S1 June 2010 Q5

EdexcelOld spec14 marksData Processing

5. A teacher selects a random sample of 56 students and records, to the nearest hour, the time spent watching television in a particular week.

Hours	1 – 10	11 – 20	21 – 25	26 – 30	31 – 40	41 – 59
Frequency	6	15	11	13	8	3
Mid-point	5.5	15.5		28		50

(a) Find the mid-points of the 21 – 25 hour and 31 – 40 hour groups. (2)

A histogram was drawn to represent these data. The 11 – 20 group was represented by a bar of width 4 cm and height 6 cm.

(b) Find the width and height of the 26 – 30 group. (3)

(d) Use linear interpolation to estimate the median length of time spent watching television by these students. (2)

The teacher estimated the lower quartile and the upper quartile of the time spent watching television to be 15.8 and 29.3 respectively.

(e) State, giving a reason, the skewness of these data. (2)

Scheme	Marks
23, 35.5 (may be in the table)	B1 B1
	(2)

Mark scheme (b)

Scheme	Marks
Width of 10 units is 4 cm so width of 5 units is 2 cm	B1
Height = \(2.6 \times 4\) =10.4 cm	M1 A1
	(3)

Notes

M1 for their width x their height=20.8.
Without labels assume width first, height second and award marks accordingly.

Mark scheme (c)

Scheme	Marks
\(\sum \mathrm{f}x = 1316.5 \Rightarrow \bar{x} = \dfrac{1316.5}{56} =\) awrt 23.5	M1 A1
\(\sum \mathrm{f}x^2 = 37378.25\) can be implied	B1
So \(\sigma = \sqrt{\dfrac{37378.25}{56} - \bar{x}^2} =\) awrt10.7 allow \(s\) = 10.8	M1 A1
	(5)

Notes

1st M1 for reasonable attempt at \(\sum x\) and /56
2nd M1 for a method for \(\sigma\) or \(s\), \(\sqrt{\phantom{x}}\) is required
Typical errors \(\sum (\mathrm{f}x)^2 = 354806.3\) M0, \(\sum \mathrm{f}^2x = 13922.5\) M0 and \((\sum \mathrm{f}x)^2 = 1733172\) M0
Correct answers only, award full marks.

Use of \(\sum \mathrm{f}(x - \bar{x})^2\) = awrt 6428.75 for B1 (this note is printed under (d) in the mark scheme)

Mark scheme (d)

Scheme	Marks
\(Q_2 = (20.5) + \dfrac{(28 - 21)}{11} \times 5 = 23.68\ldots\) awrt 23.7 or 23.9	M1 A1
	(2)

Notes

lcb can be 20, 20.5 or 21, width can be 4 or 5 and the fraction part of the formula correct for M1 - Allow 28.5 in fraction that gives awrt 23.9 for M1A1

Mark scheme (e)

Scheme	Marks
\(Q_3 - Q_2 = 5.6,\ \ Q_2 - Q_1 = 7.9\) (or \(\bar{x} \lt Q_2\))	M1
[7.9 >5.6 so ] negative skew	A1
	(2)
	(14 marks)

Notes

M1 for attempting a test for skewness using quartiles or mean and median.
Provided median greater than 22.55 and less than 29.3 award for M1 for \(Q_3 - Q_2 \lt Q_2 - Q_1\) without values as a valid reason.
SC Accept mean close to median and no skew oe for M1A1