S1 January 2007 Q4

EdexcelOld spec14 marksData Processing

4. Summarised below are the distances, to the nearest mile, travelled to work by a random sample of 120 commuters.

Distance (to the nearest mile)	Number of commuters
0–9	10
10–19	19
20–29	43
30–39	25
40–49	8
50–59	6
60–69	5
70–79	3
80–89	1

For this distribution,

(a) describe its shape, (1)

(b) use linear interpolation to estimate its median. (2)

The mid-point of each class was represented by \(x\) and its corresponding frequency by \(f\) giving

\[\Sigma fx = 3550 \quad \text{and} \quad \Sigma fx^2 = 138020\]

One coefficient of skewness is given by

\[\frac{3(\text{mean} - \text{median})}{\text{standard deviation}}.\]

(d) Evaluate this coefficient for this distribution. (3)

(e) State whether or not the value of your coefficient is consistent with your description in part (a). Justify your answer. (2)

(f) State, with a reason, whether you should use the mean or the median to represent the data in this distribution. (2)

(g) State the circumstance under which it would not matter whether you used the mean or the median to represent a set of data. (1)

Scheme	Marks
Positive skew	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(19.5 + \dfrac{(60 - 29)}{43} \times 10, = 26.7093\ldots\)	M1, A1
	(2)

Notes

A1 awrt 26.7 (N.B. Use of 60.5 gives 26.825… so allow awrt 26.8)

M1 for (19.5 or 20) \(+ \dfrac{(60 - 29)}{43} \times 10\) or better. Allow 60.5 giving awrt 26.8 for M1A1
Allow their \(0.5n\) [or \(0.5(n + 1)\)] instead of 60 [or 60.5] for M1.

Mark scheme (c)

Scheme	Marks
\(\mu = \dfrac{3550}{120} = 29.5833\ldots\) or \(29\tfrac{7}{12}\)	B1
\(\sigma^2 = \dfrac{138020}{120} - \mu^2\) or \(\sigma = \sqrt{\dfrac{138020}{120} - \mu^2}\)	M1
\(\sigma = 16.5829\ldots\) or \((s = 16.652\ldots)\)	A1
	(3)

Notes

B1 awrt 29.6

A1 awrt 16.6 (or \(s\) = 16.7)

M1 for a correct expression for \(\sigma, \sigma^2, s\) or \(s^2\). NB \(\sigma^2 = 274.99\) and \(s^2 = 277.30\)
Condone poor notation if answer is awrt16.6 (or 16.7 for \(s\))

Mark scheme (d)

Scheme	Marks
\(\dfrac{3(29.6 - 26.7)}{16.6}\)	M1A1ft
\(= 0.52\ldots\)	A1
	(3)

Notes

A1 awrt 0.520 (or with \(s\) awrt 0.518) (N.B. 60.5 in (b) …awrt 0.499 [or with \(s\) awrt 0.497])

M1 for attempt to use this formula using their values to any accuracy. Condone missing 3.
1^st A1ft for using their values to at least 3sf. Must have the 3.
2^nd A1 for using accurate enough values to get awrt 0.520 (or 0.518 if using \(s\))
NB Using only 3 sf gives 0.524 and scores M1A1A0

Mark scheme (e)

Scheme	Marks
0.520 > 0	B1ft
So it is consistent with (a)	dB1ft
	(2)

Notes

B1ft correct statement about their (d) being >0 or < 0

dB1ft ft their (d)

1^st B1 for saying or implying correct sign for their (d). B1g and B1ft. Ignore “correlation” if seen.
2^nd B1 for a comment about consistency with their (d) and (a) being positive skew, ft their (d) only
This is dependent on 1^st B1: so if (d)>0, they say yes, if (d)<0 they say no.

Scheme	Marks
Use Median	B1
Since the data is skewed or less affected by outliers/extreme values	dB1
	(2)

Mark scheme (g)

Scheme	Marks
If the data are symmetrical or skewness is zero or normal/uniform distribution (“mean =median” or “no outliers” or “evenly distributed” all score B0)	B1
	(1)
	(14 marks)