June 2022 Paper 3 Q18

AQACurrent spec11 marksData Processing Normal Distribution

18 In a particular year, the height of a male athlete at the Summer Olympics has a mean 1.78 metres and standard deviation 0.23 metres.

The heights of 95% of male athletes are between 1.33 metres and 2.22 metres.

(a) Comment on whether a normal distribution may be suitable to model the height of a male athlete at the Summer Olympics in this particular year. [3 marks]

(b) You may assume that the height of a male athlete at the Summer Olympics may be modelled by a normal distribution with mean 1.78 metres and standard deviation 0.23 metres.

(i) Find the probability that the height of a randomly selected male athlete is 1.82 metres. [1 mark]

(ii) Find the probability that the height of a randomly selected male athlete is between 1.70 metres and 1.90 metres. [1 mark]

(iii) Two male athletes are chosen at random.

Calculate the probability that both of their heights are between 1.70 metres and 1.90 metres. [1 mark]

(c) The summarised data for the heights, \(h\) metres, of a random sample of 40 male athletes at the Winter Olympics is given below.\[\sum h = 69.2 \qquad \sum (h - \bar{h})^2 = 2.81\]

Use this data to calculate estimates of the mean and standard deviation of the heights of male athletes at the Winter Olympics. [3 marks]

(d) Using your answers from part (c), compare the heights of male athletes at the Summer Olympics and male athletes at the Winter Olympics. [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Calculates 1.78 ± 2 × 0.23 or 1.78 ± 1.96 × 0.23 or calculates \(P(1.33 \lt x \lt 2.22)\) PI by 0.9469 or 0.947 or calculates \(\dfrac{1.33 - 1.78}{0.23}\) and \(\dfrac{2.22 - 1.78}{0.23}\)	M1	3.1b
Obtains 1.32 and 2.24 and states they are approximately 1.33 and 2.22 or obtains 1.33 and 2.23 and states they are approximately 1.33 and 2.22 or obtains 0.9469 or 0.947 and states is approximately 0.95 or obtains −1.96 and 1.91 and states are approximately −2 and 2	A1	2.4
Infers that the normal distribution may be suitable because height is continuous data and 95% of heights lies within two standard deviations of the mean or Infers that the normal distribution may be suitable because height is continuous data and 94.69% or 94.7% of heights lies between 1.33 and 2.22	R1	2.2b
	(3)

Typical solution

\[1.78 - 2 \times 0.23 = 1.32\]\[1.78 + 2 \times 0.23 = 2.24\]\[1.32 \approx 1.33\]\[2.24 \approx 2.22\]

Height is continuous data and 95% of heights lies within two standard deviations of the mean so normal may be a suitable model.

Mark scheme (b)

Scheme	Marks	AO
(i) States 0	B1	1.2
	(1)
(ii) Calculates the correct probability AWFW [0.335, 0.34]	B1	1.1b
	(1)
(iii) Finds the value of their answer to (b)(ii) squared Their answer must be correct to at least 2sf	B1F	3.1b
	(1)

Typical solution

(i)

(ii)

0.335

(iii)

\[0.335^2 = 0.112\]

Mark scheme (c)

Scheme	Marks	AO
Obtains 1.73 CAO Ignore missing or incorrect units	B1	1.1b
Uses the correct formula for standard deviation eg \(s = \sqrt{\dfrac{2.81}{39}}\) Do not allow variance = \(\sqrt{\dfrac{2.81}{40}}\)	M1	1.1a
Obtains the correct standard deviation AWFW [0.265, 0.27] Allow if not labelled but if labelled, must be correct Ignore missing or incorrect units	A1	1.1b
	(3)

Typical solution

Mean = 1.73

Standard deviation \(= \sqrt{\dfrac{2.81}{40}}\)

\(=\) 0.265

Mark scheme (d)

Scheme	Marks	AO
Uses their mean and 1.78 to compare heights Comparison must include on average. Follow through their answer to part (c) Do not allow ‘general’ Allow statement that they’re about the same on average	E1F	2.2b
Uses their standard deviation and 0.23 to compare heights Comparison must include ‘varies’, ‘spread’ ‘disperse’ ‘more variation’ or ‘consistent’ Follow through their answer to part (c) Allow statement that they are about the same Do not allow comparison that includes ‘range’ or ‘variety’	E1F	2.2b
	(2)
	(11 marks)

Typical solution

Summer athletes are taller on average than Winter athletes.

Summer athletes’ heights are less varied than the heights of Winter athletes.