16 The grouped frequency table gives information about the lengths of 160 pythons.
Length (\(x\) metres)
Frequency
\(0 \lt x \leqslant 1\)
4
\(1 \lt x \leqslant 2\)
8
\(2 \lt x \leqslant 3\)
16
\(3 \lt x \leqslant 4\)
32
\(4 \lt x \leqslant 5\)
72
\(5 \lt x \leqslant 6\)
28
(a) Complete the cumulative frequency table.
Length (\(x\) metres)
Cumulative frequency
\(0 \lt x \leqslant 1\)
\(0 \lt x \leqslant 2\)
\(0 \lt x \leqslant 3\)
\(0 \lt x \leqslant 4\)
\(0 \lt x \leqslant 5\)
\(0 \lt x \leqslant 6\)
(1)
(b) On the grid, draw a cumulative frequency graph for your table.(2)
(c) Use your graph to find an estimate for the median length of the pythons. (2)
Mark scheme (a)
Scheme
Marks
4, 12, 28, 60, 132, 160
B1
(1)
Notes
B1: cao
Mark scheme (b)
Scheme
Marks
Points correct
B1
Curve or line segments
B1
(2)
Notes
B1: ± ½ sq ft from sensible table ie clear attempt to add frequencies
B1: ft from points if 4 or 5 correct or if all points are plotted consistently within each interval at the correct heights Accept curve which is not joined to the origin
Mark scheme (c)
Scheme
Marks
80 (or 80.5) indicated on cf graph or stated
M1
approx 4.3
A1
(2)
Notes
M1: for 80 (or 80.5) indicated on cf axis or stated
A1: If M1 scored, ft from cf graph If no indication of method, ft only from correct curve & if answer is correct (± ½ sq tolerance) award M1 A1
