9.10 Measures of dispersion
Two sets can share a mean and still be nothing alike: \(49,50,51\) and \(10,50,90\) both average \(50\). A measure of dispersion reports how spread out the values are.
9.10.1 Range
The range is the easiest measure to find and the weakest to rely on, since it uses only two of the observations and is at the mercy of a single freak value.
9.10.2 Mean deviation
Definition 9.46. The mean deviation about the mean is \[\text {M.D}(\bar {x})=\frac {\sum _{i=1}^{n}\left |x_i-\bar {x}\right |}{n},\] and about the median \(M\), \[\text {M.D}(M)=\frac {\sum _{i=1}^{n}\left |x_i-M\right |}{n}.\] For grouped data these become \[\text {M.D}(\bar {x})=\frac {\sum f_i\left |x_i-\bar {x}\right |}{N},\qquad \text {M.D}(M)=\frac {\sum f_i\left |x_i-M\right |}{N},\qquad N=\sum f_i.\]
Note 9.47. The modulus signs are essential. Without them the deviations above the mean cancel those below, and \(\sum (x_i-\bar {x})\) is exactly zero for every set of data — which would make the measure useless. Taking sizes only, and ignoring direction, is what makes it work.
Example 9.48. Find the mean deviation about the mean for \[62,\quad 55,\quad 71,\quad 48,\quad 66,\quad 53,\quad 74,\quad 59,\quad 51,\quad 61.\]
Solution. Step 1 — find the mean. \[\bar {x}=\frac {62+55+71+48+66+53+74+59+51+61}{10}=\frac {600}{10}=60.\]
Step 2 — find the deviations and their sizes.
| \(x_i\) | \(x_i-\bar {x}\) | \(\left |x_i-\bar {x}\right |\) |
| 62 | 2 | 2 |
| 55 | \(-5\) | 5 |
| 71 | 11 | 11 |
| 48 | \(-12\) | 12 |
| 66 | 6 | 6 |
| 53 | \(-7\) | 7 |
| 74 | 14 | 14 |
| 59 | \(-1\) | 1 |
| 51 | \(-9\) | 9 |
| 61 | 1 | 1 |
| Total | 0 | 68 |
The middle column totalling zero is the check that the mean was found correctly.
Step 3 — divide. \[\text {M.D}(\bar {x})=\frac {68}{10}=6.8.\]
On average, an observation lies \(6.8\) units away from the mean.
9.10.3 Standard deviation
The mean deviation removes signs by taking sizes. The standard deviation removes them by squaring, which is more awkward to work with by hand but far better behaved, and it is the measure of spread used almost everywhere.
Definition 9.49. The standard deviation of \(\{x_1,x_2,\ldots ,x_n\}\) is \[s=\sqrt {\frac {\sum _{i=1}^{n}\left (x_i-\bar {x}\right )^2}{n}},\] and for a frequency distribution, \[s=\sqrt {\frac {\sum _{i=1}^{n}f_i\left (x_i-\bar {x}\right )^2}{N}},\qquad N=\sum f_i.\]
Note 9.50. The square root at the end is there to undo the squaring, so that \(s\) comes out in the same units as the data. A standard deviation of masses in kilograms is itself in kilograms; without the root it would be in kilograms squared, which means nothing physical.
Solution. Step 1 — the mean. \[\bar {x}=\frac {6+8+11+13+17}{5}=\frac {55}{5}=11.\]
Step 2 — tabulate the squared deviations.
| \(x_i\) | \(x_i-\bar {x}\) | \(\left (x_i-\bar {x}\right )^2\) |
| 6 | \(-5\) | 25 |
| 8 | \(-3\) | 9 |
| 11 | 0 | 0 |
| 13 | 2 | 4 |
| 17 | 6 | 36 |
| Total | 0 | 74 |
Step 3 — divide and take the root. \[s=\sqrt {\frac {74}{5}}=\sqrt {14.8}=3.85\ \text {(2 d.p.)}\]
Example 9.52. Find the standard deviation of the number of children per household, using the distribution of children per household from earlier in this section.
| Number of children, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of households, \(f\) | 4 | 9 | 15 | 12 | 7 | 3 |
Solution. The mean was found earlier to be \(\bar {x}=2.36\). Each squared deviation must now be weighted by how many households share it.
| \(x\) | \(f\) | \(x-\bar {x}\) | \(\left (x-\bar {x}\right )^2\) | \(f\left (x-\bar {x}\right )^2\) |
| 0 | 4 | \(-2.36\) | 5.5696 | 22.2784 |
| 1 | 9 | \(-1.36\) | 1.8496 | 16.6464 |
| 2 | 15 | \(-0.36\) | 0.1296 | 1.9440 |
| 3 | 12 | 0.64 | 0.4096 | 4.9152 |
| 4 | 7 | 1.64 | 2.6896 | 18.8272 |
| 5 | 3 | 2.64 | 6.9696 | 20.9088 |
| Total | \(N=50\) | 85.5200 | ||
\[s=\sqrt {\frac {\sum f\left (x-\bar {x}\right )^2}{N}}=\sqrt {\frac {85.52}{50}} =\sqrt {1.7104}=1.31\ \text {(2 d.p.)}\]
Note 9.53. The mean must be carried to full accuracy through the middle columns. Rounding \(2.36\) to \(2.4\) before subtracting shifts every squared deviation and the final answer with them. Round once, at the end.
9.10.4 Variance
Equivalently, it is the standard deviation calculation stopped one step early, before the square root: \[s^2=\frac {\sum \left (x_i-\bar {x}\right )^2}{n}.\]
For the five numbers \(6,\ 8,\ 11,\ 13,\ 17\) above, \[s^2=\frac {74}{5}=14.8,\] and for the households, \(s^2=1.7104\).
Note 9.55. Variance and standard deviation carry exactly the same information, and which one is quoted is a matter of what it is wanted for. The standard deviation is in the units of the data and is the one to report; the variance is squared and awkward to interpret, but its algebra is simpler, which is why it dominates later theory.
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