Edexcel A level Maths (9MA0) · Statistics › Data presentation and interpretation
Practise Outliers and cleaning data. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The marks out of 60 of a random sample of students in a test are listed in order. 28, 31, 32, 34, 36, 37, 38, 39, 40, 45, 53 An outlier is a value that is greater than \(Q_3 + 1.5 \times (Q_3 - Q_1)\) or less than \(Q_1 - 1.5 \times (Q_3 - Q_1)\).
(a) Find the lower quartile and the upper quartile.[2]
(b) Show that there is exactly one outlier, and state its value.[3]
A scientist measures the masses, \(x\) grams, of 20 seedlings. The results are summarised by \[\sum x = 356, \qquad \sum x^2 = 6532\]A student calculates the standard deviation as follows: \(\sigma = \sqrt{\frac{\sum x^2}{n}} - \bar{x} = \sqrt{\frac{6532}{20}} - 17.8 = 18.07 - 17.8 = 0.27\) g
(a) Explain the error in the student's method and find the correct standard deviation, giving your answer to 3 significant figures.[3]
(b) An outlier is defined as a value more than 2 standard deviations from the mean. Show that a seedling of mass 25 g is an outlier.[2]
(c) The scientist removes this seedling from the data. Find the mean and the standard deviation of the remaining 19 masses, giving your answers to 3 significant figures.[3]
(d) Give a reason why the scientist should check this seedling before deciding whether to remove it from the data.[1]
Show the answer and mark scheme
(a)Answer: 3.12 g
B1 for the correct formula is \(\sigma = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}\): the square of the mean must be subtracted inside the square root
M1 for \(\sqrt{\frac{6532}{20} - 17.8^2} = \sqrt{326.6 - 316.84}\)
A1 for 3.12 (g)
Worked solution: The variance is \(\frac{\sum x^2}{n} - \bar{x}^2\); the student took the square root before subtracting and subtracted \(\bar{x}\) instead of \(\bar{x}^2\). \(\sigma = \sqrt{326.6 - 17.8^2} = \sqrt{9.76} = 3.12\) g.
A1 for 24.05 (or 24.0 or 24.1) and a comparison with 25, with a conclusion
Worked solution: \(\bar{x} + 2\sigma = 17.8 + 2 \times 3.124 = 24.05\). Since \(25 \gt 24.05\), the mass 25 g is an outlier.
(c)Answer: Mean 17.4 g, standard deviation 2.72 g
M1 for \(\sum x = 331\) and \(\sum x^2 = 6532 - 625 = 5907\)
A1 for mean 17.4 (g)
A1 for standard deviation 2.72 (g)
Worked solution: \(\sum x = 356 - 25 = 331\), \(\sum x^2 = 6532 - 25^2 = 5907\). Mean \(= \frac{331}{19} = 17.4\) g. \(\sigma = \sqrt{\frac{5907}{19} - \left(\frac{331}{19}\right)^2} = \sqrt{7.40} = 2.72\) g.
(d)Answer: An outlier may be a genuine value (a naturally large seedling), not an error; it should only be removed if it is a mistake.
B1 for the value may be genuine (not an error in measuring or recording), in which case it should be kept
Worked solution: A seedling of 25 g may be a genuine, unusually large seedling. Outliers should only be removed if they are errors (for example in weighing or recording); removing genuine data would bias the results.
Question 3Hard8 marks
The daily maximum gust, \(g\) knots, is recorded for each of a random sample of 31 days that a student took from the large data set for Hurn in 2015. (The values are illustrative.) The largest value is 38. The values are summarised by \(\Sigma g = 689, \ \Sigma g^2 = 16{,}446\). An outlier is a value more than 2 standard deviations above or below the mean.
(a) Show that 38 is an outlier.[3]
(b) The value 38 is found to be a recording error and is removed. Find the mean and the standard deviation of the remaining values, giving your answers to 2 decimal places.[4]
(c) Comment on the changes to the mean and the standard deviation.[1]
Show the answer and mark scheme
(a)Answer: Mean 22.23, s.d. 6.04, so the upper limit is 34.31, and 38 is greater than this.
M1 for a correct method for the mean and standard deviation
A1 for mean awrt 22.2 and s.d. awrt 6.04 (or 6.14)
A1 for limit awrt 34.3 (or 34.5) and a correct conclusion
(b)Answer: Mean 21.70, standard deviation 5.40 (knots)
M1 for \(\Sigma g = 689 - 38 = 651\) and \(n = 30\)
M1 for \(\Sigma g^2 = 16{,}446 - 38^2 = 15{,}002\)
A1 for mean awrt 21.70
A1 for standard deviation awrt 5.40 (allow 5.49)
(c)Answer: Both decrease; the standard deviation decreases proportionally more, because it is based on squared deviations, so it is strongly affected by an extreme value.
B1 for both decreasing, with the standard deviation affected more (it is more sensitive to extreme values)