S1 January 2007 Q5

5. A teacher recorded, to the nearest hour, the time spent watching television during a particular week by each child in a random sample. The times were summarised in a grouped frequency table and represented by a histogram.

One of the classes in the grouped frequency distribution was 20–29 and its associated frequency was 9. On the histogram the height of the rectangle representing that class was 3.6 cm and the width was 2 cm.

(a) Give a reason to support the use of a histogram to represent these data. (1)

(b) Write down the underlying feature associated with each of the bars in a histogram. (1)

The total area under the histogram was 24 cm².

(d) Find the total number of children in the group. (2)

Scheme	Marks
Time is a continuous variable or data is in a grouped frequency table	B1
	(1)

Mark scheme (b)

Scheme	Marks
Area is proportional to frequency or \(A \propto f\) or \(A = kf\)	B1
	(1)

Notes

1^st B1 for one of these correct statements.
“Area proportional to frequency density” or “Area = frequency” is B0

Mark scheme (c)

Scheme	Marks
\(3.6 \times 2 = 0.8 \times 9\)	M1 dM1
1 child represented by 0.8	A1 cso
	(3)

Notes

1^st M1 for a correct combination of any 2 of the 4 numbers: 3.6, 2, 0.8 and 9
e.g. \(3.6 \times 2\) or \(\dfrac{3.6}{0.8}\) or \(\dfrac{0.8}{2}\) etc BUT e.g. \(\dfrac{3.6}{2}\) is M0

2^nd M1 dependent on 1^st M1 and for a correct combination of 3 numbers leading to 4^th.
May be in separate stages but must see all 4 numbers

A1cso for fully correct solution. Both Ms scored, no false working seen and comment required.

Scheme	Marks
(Total) \(= \dfrac{24}{0.8}, \ = \underline{\mathbf{30}}\)	M1, A1
	(2)
	(7 marks)