S1 June 2012 Q5

EdexcelOld spec13 marksData Processing

Figure 2: Histogram of frequency density against speed of car, 0 to 50 — Figure 2

A policeman records the speed of the traffic on a busy road with a 30 mph speed limit. He records the speeds of a sample of 450 cars. The histogram in Figure 2 represents the results.

(a) Calculate the number of cars that were exceeding the speed limit by at least 5 mph in the sample. (4)

(b) Estimate the value of the mean speed of the cars in the sample. (3)

(d) Comment on the shape of the distribution. Give a reason for your answer. (2)

(e) State, with a reason, whether the estimate of the mean or the median is a better representation of the average speed of the traffic on the road. (2)

Mark scheme (a)

Scheme	Marks
One large square \(= \dfrac{450}{\text{"}22.5\text{"}}\) or one small square \(= \dfrac{450}{\text{"}562.5\text{"}}\) (o.e. e.g. \(\dfrac{\text{"}562.5\text{"}}{450}\))	M1
One large square = 20 cars or one small square = 0.8 cars or 1 car = 1.25 squares	A1
No. > 35 mph is: \(4.5\times\text{"}20\text{"}\) or \(112.5\times\text{"}0.8\text{"}\) (or equivalent e.g. using fd)	dM1
\(=\) 90 (cars)	A1
	(4)

Notes

1st M1 for attempt to count squares (accept “22.5” in [22, 23] and “562.5” in [550, 575]) and use 450 to obtain a measure of scale. [If using fd must use 450 to obtain scale factor]

1st A1 for a correct calc. for 20 or 0.8 or 1.25 etc
[ May be fd = 4 to 1 large sq. or 0.8 to 1 small sq. May be on the diagram.]

2nd dM1 dep on 1st M1 for correctly counting squares for > 35 mph and forming suitable expr’

2nd A1 for 90 with no incorrect working seen.
e.g. \(\dfrac{4.5}{22.5}\times 450\) scores M1A1M1 and A1 when = 90 is seen. Answer only is 4/4

Mark scheme (b)

Scheme	Marks
\([\bar{x}] = \dfrac{30\times 12.5 + 240\times 25 + 90\times 32.5 + 30\times 37.5 + 60\times 42.5}{450} \quad \left[= \dfrac{12975}{450}\right]\)	M1 M1
\(= 28.83\ldots\) or \(\dfrac{173}{6}\) awrt 28.8	A1
	(3)

Notes

1st M1 for clear, sensible use of mid-points at least 3 of (12.5, 25, 32.5, 37.5, 42.5) seen

2nd M1 for an expression for \(\bar{x}\) (at least 3 correct terms on num’ and a compatible denominator). Follow through their frequencies.

You may see these fractions: \(\frac{16218.75}{562.5}\) (small squares), \(\frac{12975}{450}\) (frequencies), \(\frac{648.75}{22.5}\) (large squares)

A1 for awrt 28.8 (answer only is 3/3)

Mark scheme (c)

Scheme	Marks
\([Q_2 =]\ 20 + \dfrac{195}{240}\times 10\) (o.e.) [Allow use of \((n + 1)\) giving 195.5 instead of 195]	M1
\(= 28.125\) [Use of \((n + 1)\) gives 28.145…] awrt 28.1	A1
	(2)

Notes

M1 for a full expression for median (using their frequencies). May see e.g. \(25 + \dfrac{75}{120}\times 5\) etc. Do nor accept boundaries of 19.5 or 20.5, these are M0A0

A1 for awrt 28.1 (answer only is 2/2) [For use of \((n + 1)\) accept 28.15 but not 28.2]

Mark scheme (d)

Scheme	Marks
\(Q_2 \lt \bar{x}\) [Condone \(Q_2 \approx \bar{x}\)]	B1ft
So positive skew [ so (almost) symmetric ]	dB1ft
	(2)

Notes

1st B1ft for a correct statement about their \(Q_2\) and \(\bar{x}\) [Condone \(Q_2 \approx \bar{x}\) only if \(|Q_2 - \bar{x}| \lt 1\)]. Do not accept an argument based on the shape of the graph alone.

2nd dB1ft dependent on 1st B1 for a compatible description of skewness. F.t. their values

Quartiles If \(Q_1 = 23.4\) and \(Q_3 = 33.7 \sim 33.8\) are seen allow comparison of quartiles for 1st B1 in (d)

Mark scheme (e)

Scheme	Marks
[If chose skew in (d)] median (\(Q_2\)) [If chose symmetric in (d)] mean (\(\bar{x}\))	B1
Since the data is skewed or median not affected by extreme values [Since it uses all the data]	dB1
	(2)
	(13 marks)

Notes

1st B1 for a correct choice based on their skewness comment in (d). If no choice made in (d) only \(Q_2\)

2nd dB1 for a suitable compatible comment