Statistics: Spread, Ogives & Box Plots (Grade 11)

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

minimum · Q₁ · median · Q₃ · maximum

IQR = Q₃ − Q₁

A box-and-whisker diagram for the five-number summary 12 ; 24 ; 33 ; 42 ; 55.

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.

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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

variance = ∑(x − x̄)²n · σ = variance

Standard deviation measures spread about the mean. A small σ means the data is tightly bunched.

Worked ExampleCalculate σ for 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33

1
Mean = 1989 = 22,00.
Add them and divide by 9.
2
Variance = ∑(x − ${R(mean)})²9 = 44,00.
Square each deviation, average them.
3
σ = 44,00 = 6,63.
Use the calculator's σ_n key in the exam — it is much faster.
4
One standard deviation from the mean: [15,37 ; 28,63].
5 of the 9 values lie in this interval.

4Ogives (cumulative frequency curves)

Ogive for 40 learners' marks. Plot each point at the UPPER boundary of its interval.

An ogive always starts on the horizontal axis at the lower boundary of the first interval and ends at the total frequency — here 40.

Worked ExampleUse the ogive to estimate the median

1
The total is 40, so the median is at the 402 = 20th value.
2
Read across from 20 on the vertical axis to the curve, then down.
3
The median is approximately 25 marks.
The curve passes through (20 ; 13) and (30 ; 27), so 20 falls just after 20 marks.

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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

An outlier lies below Q₁ − 1,5×IQR or above Q₃ + 1,5×IQR

Worked ExampleFor the box plot above, find the outlier boundaries

1
IQR = 42 − 24 = 18.
2
1,5 × 18 = 27.
3
Lower: 24 − 27 = −3. Upper: 42 + 27 = 69.
4
Since the data runs 12 to 55, there are no outliers.
Everything lies inside the boundaries.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
For 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
2
Calculate the interquartile range from the diagram below.
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42 − 24 = 18
3
Calculate the mean of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
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1989 = 22,00
4
Calculate the standard deviation of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
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variance = 44,00 → σ = 6,63
5
How 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
6
Use the ogive below to estimate the median mark.
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The 20th of 40 values → ≈ 25 marks
7
Use the ogive below to write down the total number of learners.
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40
8
Determine 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