The five-number summary and box-and-whisker diagram, variance and standard deviation, drawing and reading an ogive, recognising skewness, and identifying outliers.
1The five-number summary
IQR = Q₃ − Q₁
The box holds the middle 50% of the data. The line inside it is the median — here 33, so IQR = 42 − 24 = 18.
2Skewness
- Symmetric — median sits in the middle of the box, whiskers about equal.
- Skewed right (positively) — the longer whisker is on the right; mean > median.
- Skewed left (negatively) — the longer whisker is on the left; mean < median.
The skew is named after the direction of the tail, not the bulk of the data. A long right tail means skewed right.
3Variance and standard deviation
Standard deviation measures spread about the mean. A small σ means the data is tightly bunched.
- 1Mean = 198/9 = 22,00.Add them and divide by 9.
- 2Variance = (∑(x − 22)²)/9 = 44,00.Square each deviation, average them.
- 3σ = √44,00 = 6,63.Use the calculator's σn key in the exam — it is much faster.
- 4One standard deviation from the mean: [15,37 ; 28,63].5 of the 9 values lie in this interval.
4Ogives (cumulative frequency curves)
An ogive always starts on the horizontal axis at the lower boundary of the first interval and ends at the total frequency — here 40.
- 1The total is 40, so the median is at the 40/2 = 20th value.
- 2Read across from 20 on the vertical axis to the curve, then down.
- 3The median is approximately 25 marks.The curve passes through (20 ; 13) and (30 ; 27): a cumulative frequency of 20 is reached halfway between 20 and 30 marks — about 25.
Points on an ogive go at the upper boundary of each interval, never the midpoint. Using midpoints shifts the whole curve and every reading off it.
5Outliers
- 1IQR = 42 − 24 = 18.
- 21,5 × 18 = 27.
- 3Lower: 24 − 27 = −3. Upper: 42 + 27 = 69.
- 4Since the data runs 12 to 55, there are no outliers.Everything lies inside the boundaries.
6Grouped data: the estimated mean
Grade 10 showed you grouped data; Grade 11 revises it and expects the working set out properly. Once data is in intervals the individual values are gone, so the mean can only be estimated — using the midpoint of each interval as a stand-in for every value in it.
- 1Midpoints: 10, 30, 50, 70, 90Halfway across each interval.
- 2Σ(x·f) = 10(6) + 30(10) + 50(14) + 70(9) + 90(5) = 60 + 300 + 700 + 630 + 450 = 2 140
- 3Σf = 6 + 10 + 14 + 9 + 5 = 44The total number of learners.
- 4Estimated mean = 2 140 ÷ 44 = 48,6
- 5The modal class is 40–60, and the median (22nd and 23rd values) also lies in 40–60.Cumulative: 6, 16, 30 — the middle values fall in the third interval.
7Frequency polygons
A frequency polygon joins the midpoints of the tops of the histogram bars with straight lines, and is brought down to the horizontal axis at each end. It shows the shape of a distribution more clearly than the bars do, and lets you compare two data sets on one set of axes.
- Plot each point at the midpoint of the interval, at the height of its frequency.
- Join the points with straight lines — never a smooth curve. A smooth curve is an ogive, which plots cumulative frequency.
- Bring the polygon down to the axis one half-interval before the first bar and after the last, so the area under it equals the area of the bars.
Plotting a frequency polygon at the end of each interval instead of the midpoint. The whole point of the midpoint is that it represents the interval; plotting at the boundary shifts the entire distribution half an interval to the right.
Know which picture answers which question: a histogram or polygon shows frequency and therefore the modal class; an ogive shows cumulative frequency and therefore the median, quartiles and percentiles.
Practice exercises
Work each one out, then click to reveal the answer.
- 1The mean of 64, 71, 58, 67 and x is 66. Solve for x.
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(64 + 71 + 58 + 67 + x)/5 = 66 (1)
260 + x = 66 × 5 = 330 (1)
x = 330 − 260 = 70 (1) - 2For the box-and-whisker diagram below, write down the five-number summary.
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min = 12, Q₁ = 24, median = 33, Q₃ = 42, max = 55 - 3Calculate the interquartile range from the diagram below.
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42 − 24 = 18 - 4Calculate the mean of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
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198/9 = 22,00 - 5Calculate the standard deviation of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
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variance = 44,00 → σ = 6,63 - 6How many data values lie within one standard deviation of the mean for the data above?
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Interval [15,37 ; 28,63] contains 5 values - 7Use the ogive below to estimate the median mark.
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The 20th of 40 values → ≈ 25 marks - 8Use the ogive below to write down the total number of learners.
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40 - 9Determine the outlier boundaries for the box plot below.
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IQR = 18, 1,5 × 18 = 27 → below −3 or above 69; there are no outliers - 10Estimate the mean of: 0–10 (f = 4), 10–20 (f = 9), 20–30 (f = 12), 30–40 (f = 5).
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Midpoints 5, 15, 25, 35 ✓; ∑(x·f) = 20 + 135 + 300 + 175 = 630 ✓; ∑f = 30 ✓; mean = 630 ÷ 30 = 21 ✓ - 11Why can the mean of grouped data only be ESTIMATED?
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The individual values are lost once the data is grouped ✓; the midpoint is used to stand in for every value in the interval, which is an approximation ✓ - 12At which point of each interval must a frequency polygon be plotted, and why?
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At the midpoint ✓, because the midpoint represents the interval — plotting at a boundary shifts the whole distribution ✓ - 13State ONE difference between a frequency polygon and an ogive.
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A frequency polygon plots frequency and is drawn with straight lines between midpoints ✓; an ogive plots CUMULATIVE frequency and is used to read off the median and quartiles ✓
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