Mathematics notes, chapter 8
Measures of Dispersion for Ungrouped Data
Textbook Mathematics Form 4 KSSM (DLP), pages 210–241
What this chapter covers
- Explain the meaning of dispersion (8.1.1)
- Compare and interpret the dispersion of two or more sets of data using stem-and-leaf plots and dot plots, and draw a conclusion (8.1.2)
- Find the range, interquartile range, variance and standard deviation of ungrouped data, including data in a frequency table (8.2.1)
- Explain the advantages and disadvantages of the measures of dispersion (8.2.2)
- Construct and interpret a box plot for ungrouped data (8.2.3)
- Determine the effect on the measures of dispersion when the data is changed by adding, multiplying, or by an outlier or removed value (8.2.4)
- Compare sets of ungrouped data using suitable measures of dispersion and solve problems involving measures of dispersion (8.2.5, 8.2.6)
8.1Dispersion
Dispersion tells us how spread out the values of a set of data are. If the values are close together the dispersion is small. If they are far apart the dispersion is large. (8.1.1)
The simplest way to see dispersion is the difference between the largest and the smallest value. Range, interquartile range, variance and standard deviation are the measures of dispersion you will use in this chapter.
Example 1 (textbook p. 213): the difference in mass
Masses of 20 pupils run from 47 kg to 79 kg.Difference = 79 kg − 47 kg = 32 kg. A bigger difference means the data is more widely dispersed.
Dot plots and stem-and-leaf plots
A dot plot puts one dot above a number line for each value, so repeated values pile up. It suits a small set of data. A stem-and-leaf plot splits each value into a stem (the tens) and a leaf (the units). Leaves are written in order, smallest first, and a key is always given.
| Stem | Leaf |
|---|---|
| 2 | 7 9 |
| 3 | 1 2 4 5 5 6 7 7 8 8 9 |
| 4 | 0 0 1 2 2 2 2 3 4 4 5 6 7 7 7 8 9 9 |
| 5 | 2 2 2 3 4 |
Key: 2 | 7 means 27 marks. Accounting test marks of 36 pupils, textbook p. 215. Most marks are in the 40s.
To compare two sets of data, draw them on the same scale. The set whose dots are further to the right has the higher values. The set that is more spread out has the larger dispersion. (8.1.2)
Example 2 (textbook p. 217): comparing two dot plots
Science quiz scores range from 8 to 15, a difference of 7. Mathematics quiz scores are bunched on the right.(a) The Science scores are more dispersed because their difference is larger.
(b) The Mathematics scores are better because most of its dots are on the right.
Two classes sit the same test. Class A marks range from 30 to 90 and Class B marks from 55 to 75. Which class has the greater dispersion?
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Class A: difference 60, against 20 for Class B. Class A is more dispersed.
8.2Measures of dispersion
Range
Range = largest value − smallest value. It is quick to find, but it uses only two values. (8.2.1)
From a frequency table, take the smallest and largest values of x that have a frequency. Do not subtract frequencies.
Example 3 (textbook p. 219)
34, 23, 14, 26, 40, 25, 20. Range = 40 − 14 = 26.Frequency table with scores 2 to 8: range = 8 − 2 = 6.
Interquartile range
Q1 (first or lower quartile) is the middle of the lower half of the ordered data. Q3 (third or upper quartile) is the middle of the upper half. Q2 is the median. Interquartile range = Q3 − Q1.
- Arrange the data in ascending order and find the median.
- Split the data into a part before the median and a part after it. If n is odd, the median itself is left out of both parts.
- Q1 is the median of the lower part and Q3 is the median of the upper part. With an even number of values, take the average of the two middle values.
- Interquartile range = Q3 − Q1.
Example 5 (textbook p. 220)
(a) 4, 4, 6, 8, 10, 13, 15, 16, 20, 22, 23 (n = 11). Median 13. Q1 = 6, Q3 = 20. IQR = 20 − 6 = 14.(b) 11, 14, 14, 17, 19, 20, 20, 21, 22 (n = 9). Median 19. Q1 = 14+142 = 14, Q3 = 20+212 = 20.5. IQR = 6.5.
From a frequency table
Add a cumulative frequency row. Q1 is the value at position 14N and Q3 is the value at position 34N.
Example 6 (textbook p. 221)
| Score | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|
| Frequency | 3 | 5 | 6 | 8 | 12 | 7 | 3 |
| Cumulative | 3 | 8 | 14 | 22 | 34 | 41 | 44 |
Forgetting to put the data in order before finding the quartiles. Also, in a frequency table the median and quartiles are values of x, not frequencies.
Variance and standard deviation
Variance is the mean of the squared differences from the mean. Standard deviation is its square root, so it is in the same units as the data.
| Data | Variance σ² | Standard deviation σ |
|---|---|---|
| Ungrouped list | Σ(x − x̄)²N = Σx²N − x̄² | √σ² |
| Frequency table | Σf(x − x̄)²Σf = Σfx²Σf − x̄² | √σ² |
The mean is x̄ = ΣxN, or ΣfxΣf for a frequency table. The shortcut with Σx² is usually quicker and needs fewer subtractions.
Example 7 and 9 (textbook pp. 221 to 223)
(a) 2, 4, 5, 5, 6: mean = 225 = 4.4.σ² = 4 + 16 + 25 + 25 + 365 − 4.4² = 21.2 − 19.36 = 1.84.
(b) Books read: 0, 1, 2, 3, 4 with frequencies 3, 5, 8, 2, 2 (Σf = 20).
Σfx = 35, so mean = 1.75. Σfx² = 87.
σ² = 8720 − 1.75² = 4.35 − 3.0625 = 1.2875.
σ = √1.2875 = 1.135.
Writing the variance as the answer when the question asks for the standard deviation, or forgetting to take the square root. Also do not forget to subtract x̄² after dividing by N.
Your scientific calculator can check the answer. Use SD mode, enter each value with M+ (repeat the entry for the frequency), then read σn. Use σn, not σn−1, to match the textbook.
Find the variance and the standard deviation of 3, 5, 6, 8, 8.
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Mean = 30 ÷ 5 = 6. Σx² = 9 + 25 + 36 + 64 + 64 = 198.
σ² = 198 ÷ 5 − 36 = 3.6. σ = √3.6 = 1.897.
σ² = 198 ÷ 5 − 36 = 3.6. σ = √3.6 = 1.897.
Choosing a measure
| Measure | Advantage | Disadvantage |
|---|---|---|
| Range | Easiest to find | Only two values; badly affected by an outlier |
| Interquartile range | Not affected much by outliers | Ignores the lowest and highest quarter |
| Standard deviation | Uses every value; best for comparing sets | Pulled up by an outlier; more working |
Example 10 (textbook p. 224)
10, 11, 11, 14, 18, 20, 21, 21, 25, 40. The value 40 is an outlier. Range = 30 but IQR = 21 − 11 = 10. The interquartile range is more suitable because it ignores the outlier.A low standard deviation means the data is close to the mean, so it is more consistent. When two sets have the same mean, choose the one with the smaller standard deviation for consistency. (8.2.5)
Example 11 (textbook p. 224)
Aiman and John both average 69.8 marks. Aiman has σ = 17.67 and John has σ = 7.57. John is more consistent, so John is selected.Box plots
A box plot shows five values: minimum, Q1, median, Q3 and maximum. The box covers Q1 to Q3, so its width is the interquartile range. The whiskers reach the minimum and the maximum.
Read a box plot from the number line: range = right whisker end − left whisker end, and IQR = right edge of the box − left edge of the box. If the median is near the middle of the box the data is roughly symmetric.
A box plot has minimum 10, Q1 16, median 20, Q3 31 and maximum 44. Find the range and the interquartile range.
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Range = 44 − 10 = 34. IQR = 31 − 16 = 15.
Effect of changing the data
| Change to every value | Range | IQR | Standard deviation | Variance |
|---|---|---|---|---|
| Add or subtract k | same | same | same | same |
| Multiply by k | × k | × k | × k | × k² |
| Divide by k | ÷ k | ÷ k | ÷ k | ÷ k² |
When you do two steps, such as "multiply by 4, then add 2", the adding has no effect and only the multiplying counts.
Example (like textbook Self Practice 8.2d, question 5)
Range 25 and variance 2.7. Each value is increased by 2, then multiplied by 4.New range = 25 × 4 = 100. New variance = 2.7 × 4² = 43.2.
Multiplying the variance by k instead of k². If the data is doubled, the standard deviation is doubled but the variance becomes four times as large.
Adding or removing a value
- An outlier added makes the range and the standard deviation much larger. The IQR changes little.
- A value added close to the mean makes the standard deviation smaller. A value far from the mean makes it larger.
- Removing a value far from the mean makes the standard deviation smaller.
Example 13 (textbook p. 233)
4, 6, 7, 7, 9, 11, 12, 12 has σ = 2.784. Adding 18 gives σ ≈ 3.98 (larger, because 18 is far from the mean 8.5). Adding 7 gives σ ≈ 2.67 (smaller, because 7 is near the mean).Problems with Σx and Σx²
If you are given the mean and the variance of a group, you can recover the totals: Σx = N × x̄ and Σx² = N × (σ² + x̄²). Totals of two groups can then be added to find the mean and standard deviation of the combined group.
Example 15 (textbook p. 236)
Group A: 18 pupils, mean 52, variance 2.5. Group B: 12 pupils, mean 56, variance 1.8.Σx = 936 + 672 = 1 608. Σx² = 48 717 + 37 653.6 = 86 370.6.
Combined mean = 1 608 ÷ 30 = 53.6.
σ² = 86 370.6 ÷ 30 − 53.6² = 6.06, so σ = 2.462.
Ten numbers have Σx = 80 and Σx² = 700. Find the variance.
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Mean = 8. σ² = 700 ÷ 10 − 8² = 70 − 64 = 6.