October 2020 Paper 3 Q3

EdexcelCurrent spec10 marksData Processing

3. Each member of a group of 27 people was timed when completing a puzzle.

The time taken, \(x\) minutes, for each member of the group was recorded.

These times are summarised in the following box and whisker plot.

box and whisker plot on a scale labelled Time (x minutes) from 0 to 70: lower whisker end 7, lower quartile 14, median 20, upper quartile 25, upper whisker end 40, outliers marked with crosses at 46 and 68

(a) Find the range of the times. (1)

(b) Find the interquartile range of the times. (1)

For these 27 people \(\sum x = 607.5\) and \(\sum x^2 = 17\,623.25\)

(d) calculate the standard deviation of the times taken to complete the puzzle. (2)

Taruni defines an outlier as a value more than 3 standard deviations above the mean.

(e) State how many outliers Taruni would say there are in these data, giving a reason for your answer. (1)

Adam and Beth also completed the puzzle in \(a\) minutes and \(b\) minutes respectively, where \(a \gt b\).
When their times are included with the data of the other 27 people

the median time increases
the mean time does not change

(f) Suggest a possible value for \(a\) and a possible value for \(b\), explaining how your values satisfy the above conditions. (3)

(g) Without carrying out any further calculations, explain why the standard deviation of all 29 times will be lower than your answer to part (d). (1)

Scheme	Marks	AO
[\(68 - 7 =\)] 61 (only)	B1	1.1b
	(1)

Scheme	Marks	AO
[\(25 - 14\)] = 11	B1	1.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
\(\left[\mu \text{ or } \bar{x} = \dfrac{607.5}{27} =\right]\) = 22.5	B1	1.1b
	(1)

Notes

B1: for 22.5 only (or exact equivalent such as \(\frac{45}{2}\)). Allow 22 mins and 30 secs.

Mark scheme (d)

Scheme	Marks	AO
\(\sigma = \sqrt{\dfrac{17\,623.25}{27} - \text{``}22.5\text{''}^2}\) or \(\sqrt{146.4629\ldots}\)	M1	1.1b
= 12.10218… awrt 12.1	A1	1.1b
	(2)

Notes

M1: for a correct expression including square root. Allow \(\sqrt{146}\) or better. Ft their mean

A1: for awrt 12.1 NB Allow use of \(s = 12.3327\ldots\) or awrt 12.3

Mark scheme (e)

Scheme	Marks	AO
\(\mu + 3\sigma = \text{``}22.5\text{''} + 3 \times \text{``}12.1\ldots\text{''} = \text{awrt } 59\) so only one outlier	B1ft	1.1b
	(1)

Notes

B1ft: for a correct calculation or value based on their \(\mu\) and \(\sigma\) and compatible conclusion

Mark scheme (f)

Scheme	Marks	AO
Median increases implies that both values must be > 20	M1	3.1b
Mean is the same means that \(a + b = 45\)	M1	1.1b
So possible values are: e.g. \(b = 21\) and \(a = 24\) (o.e.)	A1	2.2b
	(3)

Notes

1^st M1: Correct start to the problem and a correct statement about the values based on median
Allow if their final two values are both >20

2^nd M1: for a correct explanation leading to equation \(a + b = 45\) (o.e. e.g. equidistant from mean)
Allow if their final two values sum to 45

A1: for a correct pair of values (both > 20 with a sum of 45) and at least some attempt to explain how their values satisfy at least one of the conditions (both > 20 or \(a + b = 45\)).
Ignore \(a =\) or \(b =\) labels

NB: The values for \(a\) and \(b\) do not need to be integers.

Mark scheme (g)

Scheme	Marks	AO
Both values will be less than 1 standard deviation from the mean and so the standard deviation of all 29 values will be smaller	B1	2.4
	(1)
	(10 marks)

Notes

B1: for a correct explanation.
Must mention that both values are less than 1 sd (ft their answer to (d)) from the mean