4400 Higher June 2004 Paper 3 Q9

EdexcelHigherOld spec11 marksCumulative Frequency Grouped Data

9 The grouped frequency table gives information about the distance each of 150 people travel to work.

Distance travelled (\(d\) km)	Frequency
\(0 \lt d \leqslant 5\)	34
\(5 \lt d \leqslant 10\)	48
\(10 \lt d \leqslant 15\)	26
\(15 \lt d \leqslant 20\)	18
\(20 \lt d \leqslant 25\)	16
\(25 \lt d \leqslant 30\)	8

(a) Work out what percentage of the 150 people travel more than 20 km to work. (2)

(b) Work out an estimate for the mean distance travelled to work by the people. (4)

Distance travelled (\(d\) km)	Cumulative frequency
\(0 \lt d \leqslant 5\)
\(0 \lt d \leqslant 10\)
\(0 \lt d \leqslant 15\)
\(0 \lt d \leqslant 20\)
\(0 \lt d \leqslant 25\)
\(0 \lt d \leqslant 30\)

(1)

(d) On the grid, draw a cumulative frequency graph for your table.

Grid: cumulative frequency 0 to 160 against distance travelled d km from 0 to 30

(2)

(e) Use your graph to find an estimate for the median of the distance travelled to work by the people.
Show your method clearly. (2)

Scheme	Marks
\(\dfrac{16 + 8}{150}\) or \(\dfrac{24}{150}\) or 0.16	M1
16	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(34 \times 2.5 + 48 \times 7.5 + 26 \times 12.5\) \(+ 18 \times 17.5 + 16 \times 22.5 + 8 \times 27.5\) or 85 + 360 + 325 + 315 + 360 + 220 or 1665	M1 M1
\(\dfrac{\text{"1665"}}{150}\)	M1
11.1	A1
	(4)

Notes

M1: finds products \(f \times x\) consistently within intervals (inc end points) and sums them

M1: use of midpoints

M1: (dep on 1st M1) for division by 150

A1: Accept 11 if \(\dfrac{1665}{150}\) seen

Scheme	Marks
34, 82, 108, 126, 142, 150	B1
	(1)

Mark scheme (d)

Scheme	Marks
Points	B1
Curve	B1
	(2)

Notes

B1: ± ½ square ft from sensible table

B1: or line segments (dep on 5 pts correct or ft correctly or 5 ordinates from (c) plotted correctly and consistently within intervals but not above end points)

Scheme	Marks
cf of 75 (or 75½) used	M1
~ 9	A1
	(2)
	(11 marks)