The **standard deviation** of a dataset is a way to measure the typical deviation of individual values from the mean value.

The formula to calculate a sample standard deviation, denoted ** s**, is:

**s = √Σ(x _{i} – x̄)^{2} / (n – 1)**

where:

**Σ**: A symbol that means “sum”**x**: The i_{i}^{th}value in a dataset**x̄**: The sample mean**n**: The sample size

The are two main advantages of using the standard deviation to describe the spread of values in a dataset:

**Advantage #1: The standard deviation uses all observations in a dataset in its calculation.** In statistics, we generally say it’s a good thing when we are able to use all observations in a dataset to perform some calculation because we are using all possible “information” available in the dataset.

**Advantage #2: The standard deviation is easy to interpret**. The standard deviation is a single value that gives us a good idea of how far the “typical” observation in a dataset lies from the mean value.

However, there is one main disadvantage of using the standard deviation:

**Disadvantage #1: The standard deviation can be affected by outliers**. When extreme outliers are present in a dataset, this can inflate the value of the standard deviation and thus give a misleading idea of the spread of values in a dataset.

The following examples provide more information about these advantages and disadvantages of using the standard deviation.

**Advantage #1: The standard deviation uses all observations**

Suppose we have the following dataset that shows the distribution of exam scores for students in a class:

Scores: 68, 70, 71, 75, 78, 82, 83, 83, 85, 90, 91, 91, 92

We can use a calculator or statistical software to find that the **sample standard deviation of this dataset is 8.46.**

The nice thing about using the standard deviation in this example is that we use all possible observations in the dataset to find the typical “spread” of values.

By contrast, we could use another metric such as the interquartile range to measure the spread of values in this dataset.

We can use a calculator to find that the **interquartile range is 17.5**. This represents the spread between the middle 50% of values in the dataset.

Now suppose we change the lowest value in the dataset to be much lower:

Scores: 22, 70, 71, 75, 78, 82, 83, 83, 85, 90, 91, 91, 92

We can use a calculator to find that the **sample standard deviation is 18.37**.

However, **the interquartile range is still 17.5** because none of the middle 50% of values were affected.

This shows that the sample standard deviation considers all observations in the dataset in its calculation while other measures of dispersion do not.

**Advantage #2: The standard deviation is easy to interpret**

Recall the following dataset that shows the distribution of exam scores for students in a class:

Scores: 68, 70, 71, 75, 78, 82, 83, 83, 85, 90, 91, 91, 92

We used a calculator to find that the sample standard deviation of this dataset was **8.46**.

This is easy to interpret because it simply means the deviation of a “typical” exam score is about 8.46 away from the mean exam score.

By contrast, other measures of dispersion are not so straightforward to interpret.

For example, a **coefficient of variation** is another measure of dispersion that represents the ratio of the standard deviation to the sample mean.

Coefficient of Variation: s / x̄

In this example, the mean exam score is 81.46 so the coefficient of variation is calculated as 8.46 / 81.46 = **0.104**.

This represents the ratio of the sample standard deviation to the sample mean, which can be useful for comparing the spread of values between multiple datasets but it isn’t very straightforward to interpret as a metric by itself.

**Disadvantage #1: The standard deviation can be affected by outliers**

Suppose we have the following dataset that contains information about the salaries of 10 employees (in thousands of dollars) at some company:

Salaries: 44, 48, 57, 68, 70, 71, 73, 79, 84, 94

The sample standard deviation of salaries is about **15.57**.

Now suppose we have the exact same dataset but the largest salary is *much* larger:

Salaries: 44, 48, 57, 68, 70, 71, 73, 79, 84, 895

The sample standard deviation of salaries in this dataset is about **262.47**.

By including just one extreme outlier, the standard deviation is highly affected and now provides a misleading idea of the “typical” spread of salaries.

**Note**: When outliers are present in a dataset, the interquartile range can provide a better measure of dispersion because it is unaffected by outliers.

**Additional Resources**

The following tutorials provide additional information about using the standard deviation in statistics:

Interquartile Range vs. Standard Deviation: The Difference

Coefficient of Variation vs. Standard Deviation: The Difference

Population vs. Sample Standard Deviation: When to Use Each