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Mathematics · Ch 13 — Statistics

Standard Deviation

13.5.1

Standard Deviation

Why Standard Deviation Exists

When you calculate variance, you square the deviations (xi−xˉ)(x_i - \bar{x}). This squaring changes the units. If the original data is in centimetres, the variance is in square centimetres — a unit that has no direct physical meaning for the spread of the data. To get a measure of dispersion that is in the same units as the original observations, you take the positive square root of the variance. That square root is called the standard deviation.

The standard deviation is the most widely used measure of dispersion. It tells you, on average, how far each observation lies from the mean, but in the original units. It is denoted by the Greek letter σ\sigma (sigma).

σ=1n∑i=1n(xi−xˉ)2\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}

This is equation (1) from the textbook. Notice that the expression inside the square root is exactly the variance σ2\sigma^2. So the relationship is simple:

Standard deviation=Variance\text{Standard deviation} = \sqrt{\text{Variance}}


Worked Example: Ungrouped Data

The textbook illustrates the calculation with a concrete example. Let's walk through it step by step.

Example 8: Find the variance and standard deviation of the data:

6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Step 1 — Organise the data and choose an assumed mean.

There are n=10n = 10 observations. The data is evenly spaced, so the step-deviation method works well. Choose an assumed mean A=14A = 14. The common class size (the step) is h=2h = 2.

Step 2 — Compute the step deviations did_i.

For each xix_i, calculate di=xi−Ah=xi−142d_i = \frac{x_i - A}{h} = \frac{x_i - 14}{2}.

Step 3 — Compute the actual mean xˉ\bar{x}.

First find ∑di\sum d_i:

xix_idi=xi−142d_i = \frac{x_i - 14}{2}(xi−xˉ)(x_i - \bar{x})(xi−xˉ)2(x_i - \bar{x})^2
6-4-981
8-3-749
10-2-525
12-1-39
140-11
16111
18239
203525
224749
245981
Total∑di=5\sum d_i = 5∑(xi−xˉ)2=330\sum (x_i - \bar{x})^2 = 330

The mean using step deviation:

xˉ=A+h⋅∑din=14+2×510=14+1=15\bar{x} = A + h \cdot \frac{\sum d_i}{n} = 14 + 2 \times \frac{5}{10} = 14 + 1 = 15

Step 4 — Compute the variance. …

Table 13.7Variance of ungrouped data (6, 8, ..., 24) by step-deviation
xix_idi=xi−142d_i=\dfrac{x_i-14}{2}Deviations from mean (xi−xˉ)(x_i-\bar{x})(xi−xˉ)2(x_i-\bar{x})^2
6−4-4−9-981
8−3-3−7-749
10−2-2−5-525
12−1-1−3-39
140−1-11
16111
18239
203525