(i) Plot the cumulative frequency points:
\((3.5,\,0),\ (3.8,\,6),\ (4.0,\,18),\ (4.2,\,41),\ (4.5,\,62),\ (4.8,\,70)\),
then join with a smooth curve (or straight segments).
(ii) To estimate the percentage with nitrogen content \(>4.4\):
read \(\mathrm{CF}(4.4)\approx 55\) from the graph, so the count is
\(70-55=15\). Hence the percentage is
\(\dfrac{15}{70}\times 100 \approx 21.4\%\).
(iii) The median is the \(\dfrac{70}{2}=35\)th value.
From the cumulative frequency curve, this corresponds to
approximately \(4.15\).
(iv) Frequency table and densities (class width in parentheses):
| Nitrogen content |
3.5โ3.8 (width \(0.3\)) |
3.8โ4.0 (width \(0.2\)) |
4.0โ4.2 (width \(0.2\)) |
4.2โ4.5 (width \(0.3\)) |
4.5โ4.8 (width \(0.3\)) |
| Frequency |
6 |
12 |
23 |
21 |
8 |
Frequency density \(\text{freq}/\text{width}\) |
\(\dfrac{6}{0.3}=20\) |
\(\dfrac{12}{0.2}=60\) |
\(\dfrac{23}{0.2}=115\) |
\(\dfrac{21}{0.3}=70\) |
\(\dfrac{8}{0.3}\approx 26.7\) |
Draw a histogram using the class intervals above on the horizontal axis and the frequency densities on the vertical axis.