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1.9 · Use standard deviation to describe and compare data

Learn to use standard deviation to describe and compare data through clear examples and targeted practice.

Athabasca University MATH 215: Introduction to Statistics

Descriptive Statistics

A practical guide to understanding how much numerical data vary

A data set’s average gives a sense of its centre, but it does not tell us whether the values lie close together or are widely spread out. Standard deviation is a measure of spread: it describes the typical distance of data values from their mean. In this lesson, you will learn how to calculate standard deviation for a sample and use it to describe and compare data in context.

What you will learn

1. What standard deviation describes

A numerical variable is a characteristic recorded as a number, such as a person’s commute time in minutes or the number of items sold in a day. A data set is the collection of recorded values. The mean, often called the average, is found by adding the values and dividing by how many values there are. It gives a measure of the data’s centre.
The standard deviation describes how spread out the values are around the mean. A small standard deviation means values tend to be close to the mean. A large standard deviation means values tend to be farther from the mean. Standard deviation is measured in the same units as the original data, which helps make its meaning understandable in context.
The population is the full group we want to describe. A population parameter is a numerical summary of that full group. A sample is a smaller group of observations collected from a population, and a statistic is a numerical summary calculated from that sample. We use different standard-deviation formulas depending on whether we have the entire population or a sample.

2. Conditions and the sample formula

Before calculating a standard deviation, check that the observations are numerical measurements or counts for which differences make sense. Each observation should refer to the same variable and use the same units. For example, it makes sense to find a standard deviation for commute times all measured in minutes. It would not make sense to mix minutes and kilometres as though they were measurements of the same variable.
The sample standard deviation is used when the data are a sample and we want to describe their spread or use them to learn about a larger group. In its formula, xix_i means the value of observation number ii, xˉ\bar{x} is the sample mean, and nn is the number of observations. The symbol ∑\sum means to add the indicated quantities for all observations. The denominator is n−1n-1, not nn, for the sample standard deviation.
Here is the algebra behind the calculation. First find each value’s difference from the mean. Square each difference so that negative and positive differences do not cancel. Add the squared differences, divide by one less than the sample size, and take the square root. The square root reverses the squaring and returns the result to the data’s original units. You can use a calculator for arithmetic, but the steps show what the result summarizes.
If the data include every member of the population of interest, use the population formula instead. Its denominator is the population size, often written as NN. These two formulas answer different questions: one summarizes a sample, and the other summarizes a complete population.
s=∑i=1n(xi−xˉ)2n−1s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}

3. Interpreting and comparing standard deviations

A calculated standard deviation is most useful when stated in context. For example, a standard deviation of about 3.163.16 minutes indicates that observations typically differ from their mean by a few minutes. This is a description of spread, not a claim that every observation is exactly that distance from the mean.
To compare two standard deviations, first confirm that both describe the same kind of variable in the same units. Then compare their sizes alongside the data’s centres and the setting. If the standard deviations are different, the larger one indicates more spread around its own data set’s mean. A comparison is particularly clear when the means are similar, but the standard deviations still describe spread even when the means differ.
Standard deviation can be affected by unusually high or low values because it uses squared distances from the mean. A single value far from the rest can increase the standard deviation. Therefore, do not interpret the number without considering the values and their context. Also, a larger standard deviation does not mean a larger average; centre and spread are different features of data.

Worked example

Comparing the spread of two sets of times

Two small samples record how many minutes students spend travelling to campus. Sample A has values 4, 6, 8, 10, and 12 minutes. Sample B has values 6, 7, 8, 9, and 10 minutes. Calculate each sample standard deviation and compare the spreads.
  1. Check the data and identify the summaries
    Both samples measure the same numerical variable, travel time, in minutes. There are five observations in each sample. The samples are not described as the full population of students, so use the sample standard deviation formula. We will calculate a separate mean and standard deviation for each sample.
  2. Find each sample mean
    Add the five observations in each sample and divide by five. Both samples have the same mean, which makes their spreads especially easy to compare.
    xˉA=4+6+8+10+125=8,xˉB=6+7+8+9+105=8\bar{x}_A=\frac{4+6+8+10+12}{5}=8,\qquad \bar{x}_B=\frac{6+7+8+9+10}{5}=8
  3. Calculate Sample A’s squared deviations
    Subtract the mean of 88 from each Sample A value, square each difference, and add the results. For instance, the first deviation is 4−8=−44-8=-4, and its square is 1616. The squared deviations add to 4040.
    (4−8)2+(6−8)2+(8−8)2+(10−8)2+(12−8)2=40(4-8)^2+(6-8)^2+(8-8)^2+(10-8)^2+(12-8)^2=40
  4. Calculate Sample A’s standard deviation
    Divide the sum of squared deviations by n−1=4n-1=4, then take the square root. Keep the unrounded value for the next interpretation; round the final reported standard deviation to two decimal places.
    sA=405−1=10=3.162278… minutess_A=\sqrt{\frac{40}{5-1}}=\sqrt{10}=3.162278\ldots\text{ minutes}
  5. Calculate Sample B’s standard deviation
    For Sample B, the deviations from 88 are −2,−1,0,1,2-2,-1,0,1,2. Their squares add to 1010. Divide by 44 and take the square root, following the same sample formula.
    sB=(6−8)2+(7−8)2+(8−8)2+(9−8)2+(10−8)25−1=2.5=1.581139… minutess_B=\sqrt{\frac{(6-8)^2+(7-8)^2+(8-8)^2+(9-8)^2+(10-8)^2}{5-1}}=\sqrt{2.5}=1.581139\ldots\text{ minutes}
  6. Compare in context
    Rounded to two decimal places, Sample A’s standard deviation is 3.163.16 minutes and Sample B’s is 1.581.58 minutes. Both means are 88 minutes, but Sample A has the larger standard deviation. Its travel times are more spread out around the mean than Sample B’s. The results describe these samples; they do not say that every student’s travel time differs from the mean by exactly the standard deviation.
Answer: Sample A’s standard deviation is 3.16 minutes; Sample B’s is 1.58 minutes. Sample A has greater spread around the shared mean of 8 minutes.
Check: Each mean is 8 minutes, and Sample A includes values farther from 8 than Sample B does. The larger standard deviation for Sample A is consistent with that pattern.

Common mistakes and how to avoid them

Dividing by the sample size nn when calculating a sample standard deviation.
Correction: For a sample standard deviation, divide the sum of squared deviations by n−1n-1. Use NN in the population formula only when the data include the entire population of interest.
Treating standard deviation as the distance of every observation from the mean.
Correction: Standard deviation summarizes the overall spread. Individual observations can be closer to or farther from the mean.
Saying that the group with the larger standard deviation has the larger average.
Correction: The mean describes centre, while standard deviation describes spread. Compare each measure for the feature it represents.
Comparing standard deviations for variables with different units as if their sizes had a direct meaning.
Correction: Check that the variables and units are comparable before using the numerical sizes to compare spread.

Lesson summary

Check your understanding

Question 1

Two samples measure the same variable in the same units. Sample X has standard deviation 2.4, and Sample Y has standard deviation 5.1. Which statement is best?
  1. Sample Y has greater spread around its mean.
  2. Sample Y must have a greater mean.
  3. Every value in Sample Y is 5.1 units from its mean.
  4. Sample X has greater spread because its standard deviation is smaller.
Show answer and explanation
Sample Y has greater spread around its mean.
A larger standard deviation indicates more spread around the data set’s mean. It does not determine which mean is larger, nor does it give the distance of every observation from the mean.

Question 2

A sample has four observations. What number belongs in the denominator when calculating its sample standard deviation?
  1. 2
  2. 3
  3. 4
  4. 5
Show answer and explanation
3
The sample formula divides by n−1n-1. With four observations, the denominator is 4−1=34-1=3.

Key terms

Data set
A collection of recorded observations.
Mean
The sum of the data values divided by the number of values.
Spread
How much the values in a data set vary or differ from one another and from their centre.
Standard deviation
A measure of the spread of numerical data around the mean, expressed in the original data units.
Population
The full group of people or items that a study aims to describe.
Sample
A subset of a population from which observations are collected.
Parameter
A numerical summary describing a population.
Statistic
A numerical summary calculated from a sample.

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Published by DoAssignment. This AI-assisted lesson follows Athabasca University MATH 215: Introduction to Statistics, study topic 1.9. It is a study resource, not an official curriculum publication.

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