S1 June 2006 Q2

EdexcelOld spec10 marksData Processing Probability

2. Sunita and Shelley talk to one another once a week on the telephone. Over many weeks they recorded, to the nearest minute, the number of minutes spent in conversation on each occasion. The following table summarises their results.

Time (to the nearest minute)	Number of Conversations
5–9	2
10–14	9
15–19	20
20–24	13
25–29	8
30–34	3

Two of the conversations were chosen at random.

(a) Find the probability that both of them were longer than 24.5 minutes. (2)

The mid-point of each class was represented by \(x\) and its corresponding frequency by \(f\), giving \(\Sigma fx = 1060\).

(b) Calculate an estimate of the mean time spent on their conversations. (2)

During the following 25 weeks they monitored their weekly conversations and found that at the end of the 80 weeks their overall mean length of conversation was 21 minutes.

(d) Comment on these two mean values. (2)

Mark scheme (a)

Scheme	Marks
\(\mathrm{P}(\text{both longer than } 24.5) = \dfrac{11}{55} \times \dfrac{10}{54} = \dfrac{1}{27}\) or \(0.\dot{0}3\dot{7}\) or 0.037	M1A1
	(2)

Notes

M1 2 fracs × w/o rep.

A1 awrt 0.037

Mark scheme (b)

Scheme	Marks
Estimate of mean time spent on their conversations is \(\bar{x} = \dfrac{1060}{55} = 19\tfrac{3}{11}\) or \(19.\dot{2}\dot{7}\) or 19.3	M1A1
	(2)

Notes

M1A1 1060/total, awrt 19.3 or 19mins 16s

Mark scheme (c)

Scheme	Marks
\(\dfrac{1060 + \sum fy}{80} = 21\)	B1
\(\sum fy = 620\)	M1
\(\therefore \bar{y} = \dfrac{620}{25} = 24.8\)	M1A1
	(4)

Notes

B1 \(21 \times 80 = 1680\)

M1 Subtracting ‘their 1060’

M1A1 Dividing their 620 by 25

Scheme	Marks
Increase in mean value.	B1
Length of conversations increased considerably during 25 weeks relative to 55 weeks	B1ft
	(2)
	(10 marks)