S1 January 2005 Q3

3. The following table shows the height \(x\), to the nearest cm, and the weight \(y\), to the nearest kg, of a random sample of 12 students.

\(x\)	148	164	156	172	147	184	162	155	182	165	175	152
\(y\)	39	59	56	77	44	77	65	49	80	72	70	52

(a) On graph paper, draw a scatter diagram to represent these data. (3)

(b) Write down, with a reason, whether the correlation coefficient between \(x\) and \(y\) is positive or negative. (2)

The data in the table can be summarised as follows.

\[\Sigma x = 1962, \quad \Sigma y = 740, \quad \Sigma y^2 = 47\,746, \quad \Sigma xy = 122\,783, \quad S_{xx} = 1745.\]

The equation of the regression line of \(y\) on \(x\) is \(y = -106.331 + bx\).

(d) Find, to 3 decimal places, the value of \(b\). (2)

(e) Find, to 3 significant figures, the mean \(\bar{y}\) and the standard deviation \(s\) of the weights of this sample of students. (3)

(f) Find the values of \(\bar{y} \pm 1.96s\). (2)

(g) Comment on whether or not you think that the weights of these students could be modelled by a normal distribution. (1)

Scheme	Marks
Scales and labels	B1
10 or 11 points correct, all correct	B1,B1
	(3)

Scheme	Marks
Positive; as \(x\) increases, \(y\) increases	B1;B1
	(2)

Scheme	Marks
\(S_{xy} = 122783 - \dfrac{1962 \times 740}{12} = 1793\)	M1A1
	(2)

Scheme	Marks
\(b = \dfrac{S_{xy}}{S_{xx}} = \dfrac{1793}{1745} = 1.027507\ldots\)	M1A1
	(2)

Mark scheme (e)

Scheme	Marks
\(\bar{y} = \dfrac{740}{12} = 61\tfrac{2}{3}\)	B1
\(s = \sqrt{\dfrac{47746}{12} - \left(\dfrac{740}{12}\right)^2} = 13.26859\ldots\)	M1A1
	(3)

B1 \(61\tfrac{2}{3}\) or \(61.\dot{6}\) or 61.7

M1A1 use of formula including root, 13.3

Scheme	Marks
35.7, 87.7	B1B1
	(2)

Scheme	Marks
All values between 35.7 and 87.7 so could be normal.	B1
	(1)
	(15 marks)