Statistics: Data Presentation and Interpretation
Data presentation and interpretation covers the methods used to summarise, display and describe a set of data, including measures of location and spread, correlation, and diagrams such as histograms and box plots. A Level Statistics also covers identifying outliers, coding data, and interpreting scatter diagrams and box plots in context.
Method
- Calculate the mean using (sum of the values) / n, and for grouped data use the midpoint of each class in place of the actual values.
- Calculate the standard deviation using sqrt[(sum of x^2)/n - mean^2], using the midpoint of each class for grouped data.
- To find the median and quartiles from a list, use their position in the ordered data; for grouped data, use linear interpolation within the class that contains the median or quartile.
- Identify outliers using a stated rule, e.g. more than 2 standard deviations from the mean, or more than 1.5 x IQR beyond the nearer quartile, and check each value against the calculated boundary.
- Use a scatter diagram to describe correlation as positive, negative or (approximately) zero, and comment on how an unusual point would affect the product moment correlation coefficient.
- When comparing two data sets (e.g. using box plots), compare a measure of location (mean or median) and a measure of spread (range or IQR), always writing the comparison in context.
Worked example
The times, in minutes, taken by 8 runners to complete a 5 km route are: 28, 31, 26, 35, 30, 27, 33, 29. (a) Calculate the mean time. (b) Calculate the standard deviation of these times. (c) A ninth runner joins with a time of 50 minutes, a clear outlier. Without recalculating, state and justify the effect on the mean and standard deviation.
- Sum of the 8 times = 28+31+26+35+30+27+33+29 = 239. Mean = 239/8 = 29.875 minutes.
- Sum of squares = 28^2+31^2+26^2+35^2+30^2+27^2+33^2+29^2 = 784+961+676+1225+900+729+1089+841 = 7205.
- Variance = 7205/8 - 29.875^2 = 900.625 - 892.515625 = 8.109375. Standard deviation = sqrt(8.109375) = 2.848 (3dp).
- Adding 50 (well above the current mean of 29.875) would increase the mean, since it pulls the average up.
- The standard deviation would also increase, since 50 is a long way from the mean and adds a large deviation, increasing the overall spread.
Practice questions
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Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
The mass, in kg, of 10 sacks of potatoes are: 24, 26, 22, 29, 25, 23, 28, 27, 26, 30. (a) Calculate the mean mass. (2) (b) Calculate the standard deviation of the masses. (2)
The weekly hours worked by 11 part-time staff, listed in ascending order, are: 10, 12, 14, 15, 17, 18, 19, 21, 23, 25, 48. (a) Find the median, the lower quartile (Q1) and the upper quartile (Q3). (3) (b) Show that 48 hours is an outlier, using the rule that a value is an outlier if it lies more than 1.5 x IQR beyond the nearer quartile. (2)
Two classes, C and D, sat the same physics test out of 60 marks. Box plots of their results gave the following five-figure summaries. Class C: minimum = 15, Q1 = 24, median = 33, Q3 = 44, maximum = 52. Class D: minimum = 20, Q1 = 30, median = 34, Q3 = 37, maximum = 40. (a) Compare the medians and the interquartile ranges of the two classes, in the context of the test. (3) (b) Using the quartiles, comment on the skewness of each class's distribution of marks. (3)
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