Outliers and cleaning data - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 16 exam-style questions on Outliers and cleaning data for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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3–9 marks each
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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A council recorded the number of visitors to a park in a UK town on each of 30 days in June. On 29 of the days there were between 300 and 700 visitors. On the other day there was a music festival in the park, and there were 4200 visitors.

The council wants to summarise the number of visitors on a typical day.

State, with a reason, which measure of location and which measure of spread the council should use.

Describe the effect on the mean and on the standard deviation of removing the festival day from the data.

Model answer

The council should use the median and the interquartile range, because the festival day (4200 visitors) is an extreme value which is not typical, and these measures are not affected by it.
Removing the festival day would make the mean much smaller and the standard deviation much smaller, because 4200 is far above all the other values.

Explanation

The mean and standard deviation use every value, so one extreme value can change them greatly, while the median and interquartile range are resistant to it. Whether to remove the festival day depends on the purpose: it is a genuine value, but it is not a typical day. Either way, the summary should say how it was treated.

Mark scheme

1 mark
B1 The median (for location) and the interquartile range (for spread)
Both needed. Accept the 10th to 90th interpercentile range for spread.
1 mark
B1 Because they are not affected by the extreme value (the festival day), which is not typical
Must refer to the outlier or extreme value (4200, the festival day).
1 mark
B1 Removing the festival day would reduce both the mean and the standard deviation (by a large amount)
Both effects needed.

Question 2

4 marks

A cycling club asked its 30 members how far they cycled last week. Only 26 members replied. Their distances have mean 84 km and standard deviation 23 km.

The club secretary replaces each of the 4 missing values by 84 km and then calculates the mean and the standard deviation of all 30 values.

Find the mean and the standard deviation that the secretary obtains.

Comment on the secretary's method.

[Standard deviation: \(\sqrt{\dfrac{S_{xx}}{n}}\) or \(\sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}\), where \(S_{xx} = \sum x^2 - \dfrac{(\sum x)^2}{n}\)]

Model answer

The 4 new values all equal the mean, so the mean is still 84 km.
They add nothing to \(S_{xx}\): \(S_{xx} = 26 \times 23^2 = 13\,754\)
Standard deviation \(= \sqrt{\dfrac{13\,754}{30}} = 21.4\) km (3 s.f.)
This makes the distances appear less variable than they really are, and the members who did not reply may have cycled different distances (for example, less) from those who replied, so the results may be biased.

Explanation

Values equal to the mean have zero deviation, so they leave the mean and \(S_{xx}\) unchanged but increase n, which reduces the standard deviation. Replacing missing values by the mean therefore understates the spread. It also assumes the non-responders are typical, which may not be true.

Mark scheme

1 mark
B1 Mean \(= 84\) km (unchanged); answers of 84 and 21.4 with no working score 3 marks
Answers of 84 km and 21.4 km with no working score 3 marks (this B1, the M1 and the A1).
1 mark
M1 \(S_{xx}\) is unchanged by adding values equal to the mean: \(S_{xx} = 26 \times 23^2 = 13\,754\), and standard deviation \(= \sqrt{\dfrac{13\,754}{30}}\)
Or reconstructs \(\sum x^2 = 26(23^2 + 84^2) + 4 \times 84^2\) and uses \(\sqrt{\dfrac{\sum x^2}{30} - 84^2}\).
1 mark
A1 awrt 21.4 km, i.e. in \([21.35, 21.45)\)
Allow 21.8 (awrt 21.8, i.e. in [21.75, 21.85)) if s is used for the 30 values.
1 mark
B1 The method makes the data look less spread out than they are (the standard deviation is too small), and the 4 members who did not reply may not be typical (e.g. they may have cycled little), so the mean may be biased
Needs at least one valid criticism: the standard deviation is underestimated, or the missing members may differ from those who replied.

Question 3

6 marks

The best long jump distances, \(x\) m, of 25 athletes in a competition were summarised by
\[n = 25 \qquad \sum x = 128.4 \qquad \sum x^2 = 691.1736\]
One athlete fouled every jump, and this was recorded as a distance of 0 m.

Explain why the value of 0 should be removed before summary statistics are calculated.

Find the mean and the standard deviation of the distances of the other 24 athletes, and describe the effect that including the 0 had on the mean and on the standard deviation.

[Standard deviation: \(\sqrt{\dfrac{S_{xx}}{n}}\) or \(\sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}\), where \(S_{xx} = \sum x^2 - \dfrac{(\sum x)^2}{n}\)]

Model answer

A foul is not a distance jumped, so the 0 is not a valid value of x and would distort the summary of the distances.
Without the 0: \(n = 24\), \(\sum x = 128.4\), \(\sum x^2 = 691.1736\)
Mean \(= \dfrac{128.4}{24} = 5.35\) m
\(\sigma = \sqrt{\dfrac{691.1736}{24} - 5.35^2} = \sqrt{28.7989 - 28.6225} = \sqrt{0.1764} = 0.42\) m
With the 0 included, the mean was \(\dfrac{128.4}{25} = 5.14\) m and the standard deviation was \(\sqrt{\dfrac{691.1736}{25} - 5.136^2} = 1.13\) m, so including the 0 lowered the mean and nearly tripled the standard deviation.

Explanation

Removing a value of 0 leaves \(\sum x\) and \(\sum x^2\) unchanged but reduces n by 1. A single invalid value far from the rest pulls the mean towards it and inflates the standard deviation greatly. This is why data should be cleaned of values that are not genuine observations before summarising.

Mark scheme

1 mark
B1 0 is not a genuine jump distance (the athlete did not record a valid jump), so it is not part of the data on distances jumped; answers of 5.35 and 0.42 with no working score 4 marks
Answers of 5.35 m and 0.42 m (or s = 0.429) with no working score 4 marks (the two M marks and two A marks). Accept "it is an invalid value or an error".
1 mark
M1 Removes the 0: \(n = 24\), \(\sum x = 128.4\) and \(\sum x^2 = 691.1736\) unchanged, mean \(= \dfrac{128.4}{24}\)
Dividing by 25 is M0.
1 mark
A1 Mean \(= 5.35\) m
cao.
1 mark
M1 \(\sqrt{\dfrac{691.1736}{24} - 5.35^2}\)
Ft their mean. Square root needed.
1 mark
A1 Standard deviation \(= 0.42\) m
Accept 0.420. Allow s = 0.429.
1 mark
B1 Including the 0 reduced the mean (to 5.14 m) and greatly increased the standard deviation (to 1.13 m)
Both effects needed. The values with the 0 included (mean 5.136, standard deviation 1.126) need not be calculated, but if given must be correct.

Question 4

8 marks

In a reaction test, a ruler is dropped and a student catches it as quickly as possible. The distance, \(x\) cm, that the ruler falls before it is caught is recorded for each of 11 students. For these data
\[n = 11 \qquad \sum x = 207.8 \qquad \sum x^2 = 4131.36\]
The lower quartile is 16.5 cm and the upper quartile is 19.6 cm. The largest distance is 31.5 cm and the second largest is 20.4 cm.

Using the rule that an outlier is a value more than \(1.5 \times\) the interquartile range above the upper quartile, show that 31.5 is an outlier and that no other distance is an outlier under this rule.

Find the mean and the standard deviation of \(x\).

Ryan uses instead the rule that an outlier is a value more than 3 standard deviations above the mean. Determine whether 31.5 is an outlier under Ryan's rule.

Explain why the two rules give different conclusions about the value 31.5.

[Standard deviation: \(\sqrt{\dfrac{S_{xx}}{n}}\) or \(\sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}\), where \(S_{xx} = \sum x^2 - \dfrac{(\sum x)^2}{n}\)]

Model answer

IQR \(= 19.6 - 16.5 = 3.1\)
Upper limit \(= 19.6 + 1.5 \times 3.1 = 24.25\)
\(31.5 > 24.25\), so 31.5 is an outlier; the next largest distance is 20.4, which is less than 24.25, so no other distance is an outlier under this rule.
\(\bar{x} = \dfrac{207.8}{11} = 18.89\) cm, so the mean is 18.9 cm (3 s.f.)
\(\sigma = \sqrt{\dfrac{4131.36}{11} - 18.89^2} = \sqrt{18.71} = 4.33\) cm (3 s.f.)
Ryan's limit \(= 18.89 + 3 \times 4.326 = 31.9\)
\(31.5 < 31.9\), so 31.5 is not an outlier under Ryan's rule.
The value 31.5 is included when the mean and standard deviation are found. Because it is so far from the other distances, it makes the standard deviation much larger (without it the other 10 distances have a standard deviation of only 1.76 cm), so Ryan's limit rises above 31.5. The quartiles depend only on the middle of the ordered data, so one extreme value hardly affects the first rule.

Explanation

The interquartile-range rule uses the quartiles, which do not depend on how large the most extreme value is, so a single very large value does not move the limit. Ryan's rule uses the mean and standard deviation, and the extreme value itself inflates the standard deviation, so in a small data set an outlier can hide itself: here the limit rises to 31.9 cm, just above 31.5 cm. A common error is to apply the standard deviation rule with the median, or to forget the square root when finding the standard deviation.

Mark scheme

1 mark
M1 Finds the upper limit: \(19.6 + 1.5 \times (19.6 - 16.5)\) \((= 19.6 + 4.65)\). An answer with no working scores 0 marks for showing that 31.5 is an outlier; a mean of 18.9 cm and a standard deviation of 4.33 cm with no working score the B1, M1 and A1 marks for them.
An answer with no working scores 0 marks for the outlier part, which must be shown; a mean of 18.9 cm and a standard deviation of 4.33 cm with no working score 3 marks (B1 M1 A1). IQR = 3.1.
1 mark
A1* Limit 24.25: \(31.5 > 24.25\), so 31.5 is an outlier; the next largest distance, 20.4, is less than 24.25, so no other distance is above the limit
cso. Must refer to 20.4 (or to every other distance being at most 20.4). Only the rule for large values is used here, as the question states.
1 mark
B1 Mean \(= \dfrac{207.8}{11} =\) awrt 18.9 cm, i.e. in [18.85, 18.95)
\(\dfrac{207.8}{11} = 18.890...\)
1 mark
M1 \(\sqrt{\dfrac{4131.36}{11} - 18.89...^2}\) (or \(\sqrt{\dfrac{S_{xx}}{11}}\) with \(S_{xx} = 4131.36 - \dfrac{207.8^2}{11}\))
Square root needed. Ft their mean.
1 mark
A1 Standard deviation \(=\) awrt 4.33 cm, i.e. in [4.325, 4.335)
\(\sqrt{18.711...} = 4.3257...\). Allow s = 4.54 (awrt 4.54, i.e. in [4.535, 4.545)): the conclusion under Ryan's rule is the same.
1 mark
M1 Ryan's limit: \(18.89 + 3 \times 4.326\) (or \(18.89 + 3 \times 4.537\) with s)
Ft their mean and standard deviation. Using 2 standard deviations, or the median, scores M0.
1 mark
A1 Limit awrt 31.9, i.e. in [31.85, 31.95) (or awrt 32.5, i.e. in [32.45, 32.55), with s): \(31.5 < 31.9\), so 31.5 is not an outlier under Ryan's rule
cso. Needs the limit and the conclusion. \(18.890... + 3 \times 4.3257... = 31.868...\); with s, 32.50...
1 mark
B1 The value 31.5 is itself used to find the mean and standard deviation, and because it is so far from the other values it makes the standard deviation (and the mean) much larger, which raises Ryan's limit above 31.5; the quartiles hardly depend on the size of one extreme value, so the first rule is not affected in this way
Must refer to the extreme value itself inflating the standard deviation (or the mean and standard deviation). Accept, e.g., "without 31.5 the other 10 distances have a standard deviation of only 1.76 cm". "The rules are different" with no reason scores B0.

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