Variance and standard deviation – Measures of Dispersion (Senior 4 Mathematics) | Ganzaa

Measures of Dispersion

Lessons

Senior 4 · Mathematics

Measures of Dispersion

3 lessons

Lesson 2 of 3

Variance and standard deviation

25 min

Measuring how far the values stray

The range uses only the two most extreme values, and the IQR ignores half the data. A better measure of spread should use every value. The natural idea is to measure how far each value lies from the mean. For Company A's 11 journey times from the last lesson (mean 150 minutes), these distances, called deviations, are drawn below.

The deviations are $- 8, - 5, - 3, - 2, - 1, 0, 1, 2, 3, 5, 8$ . Their sum is 0, and this is no accident: for any data set,

\sum_{i=1}^{n} (x_i - \bar{x}) = \sum x_i - n\bar{x} = n\bar{x} - n\bar{x} = 0.

So the mean deviation is always zero and tells us nothing. Squaring each deviation removes the signs, and gives the most important measures of spread in statistics.

Definition

Variance: the mean of the squared deviations from the mean. For values $x_1, \dots, x_n$ with mean $\bar{x}$ ,

\sigma^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2 .

Standard deviation: the positive square root of the variance, $\sigma = \sqrt{\sigma^2}$ .

Notation: $\sigma$ is the Greek letter sigma; some books write $\operatorname{Var}(x)$ for the variance and $s$ for the standard deviation. The variance is in squared units ( $\mathrm{min}^2$ ); the standard deviation is in the units of the data (minutes), which is why it is the one we usually quote.

Example: for 2, 4, 6 the mean is 4, the deviations are $- 2, 0, 2$ , so $\sigma^2 = \frac{4 + 0 + 4}{3} = \frac{8}{3}$ and $\sigma = \sqrt{\frac{8}{3}} \approx 1.63$ .

Worked example 1: the two bus companies

Find the standard deviation of the journey times of each company.

Solution

Company A: squared deviations $64, 25, 9, 4, 1, 0, 1, 4, 9, 25, 64$ , with sum 206.

\sigma_A^2 = \frac{206}{11} \approx 18.73\ \mathrm{min}^2, \qquad \sigma_A = \sqrt{18.73} \approx 4.33\ \text{minutes}.

Company B: deviations $- 30, - 22, - 15, - 10, - 4, 0, 6, 10, 15, 22, 28$ ; squares $900, 484, 225, 100, 16, 0, 36, 100, 225, 484, 784$ , with sum 3354.

\sigma_B^2 = \frac{3354}{11} \approx 304.9\ \mathrm{min}^2, \qquad \sigma_B = \sqrt{304.9} \approx 17.46\ \text{minutes}.

Company B's standard deviation is about four times Company A's: a typical B trip is some 17 minutes away from the mean, a typical A trip only about 4.

A quicker formula

Working out every deviation is slow when the mean is not a whole number. Expanding the square gives a formula that needs only $\sum x$ and $\sum x^2$ .

Important

Computing formula. $\sigma^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2 , \qquad \text{and for a frequency table} \qquad \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \bar{x}^2 .$

Proof. $\sum (x_i - \bar{x})^2 = \sum x_i^2 - 2\bar{x}\sum x_i + n\bar{x}^2 = \sum x_i^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2 = \sum x_i^2 - n\bar{x}^2$ . Divide by $n$ . $\blacksquare$

In words: variance = (mean of the squares) − (square of the mean).

Worked example 2: a frequency table

The numbers of goals scored in 30 matches of a school league were (illustrative data):

Goals, $x$	0	1	2	3	4	5	Total
Matches, $f$	4	8	9	5	3	1	30
$f x$	0	8	18	15	12	5	58
$f x^{2}$	0	8	36	45	48	25	162

Find the mean and the standard deviation.

Solution

\bar{x} = \frac{58}{30} \approx 1.933, \qquad \sigma^2 = \frac{162}{30} - \left(\frac{58}{30}\right)^2 = 5.4 - 3.7378 = 1.6622, \qquad \sigma \approx 1.29 \text{ goals}.

Keep $\bar{x}$ unrounded (as the fraction $\frac{58}{30}$ ) until the end: squaring a rounded mean can spoil the answer.

Worked example 3: grouped data

A dairy cooperative recorded the morning milk yield of 40 cows (illustrative data). For grouped data, each class is represented by its midpoint $x$ .

Yield (litres)	$0 \le y \lt 5$	$5 \le y \lt 10$	$10 \le y \lt 15$	$15 \le y \lt 20$	$20 \le y \lt 25$
Cows, $f$	4	10	14	8	4
Midpoint, $x$	2.5	7.5	12.5	17.5	22.5
$f x$	10	75	175	140	90
$f x^{2}$	25	562.5	2187.5	2450	2025

Solution

$\sum f = 40$ , $\sum fx = 490$ , $\sum fx^2 = 7250$ .

\bar{x} = \frac{490}{40} = 12.25\ \text{L}, \qquad \sigma^2 = \frac{7250}{40} - 12.25^2 = 181.25 - 150.0625 = 31.1875, \qquad \sigma \approx 5.58\ \text{L}.

These are estimates: using midpoints assumes the values in each class are centred on its middle.

Warning

Two slips that cost marks. (1) Forgetting the square root: the variance (31.19 L²) is not the standard deviation (5.58 L). (2) Using the wrong calculator key: most calculators show both $\sigma_n$ (divide by $n$ ) and $s_{n-1}$ (divide by $n - 1$ ). In this unit we describe a whole data set, so we divide by $n$ and use $\sigma_n$ . Dividing by $n - 1$ is used when a sample is used to estimate the spread of a larger population; you will meet it later.

Combining two data sets

Two classes, or two weeks of data, often need to be pooled. The combined mean is a weighted mean, but the combined standard deviation is not the average of the two standard deviations. The trick is to rebuild each group's $\sum x^2$ from its summary: since $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$ ,

\sum x^2 = n\left(\sigma^2 + \bar{x}^2\right).

Important

Combined data sets. For groups of sizes $n_{1}, n_{2}$ , means $\bar{x}_1, \bar{x}_2$ and standard deviations $\sigma_1, \sigma_2$ :

\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}, \qquad \sigma^2 = \frac{n_1\left(\sigma_1^2 + \bar{x}_1^2\right) + n_2\left(\sigma_2^2 + \bar{x}_2^2\right)}{n_1 + n_2} - \bar{x}^2 .

Worked example 4: two classes, one result

In a mathematics test, the 30 students of S4 A had mean 62 and standard deviation 8; the 20 students of S4 B had mean 70 and standard deviation 10. Find the mean and the standard deviation of all 50 marks.

Solution

\bar{x} = \frac{30(62) + 20(70)}{50} = \frac{1860 + 1400}{50} = 65.2 .

Rebuild the sums of squares: S4 A: $30(8^2 + 62^2) = 30(3908) = 117\,240$ ; S4 B: $20(10^2 + 70^2) = 20(5000) = 100\,000$ .

\sigma^2 = \frac{117\,240 + 100\,000}{50} - 65.2^2 = 4344.8 - 4251.04 = 93.76, \qquad \sigma = \sqrt{93.76} \approx 9.68 .

The combined standard deviation (9.68) is larger than both 8 and 10 would suggest on average, because the gap between the two class means adds extra spread.

Tip

Shifting and scaling. If every value is increased by a constant $a$ , the mean increases by $a$ but the standard deviation does not change: every deviation stays the same. If every value is multiplied by $k$ , both the mean and the standard deviation are multiplied by $∣ k ∣$ , and the variance by $k^{2}$ . So you can subtract a convenient number (such as 150 from the journey times) before calculating, and add it back to the mean only.

1st Class Maths: Variance and Standard Deviation (A-level statistics). Watch how the formula 'mean of the squares minus the square of the mean' is used with a frequency table.

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Practice

Try each one before opening the answers. Give answers to 3 significant figures where they are not exact.

Find the variance and standard deviation of 5, 7, 8, 10, 15.
Use the computing formula to find the standard deviation of 12, 15, 11, 18, 14, 20.
A shop sold the following numbers of phones per day over 20 days: 0 phones (3 days), 1 (6), 2 (7), 3 (4). Find the mean and the standard deviation.
For 25 values, $\sum x = 400$ and $\sum x^2 = 6800$ . Find $\bar{x}$ and $\sigma$ .
Estimate the mean and standard deviation of the times (minutes) 16 students took to walk to school: $0 \le t \lt 10$ : 3; $10 \le t \lt 20$ : 7; $20 \le t \lt 30$ : 4; $30 \le t \lt 40$ : 2.
Every value in a data set with mean 40 and standard deviation 6 is doubled and then 5 is added. Find the new mean and standard deviation.
Group P has 10 values with mean 12 and standard deviation 3; group Q has 15 values with mean 17 and standard deviation 4. Find the mean and standard deviation of all 25 values.

Show answersHide answers

Mean 9; deviations $- 4, - 2, - 1, 1, 6$ ; squares sum to 58; $\sigma^2 = 11.6$ , $\sigma \approx 3.41$ .
$\sum x = 90$ , $\bar{x} = 15$ , $\sum x^2 = 1410$ ; $\sigma^2 = \frac{1410}{6} - 225 = 235 - 225 = 10$ ; $\sigma = \sqrt{10} \approx 3.16$ .
$\sum fx = 6 + 14 + 12 = 32$ , $\bar{x} = 1.6$ ; $\sum fx^2 = 6 + 28 + 36 = 70$ ; $\sigma^2 = 3.5 - 2.56 = 0.94$ ; $\sigma \approx 0.970$ phones.
$\bar{x} = 16$ ; $\sigma^2 = \frac{6800}{25} - 256 = 272 - 256 = 16$ ; $\sigma = 4$ .
Midpoints 5, 15, 25, 35: $\sum fx = 15 + 105 + 100 + 70 = 290$ , $\bar{x} = 18.125$ min; $\sum fx^2 = 75 + 1575 + 2500 + 2450 = 6600$ ; $\sigma^2 = 412.5 - 328.52 = 83.98$ ; $\sigma \approx 9.16$ min.
Mean $2 (40) + 5 = 85$ ; standard deviation $2 \times 6 = 12$ (adding 5 changes nothing).
$\bar{x} = \frac{120 + 255}{25} = 15$ . Sums of squares: $10 (9 + 144) = 1530$ and $15 (16 + 289) = 4575$ . $\sigma^2 = \frac{6105}{25} - 225 = 244.2 - 225 = 19.2$ ; $\sigma \approx 4.38$ .

Quick check

Why is the variance built from squared deviations rather than plain deviations from the mean?

Key points

Variance $\sigma^2 = \frac{1}{n}\sum (x - \bar{x})^2 = \frac{\sum x^2}{n} - \bar{x}^2$ ; standard deviation $\sigma = \sqrt{\sigma^2}$ , in the units of the data.
For a frequency table use $\frac{\sum fx^2}{\sum f} - \bar{x}^2$ ; for grouped data use class midpoints (an estimate).
Combine groups through $\sum x^2 = n(\sigma^2 + \bar{x}^2)$ , never by averaging standard deviations.
Adding a constant leaves $\sigma$ unchanged; multiplying by $k$ multiplies $\sigma$ by $∣ k ∣$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions. The league, milk-yield and class data are illustrative.
Video: Variance and Standard Deviation - Year 1 Statistics [A-level], 1st Class Maths, embedded from YouTube.

1Measuring how far the values stray

2A quicker formula

3Combining two data sets

Measuring how far the values stray

A quicker formula

Combining two data sets