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Calculating and interpreting measures of location (mean, median, mode) and spread (range, interquartile and interpercentile ranges, variance and standard deviation) for raw, grouped and summarised data, including coding. These skills are used throughout the statistics on Paper 2, often in the first part of a longer question.
Key facts
\(\bar{x} = \dfrac{\Sigma x}{n}\) or \(\dfrac{\Sigma fx}{\Sigma f}\)
Mean \(\bar{x} = \dfrac{\Sigma x}{n}\), or \(\dfrac{\Sigma fx}{\Sigma f}\) for a frequency table. For grouped data use the class midpoints: the result is an estimate.
The mean uses all the data but is affected by extreme values; the median is not. The mode (or modal class) suits discrete or non-numerical data.
Quartiles and percentiles
diagram
For a list of \(n\) ordered values, to find \(Q_1\) work out \(\tfrac{n}{4}\). If it is a whole number, \(Q_1\) is halfway between that value and the next one; if not, round up and take that value. Use \(\tfrac{n}{2}\) for the median and \(\tfrac{3n}{4}\) for \(Q_3\) in the same way.
For grouped continuous data use linear interpolation: \(Q_1\) is the \(\tfrac{n}{4}\)th value, the median the \(\tfrac{n}{2}\)th, \(Q_3\) the \(\tfrac{3n}{4}\)th and the \(k\)th percentile the \(\tfrac{kn}{100}\)th.
Estimate = lower class boundary + \(\dfrac{d}{f}\) × class width, where \(d\) is the position minus the cumulative frequency before the class, and \(f\) is the frequency of the class.
The 12 values in the class are assumed evenly spread: the 25th value is 5 of the 12 steps along
IQR \(= Q_3 - Q_1\); the 10th to 90th interpercentile range is \(P_{90} - P_{10}\)
For a frequency table, \(\sigma^2 = \dfrac{\Sigma fx^2}{\Sigma f} - \bar{x}^2\)
Standard deviation \(\sigma = \sqrt{\text{variance}}\): the spread about the mean, in the same units as the data.
Deviations x − x̄ always add to 0, so they are squared: σ² = Σ(x − x̄)² ÷ n
Edexcel's formula uses the divisor \(n\) (\(\sigma_x\) on a calculator); answers using \(n - 1\) (\(s_x\)) are normally accepted too.
Coding
If \(y = \dfrac{x - a}{b}\) (with \(b \gt 0\)), then \(\bar{y} = \dfrac{\bar{x} - a}{b}\) and \(\sigma_y = \dfrac{\sigma_x}{b}\). Decode with \(\bar{x} = a + b\bar{y}\) and \(\sigma_x = b\sigma_y\)
Adding or subtracting a constant does not change the spread; multiplying or dividing scales it.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the mean and standard deviation from summary statistics
3 marksGrade C
Mean \(= \Sigma x \div n\)
Variance \(= \Sigma x^2 \div n - \bar{x}^2\), using the unrounded mean.
Square root, and give units and 3 significant figures.
Example. For a random sample of 10 days, the daily total sunshine, \(h\) hours, gives \(\Sigma h = 68.0\) and \(\Sigma h^2 = 518.6\). Find the mean and the standard deviation of \(h\)
Decode the mean: apply both the multiplying and the adding.
Decode the standard deviation: apply only the multiplying.
Example. The masses, \(m\) grams, of 40 apples are coded using \(y = \dfrac{m - 150}{5}\). For the coded data, \(\bar{y} = 2.4\) and \(\sigma_y = 1.8\). Find the mean and the standard deviation of the masses.
Show the model answerHide the model answer
\(m = 5y + 150\), so \(\bar{m} = 5 \times 2.4 + 150\) (M1) \(= 162\) g
(A1)
\(\sigma_m = 5 \times 1.8 = 9\) g
(B1)
Combine two sets of data
5 to 6 marksGrade A
For each set find \(n\), \(\Sigma x\) and \(\Sigma x^2\), using \(\Sigma x = n\bar{x}\) and \(\Sigma x^2 = n(\sigma^2 + \bar{x}^2)\) if needed.
Add the values of \(n\), \(\Sigma x\) and \(\Sigma x^2\)
Calculate the combined mean and standard deviation.
Example. Sample A has 12 values with mean 20 and standard deviation 3. Sample B has 8 values with \(\Sigma x = 176\) and \(\Sigma x^2 = 3960\). Find the mean and the standard deviation of all 20 values.
Show the model answerHide the model answer
Sample A: \(\Sigma x = 12 \times 20 = 240\) (M1) and \(\Sigma x^2 = 12(3^2 + 20^2) = 4908\)
(M1)
Combined: \(\Sigma x = 416\),\(\Sigma x^2 = 8868\), so the mean is \(\dfrac{416}{20} = 20.8\)
Forgetting to square root the variance, or subtracting \(\bar{x}\) instead of \(\bar{x}^2\)
Rounding the mean before using it in the variance formula.
Using class limits instead of class boundaries, or the wrong class width, when interpolating.
Applying the added constant to the standard deviation when decoding.
Dividing by the number of classes instead of \(\Sigma f\)
More tips
Memory tricks
Variance is ‘the mean of the squares minus the square of the mean’.
Use your calculator's statistics mode (with a frequency column) to check \(\bar{x}\),\(\sigma_x\), \(\Sigma x\) and \(\Sigma x^2\)
Coding: an added or subtracted constant moves the mean but never changes the standard deviation.
Working backwards from a mean and standard deviation: \(\Sigma x = n\bar{x}\) and \(\Sigma x^2 = n(\sigma^2 + \bar{x}^2)\)
Exam technique
Give answers to 3 significant figures unless told otherwise, and keep full calculator values between steps.
When asked to choose a measure, justify it: the median and IQR are not affected by outliers or skew; the mean and standard deviation use all the data.
Show the formula with the numbers substituted: the method mark is for that line, even if you then use your calculator.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Find the mean, median and mode of dataMean \(= \dfrac{\Sigma x}{n}\); the median is the middle value of the ordered data; the mode is the most common value.
Grade E
Find quartiles and the interquartile rangeUse the rule for positions in an ordered list, then IQR \(= Q_3 - Q_1\)
Grade D
Estimate the mean from grouped dataUse the class midpoints: \(\bar{x} \approx \dfrac{\Sigma fx}{\Sigma f}\)
Grade C
Calculate standard deviation from summary statisticsUse \(\sigma = \sqrt{\dfrac{\Sigma x^2}{n} - \bar{x}^2}\), or \(\sqrt{\dfrac{S_{xx}}{n}}\)
Grade C
Interpolate medians and percentiles from grouped dataFind the position, locate its class, then go the right fraction of the way through that class.
Grade B
Use coding to find mean and standard deviationReverse the coding for the mean; for the standard deviation only the scaling matters.
Grade A
Combine data sets using \(\Sigma x\) and \(\Sigma x^2\)Rebuild \(\Sigma x = n\bar{x}\) and \(\Sigma x^2 = n(\sigma^2 + \bar{x}^2)\) for each set, add them, then recalculate.
Grade A*
Find unknown data values from summary statisticsForm and solve equations (sometimes quadratics) using \(\Sigma x\),\(\Sigma x^2\), the mean or the variance.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The numbers of goals scored in a random sample of football matches are: 1, 3, 4, 4, 4, 5, 4, 3, 2, 4
(a) Find the mean, giving your answer to 3 significant figures.[2]
(b) Find the median and the mode.[2]
Show the answer and mark scheme
(a)Answer: 3.40
M1 for \(\dfrac{34}{10}\)
A1 for awrt 3.40
(b)Answer: Median 4, mode 4
B1 for median 4
B1 for mode 4
Question 2Medium4 marks
The table shows the number of cups of coffee drunk yesterday by each of a random sample of office workers.
[object Object]
(a) Calculate the mean and the standard deviation of these data, giving your answers to 3 significant figures.[3]
(b) One more office worker, who drank no coffee yesterday, is added to the sample. Without calculating the new mean, state whether the mean number of cups of coffee increases, decreases or stays the same. Give a reason for your answer.[1]
Show the answer and mark scheme
(a)Answer: Mean 1.59, standard deviation 1.17
M1 for \(\dfrac{\Sigma fx}{\Sigma f} = \dfrac{92}{58}\) or a correct method for the standard deviation, e.g. \(\sqrt{\dfrac{226}{58} - \bar{x}^2}\)
A1 for mean awrt 1.59
A1 for standard deviation awrt 1.17 (allow 1.19 using \(n - 1\))
(b)Answer: The mean decreases, because the new value, 0, is less than the current mean of 1.59.
B1 for “decreases” with a reason: 0 is less than the mean (1.59)
Question 3Hard4 marks
The table shows the daily mean pressures for a random sample of days that a student took from the large data set for Heathrow in 2015 (illustrative values). One of the frequencies is unknown. Using linear interpolation, with the median at position \(\tfrac{n}{2}\), where \(n\) is the total frequency, an estimate of the median daily mean pressure is 1016.1 hPa.
[object Object]
Find the value of \(x\).[4]
Show the answer and mark scheme
Answer: \(x = 12\)
M1 for a total frequency of \(85 + x\) and the median position \(\tfrac{85 + x}{2}\)
M1 for a correct interpolation equation, e.g. \(1015 + \dfrac{\tfrac{1}{2}(85 + x) - (31 + x)}{25} \times 5 = 1016.1\)
dM1 for solving their linear equation
A1 for \(x = 12\)
Worked solution: The median lies in the class 1015 ≤ p < 1020, so \(1015 + \dfrac{\tfrac{1}{2}(85 + x) - (31 + x)}{25} \times 5 = 1016.1\), which gives \(x = 12\).