text{fd} = \dfrac{\text{frequency}}{\text{class width}} \big) \):

\[ \begin{aligned} &<20:\ \text{fd} = \dfrac{0}{10} = 0,\\[4pt] &20\text{–}30:\ \text{fd} = \dfrac{20}{10} = 2,\\[4pt] &30\text{–}40:\ \text{fd} = \dfrac{30}{10} = 3,\\[4pt] &40\text{–}45:\ \text{fd} = \dfrac{50}{5} = 10,\\[4pt] &45\text{–}50:\ \text{fd} = \dfrac{60}{5} = 12,\\[4pt] &50\text{–}60:\ \text{fd} = \dfrac{50}{10} = 5,\\[4pt] &60\text{–}70:\ \text{fd} = \dfrac{10}{10} = 1. \end{aligned} \]

Plot these frequency densities against the corresponding weight intervals, with bar widths equal to class widths and no gaps.