S1 June 2017 Q2

EdexcelOld spec14 marksData Processing Normal Distribution

2. An estate agent is studying the cost of office space in London. He takes a random sample of 90 offices and calculates the cost, £\(x\) per square foot. His results are given in the table below.

Cost (£\(x\))	Frequency (f)	Midpoint (£\(y\))
\(20 \leqslant x \lt 40\)	12	30
\(40 \leqslant x \lt 45\)	13	42.5
\(45 \leqslant x \lt 50\)	25	47.5
\(50 \leqslant x \lt 60\)	32	55
\(60 \leqslant x \lt 80\)	8	70

(You may use \(\sum \mathrm{f}y^2 = 226\,687.5\))

A histogram is drawn for these data and the bar representing \(50 \leqslant x \lt 60\) is 2 cm wide and 8 cm high.

(a) Calculate the width and height of the bar representing \(20 \leqslant x \lt 40\) (3)

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

(d) Estimate the standard deviation for these data. (2)

(e) Describe, giving a reason, the skewness. (1)

Rika suggests that the cost of office space in London can be modelled by a normal distribution with mean £50 and standard deviation £10

(f) With reference to your answer to part (e), comment on Rika’s suggestion. (1)

(g) Use Rika’s model to estimate the 80th percentile of the cost of office space in London. (3)

Mark scheme (a)

Scheme	Marks
Width (\(w\)) = 4 cm	B1
Areas: 16 cm\(^2\) represents 32 offices (o.e.) or their \(h = \dfrac{6}{\text{their } w}\) (3sf) or \(\dfrac{8}{3.2} \times 0.6\)	M1
So height (\(h\)) = 1.5 cm	A1
	(3)

Notes

M1 for a correct calculation of areas 1 cm\(^2\) = 2 offices (o.e.)

A1 for \(h\) = 1.5 cm (Correct answer only 2/2)

Mark scheme (b)

Scheme	Marks
e.g. \((45) + \dfrac{20}{25} \times 5\) or \((50) - \dfrac{5}{25} \times 5\) (o.e.) ; = (£) 49	M1; A1
	(2)

Notes

M1 for a correct expression without end point. Allow “\(n\) + 1” so e.g. \((45) + \frac{20.5}{25} \times 5\)

A1 for 49 or, if ( \(n\) + 1) used, allow 49.1 (Correct answer of 49 only 2/2)

Mark scheme (c)

Scheme	Marks
\(\dfrac{\sum \mathrm{f}y}{90} = \dfrac{4420}{90}\), = (£) 49.11 (or better) (Allow \(\dfrac{442}{9}\) or \(49\tfrac{1}{9}\))	M1, A1
	(2)

Notes

M1 for an attempt at \(\frac{\sum \mathrm{f}y}{90}\) with at least 3 correct products of \(\sum \mathrm{f}y\) or \(4000 \leqslant \sum \mathrm{f}y \leqslant 5000\)

A1 for 49.11 (Allow 49.1 from correct working) (Correct answer only 2/2, 49.1 only M1A0)

Mark scheme (d)

Scheme	Marks
\(\sqrt{\dfrac{226687.5}{90} - \bar{x}^2} = \sqrt{106.8487\ldots}\) , = 10.3367 = awrt (£) 10.3	M1, A1
	(2)

Notes

M1 for a correct expression including \(\sqrt{\ }\), ft their mean. Allow use of \(s\)

A1 for awrt 10.3 Allow \(s\) = awrt 10.4 if clearly used. [NB use of 49.1 gives 10.389 \(\Rightarrow\) A0]
(Correct answer of 10.3 with no working is 2/2)

Mark scheme (e)

Scheme	Marks
Mean \(\approx\) median so distribution is symmetric (no skew or very little skew) [Allow mean > median or \(k(\bar{x} - Q_2)\) (\(k\)>0) so +ve skew if compatible with their figures] [If using quartiles we must see \(Q_1 = 44.0\) and \(Q_3 = 55.5\) used]	B1ft
	(1)

Notes

B1ft for reason and “symmetric” (or other correct) statement [Allow positive skew]
Allow ft of their (b) and their (c). For “symmetric” need \(|\bar{x} - Q_2| \lt 1\) “correlation” is B0

Mark scheme (f)

Scheme	Marks
Symmetric ( or little skew) so normal (or Rika’s suggestion) may be suitable	B1ft
	(1)

Notes

B1ft Suggest normal is or isn’t suitable with suitable reason based on (e) or mean and med

Mark scheme (g)

Scheme	Marks
\(\dfrac{c - 50}{10} = 0.8416\) [N.B. use of (1 – 0.8416) is B0]	M1, B1
\(c = 58.416\) = (£) 58.42 awrt 58.4	A1
	(3)
	(14 marks)

Notes

M1 for stand’ing using “\(c\)”, 50 and 10 and setting equal to \(\pm z\) value where \(0.84 \leqslant z \leqslant 0.85\)

B1 for using \(z = \pm 0.8416\) or better (calc gives 0.8416212…) in standard’ attempt e.g. \(\sqrt{10}\) for 10

A1 for awrt 58.4 (accept 3sf here) (Ans only of awrt 58.4 is M1B0A1 but 58.416 or better is 3/3)