Measures of dispersion | Grade 10

Two basketball players both average $8$ points a game. One scores $7, 8, 9$ night after night; the other scores $0, 8, 16$ . Same mean, completely different players: one is steady, one is a gamble. The mean alone hides this. To describe data honestly we also need a number for how spread out it is.

Range and interquartile range

The range is easy but depends only on two values, so one extreme value ruins it. The interquartile range ignores the extremes.

Variance and standard deviation

Both measures above ignore most of the data. The variance uses every value: it is the average of the squared distances from the mean.

The standard deviation $s$ has the same unit as the data, which makes it easier to interpret. A small $s$ means the values hug the mean; a large $s$ means they are scattered.

Example Two ways to get the variance

For $4, 6, 7, 7, 8, 9, 10, 11$ the mean is $\bar{x}=\frac{62}{8}=7.75$ . Find $s^2$ and $s$ .

Solution.

Direct formula: the squared deviations are $14.0625,\ 3.0625,\ 0.5625,\ 0.5625,\ 0.0625,\ 1.5625,\ 5.0625,\ 10.5625$ , with sum $35.5$ . So $s^2=\frac{35.5}{8}=4.4375\approx 4.44$ .
Second formula: $\frac{16+36+49+49+64+81+100+121}{8}=\frac{516}{8}=64.5$ , and $64.5-7.75^2=64.5-60.0625=4.4375$ . ✓
$s=\sqrt{4.4375}\approx 2.11$ .

The box plot

A box plot draws five numbers: the minimum, $Q_1$ , the median, $Q_3$ and the maximum. The box covers the middle half of the data and the "whiskers" reach out to the extremes.

Box plot of

4, 6, 7, 7, 8, 9, 10, 11

Outliers

A value is an outlier if it lies far from the bulk of the data. A common rule: $x$ is an outlier if $x<Q_1-1.5\,\Delta_Q\qquad\text{or}\qquad x>Q_3+1.5\,\Delta_Q.$

Key terms

English	Tiếng Việt
range	khoảng biến thiên
interquartile range	khoảng tứ phân vị
variance	phương sai
standard deviation	độ lệch chuẩn
deviation	độ lệch
box plot	biểu đồ hộp
outlier	giá trị ngoại lệ
spread	độ phân tán

Summary

$R=x_{\max}-x_{\min}$ and $\Delta_Q=Q_3-Q_1$ .
$s^2=\frac1n\sum(x_i-\bar{x})^2=\overline{x^2}-\bar{x}^2$ and $s=\sqrt{s^2}$ .
A smaller $s$ (or $\Delta_Q$ ) means more consistent data.
$x$ is an outlier if $x<Q_1-1.5\Delta_Q$ or $x>Q_3+1.5\Delta_Q$ .

Practice

Find $R$ and $\Delta_Q$ for $3, 5, 7, 9, 11$ .
Find $s^2$ and $s$ for $2, 4, 4, 4, 5, 5, 7, 9$ .
Find the variance of $6, 6, 6, 6$ . What does the answer tell you?
For $3, 5, 7, 9, 11$ use $s^2=\overline{x^2}-\bar{x}^2$ to find $s^2$ and $s$ (to $0.01$ ).
Which group is more consistent: A $=4, 6, 7, 7, 8, 9, 10, 11$ or B $=7, 7, 7, 8, 8, 8, 9, 9$ ? Give a reason using $s$ .
Is $40$ an outlier in $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 40$ ?
Add $5$ to every value of $3, 5, 7, 9, 11$ . How do the mean and $s$ change?
Multiply every value of $3, 5, 7, 9, 11$ by $2$ . Find the new variance and $s$ (to $0.01$ ).
Challenge. The numbers $4, 6, 8, 10, x$ have mean $8$ . Find $x$ and $s^2$ .
Challenge. Four numbers have $\sum x_i=20$ and $\sum x_i^2=120$ . Find $s^2$ .

Show solutions

$R=11-3=8$ . $Q_1=\frac{3+5}{2}=4$ , $Q_3=\frac{9+11}{2}=10$ , so $\Delta_Q=6$ .
$\bar{x}=5$ . Squared deviations: $9, 1, 1, 1, 0, 0, 4, 16$ , sum $32$ . $s^2=\frac{32}{8}=4$ and $s=2$ .
$\bar{x}=6$ and every deviation is $0$ , so $s^2=0$ . All values are equal: there is no spread.
$\bar{x}=7$ and $\overline{x^2}=\frac{9+25+49+81+121}{5}=\frac{285}{5}=57$ . $s^2=57-49=8$ and $s=\sqrt8\approx 2.83$ .
For A, $s\approx 2.11$ ; for B, $s\approx 0.78$ . B has the smaller standard deviation, so B is more consistent.
$Q_1=3$ , $Q_3=9$ (median $6$ left out), $\Delta_Q=6$ , $1.5\Delta_Q=9$ . Upper fence $=9+9=18$ . Since $40>18$ , it is an outlier.
The new data is $8, 10, 12, 14, 16$ : the mean becomes $12$ (it increases by $5$ ), while $s=\sqrt8\approx 2.83$ does not change. Shifting all values does not change the spread.
The new data is $6, 10, 14, 18, 22$ with mean $14$ . $s^2=\frac{64+16+0+16+64}{5}=32$ and $s=\sqrt{32}\approx 5.66$ . The variance is multiplied by $4$ and $s$ by $2$ .
Mean $8$ gives $4+6+8+10+x=40$ , so $x=12$ . Deviations $-4, -2, 0, 2, 4$ , squares sum to $40$ . $s^2=\frac{40}{5}=8$ .
$\bar{x}=\frac{20}{4}=5$ . $s^2=\frac{120}{4}-5^2=30-25=5$ .