S1 June 2013 Q4

EdexcelOld spec14 marksData Processing

4. The following table summarises the times, \(t\) minutes to the nearest minute, recorded for a group of students to complete an exam.

Time (minutes) \(t\)	11 – 20	21 – 25	26 – 30	31 – 35	36 – 45	46 – 60
Number of students f	62	88	16	13	11	10

[You may use \(\sum \mathrm{f}t^2 = 134281.25\)]

(a) Estimate the mean and standard deviation of these data. (5)

(b) Use linear interpolation to estimate the value of the median. (2)

(d) Estimate the interquartile range of this distribution. (2)

(e) Give a reason why the mean and standard deviation are not the most appropriate summary statistics to use with these data. (1)

The person timing the exam made an error and each student actually took 5 minutes less than the times recorded above. The table below summarises the actual times.

Time (minutes) \(t\)	6 – 15	16 – 20	21 – 25	26 – 30	31 – 40	41 – 55
Number of students f	62	88	16	13	11	10

(f) Without further calculations, explain the effect this would have on each of the estimates found in parts (a), (b), (c) and (d). (3)

Mark scheme (a)

Scheme	Marks
\(\sum \mathrm{f}t = 4837.5\) (allow 4838 or 4840)	B1
Mean \(= \dfrac{\text{"}4837.5\text{"}}{200} = 24.1875\) awrt 24.2 or \(\dfrac{387}{16}\)	M1 A1
\(\sigma = \sqrt{\dfrac{134281.25}{200} - \left(\dfrac{4837.5}{200}\right)^2}\)	M1
\(= 9.293\ldots\) (accept \(s = 9.32\)) awrt 9.29	A1
	(5)

Notes

Correct answers only score full marks in each part except (c)

B1 for 4837.5 or 4838 or 4840 seen. If no \(\sum \mathrm{f}t\) seen (or attempt at \(\sum \mathrm{f}t\) seen), B1 can be implied by a correct mean of awrt 24.2

1st M1 for attempt at their \(\frac{\sum \mathrm{f}t}{\sum \mathrm{f}}\) allow 1sf so \(\sum \mathrm{f}\) = awrt 200 and \(\sum \mathrm{f}t\) = awrt 5000.
Or award M1 for a clear attempt at mean where at least 4 correct products of \(\sum \mathrm{f}t\) are seen

2nd M1 for correct expression including square root seen. Follow through their mean. Allow a transcription error in 134281.25 but not an incorrect re-calculation.

Mark scheme (b)

Scheme	Marks
\(\mathrm{Q}_2 = [20.5] + \dfrac{(100/100.5 - 62)}{88}\times 5 = 22.659\ldots\) awrt 22.7	M1 A1
	(2)

Notes

M1 for a correct fraction \(\times 5\). Ignore end point but must be +. Allow use of \((n + 1)\) giving 100.5…

Mark scheme (c)

Scheme	Marks
\(\mathrm{Q}_1 = 10.5 + \dfrac{(50/50.25)}{62}\times 10\ [= 18.56]\) (*) (\(n + 1\) gives 18.604…)	B1 cso
	(1)

Notes

B1cso for a fully correct expression including end point. NB Answer is given. Allow use of \((n + 1)\) giving 50.25…but use of 50.5 scores B0

Mark scheme (d)

Scheme	Marks
\(\mathrm{Q}_3 = 25.5\) (Use of \(n + 1\) gives 25.734…)	B1
IQR = 6.9 (Use of \(n + 1\) gives 7.1)	B1 ft
	(2)

Notes

1st B1 for 25.5 (or awrt 25.7 using \(n + 1\))

2nd B1ft for their \(Q_3 -\) their \(Q_1\) (or 18.6) (provided > 0) Accept awrt 2sf . Correct ans. only scores 2/2

Scheme	Marks
The data is skewed (condone “negative skew”)	B1
	(1)

Mark scheme (f)

Scheme	Marks
Mean decreases and st. dev. remains the same. [Must mention mean and st. dev.] (from(a))	B1
The median and quartiles would decrease. [Must refer to median and at least \(Q_1\).] ((b)(c))	B1
The IQR would remain unchanged (from (d))	B1
	(3)
	(14 marks)

Notes

1st B1 for one correct comment from the above. May refer to parts (a), (b), (c) or (d)

2nd B1 for two correct comments from the above

3rd B1 for all 3 correct comments from the above