Mathematics notes, chapter 7

Measures of Dispersion for Grouped Data

Textbook Mathematics Form 5 (DLP), pages 196–225

What this chapter covers

Construct a histogram and a frequency polygon for grouped data, and compare the shape and dispersion of two or more distributions (7.1)
Construct an ogive for grouped data and use it to find quartiles and percentiles (7.1)
Determine the range, interquartile range, variance and standard deviation for grouped data (7.2)
Construct and interpret a box plot for grouped data (7.2)
Compare two or more sets of grouped data using measures of dispersion and make a conclusion (7.2)
Solve problems involving measures of dispersion for grouped data, and plan a statistical mini-project (7.2)

7.1Dispersion

Organising grouped data

Grouped data are raw data sorted into class intervals in a frequency table. Histograms and frequency polygons are drawn for continuous data (measured, such as height or time).

Term	How to find it	Class 150 – 154
Lower limit, upper limit	Smallest and largest value written in the class	150 and 154
Midpoint	(lower limit + upper limit) ÷ 2	152
Lower boundary	(upper limit of class before + lower limit of this class) ÷ 2	149.5
Upper boundary	(upper limit of this class + lower limit of class after) ÷ 2	154.5
Size of class interval	Largest − smallest value ÷ number of classes; or upper boundary − lower boundary	154.5 − 149.5 = 5

Do not find the class size from the limits of one class (154 − 150 = 4 is wrong). Use the difference between two lower limits, or between the boundaries.

The cumulative frequency of a class is its frequency added to the total of all the classes before it. It tells you how many values are at or below the upper boundary.

Example (textbook p. 199, heights of 20 pupils): build the table

Largest 173, smallest 145, 6 classes: size = (173 − 145) ÷ 6 = 4.7, so use 5. Classes: 145 – 149, 150 – 154, 155 – 159, 160 – 164, 165 – 169, 170 – 174 with frequencies 2, 3, 6, 5, 3, 1.
For 145 – 149: midpoint 147, boundaries 144.5 and 149.5.

Histogram and frequency polygon

Histogram: bars touch each other. Bar width = class size, drawn from lower boundary to upper boundary. Bar height = frequency.
Frequency polygon: mark the midpoint at the top of each bar and join them with straight lines. Add one class with frequency 0 before the first class and after the last class so the polygon touches the horizontal axis.

Histogram of car speeds with the frequency polygon drawn over it. Frequencies 5, 10, 20, 30, 25, 10; the tallest bar is 99.5 to 109.5 — Speeds of cars (km h⁻¹): histogram with its frequency polygon. The bar edges are class boundaries.

Redrawn from textbook Example 3, p. 201.

A class is 20 – 24 and the next class is 25 – 29. State its lower boundary, upper boundary and midpoint.

Show answerHide answer

Lower boundary = (19 + 20) ÷ 2 = 19.5. Upper boundary = (24 + 25) ÷ 2 = 24.5. Midpoint = (20 + 24) ÷ 2 = 22.

Shape of a distribution

Three histograms: symmetric, right-skewed with a long tail to the right, and left-skewed with a long tail to the left — A skewed distribution is named after the side of its **longer tail**.

Shape	What you see	Typical meaning
Symmetric (bell or uniform)	Left and right halves look alike	Values spread evenly about the middle
Skewed to the right	Peak on the left, long tail to the right	Most values are low, a few are very high
Skewed to the left	Peak on the right, long tail to the left	Most values are high, a few are very low

To compare dispersion of two distributions, compare how far each is spread out on the horizontal axis (the wider spread has the greater dispersion). Then say which one performed better by looking at where most of the values lie. Always give a reason that refers to the graph.

Ogive, quartiles and percentiles

An ogive is a graph of cumulative frequency against upper boundary. It is an S-shaped smooth curve that starts at 0 on the upper boundary of an added class with frequency 0.

For N values the positions are: Q₁ at N/4, median Q₂ at N/2, Q₃ at 3N/4, and the k-th percentile P_k at (k/100) × N. Go across from the position on the cumulative frequency axis to the curve, then down to read the value. P₂₅ = Q₁, P₅₀ = Q₂, P₇₅ = Q₃.

Ogive of salt content of 60 foods with dashed lines at cumulative frequency 15, 30 and 45 reading about 199.5, 249.5 and 285 mg — Salt content in 60 foods: N = 60, so the positions are 15, 30 and 45.

Redrawn from textbook Example 7, p. 207. Reading the curve gives Q₁ = 199.5 mg, Q₂ = 249.5 mg and Q₃ ≈ 285 mg (the book reads 284.5 mg).

Plot the cumulative frequency against the upper boundary, not the midpoint or the upper limit. The starting point is (lower boundary of the first class, 0).

An ogive is drawn for 200 values. At which cumulative frequencies do you read Q₃ and P₃₅?

Show answerHide answer

Q₃: ¾ × 200 = 150. P₃₅: 0.35 × 200 = 70.

7.2Measures of dispersion

Range and interquartile range

Range = midpoint of the highest class − midpoint of the lowest class.
Interquartile range = Q₃ − Q₁ (read from the ogive). It is the spread of the middle 50% of the values and is not affected by extreme values.

Example (textbook p. 211, pineapple masses)

Highest class 900 – 999, lowest class 400 – 499. Range = 949.5 − 449.5 = 500 g.
N = 80: Q₁ at position 20, Q₃ at position 60. From the ogive Q₁ = 614.5 g and Q₃ = 809.5 g, so IQR = 195 g.

Variance and standard deviation

Variance σ² = Σfx² ÷ Σf − x̄², where x is the class midpoint, f the frequency and x̄ = Σfx ÷ Σf the mean. Standard deviation σ = √variance, in the same unit as the data.

Worked example: marks of 20 pupils

Marks	f	Midpoint x	fx	fx²
10 – 19	6	14.5	87	1 261.5
20 – 29	10	24.5	245	6 002.5
30 – 39	3	34.5	103.5	3 570.75
40 – 49	1	44.5	44.5	1 980.25
Total	20		480	12 815

Mean = 480 ÷ 20 = 24. Variance = 12 815 ÷ 20 − 24² = 640.75 − 576 = 64.75. Standard deviation = √64.75 = 8.05.

Check on your calculator: use the statistics (SD) mode, enter each midpoint with its frequency, then read x̄ and σ (the population one, σ_n).

Use the midpoints in the table, never the limits. Square the midpoint first (x²) and then multiply by f. Also subtract the square of the mean, not the mean itself.

For 50 values, Σfx = 2 000 and Σfx² = 82 000. Find the variance and standard deviation.

Show answerHide answer

Mean = 40. Variance = 82 000 ÷ 50 − 40² = 1 640 − 1 600 = 40. σ = √40 = 6.32.

Box plot

A box plot shows the five-number summary: minimum, Q₁, median, Q₃ and maximum. For grouped data, read Q₁, Q₂ and Q₃ from the ogive, and take the minimum and maximum from the ogive axis ends.

Redrawn from textbook Example 11, p. 214.

Box plot looks like	Distribution
Median in the middle, whiskers about equal	Symmetric
Left side (box or whisker) longer	Skewed to the left
Right side (box or whisker) longer	Skewed to the right

Comparing two sets of data

Compare the mean first (which set is larger or better on average).
Then compare the standard deviation: the smaller one means the data are more consistent (closer to the mean).
Give a conclusion in words that fits the question, such as "larger and more consistent".

Example (textbook p. 215, hibiscus hybrids A and B)

Hybrid A: mean 14.35 cm, σ = 0.64 cm. Hybrid B: mean 14.1 cm, σ = 0.7 cm.
Hybrid A produces larger petals (greater mean) and they are more consistent (smaller σ).

Going from an ogive to a frequency table: subtract each cumulative frequency from the one after it to get the class frequency. Then use midpoints to estimate the mean and standard deviation.

Class P has mean 62 and σ = 9. Class Q has mean 58 and σ = 6. Which class is better on average and which is more consistent?

Show answerHide answer

Class P has the higher mean (62 > 58). Class Q is more consistent because its standard deviation is smaller (6 < 9).