2.1 Understanding central tendencies

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Understanding central tendencies

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Measures of central tendency describe a typical value of a data set, while measures of spread describe how much the data varies. These summaries help you compare distributions and understand data at a glance. Here are the key statistical measures you need to know for data analysis.

Definitions

Mean (arithmetic average)

Formula

\overset{x}{ˉ} = \frac{\sum _{i = 1}^{n} x _{i}}{n}

. The balance point of the data. Sensitive to extreme values.

Median (middle value)

Order data from smallest to largest.

If $n$ is odd, the median is the $\frac{n + 1}{2}$ th value.
If $n$ is even, the median is the average of the $\frac{n}{2}$ th and $(\frac{n}{2} + 1)$ th values. Resistant to extreme values.

Mode (most frequent value)

The value or values that occur most often. A data set can be unimodal, bimodal, multimodal, or have no mode.

Range

The difference between the highest and lowest values in a data set. It gives a quick sense of how spread out the data are.

Interquartile range (IQR)

The spread of the middle 50% of the data, found by subtracting the first quartile (

Q_{1}

) from the third quartile (

Q_{3}

IQR = Q_{3} - Q_{1}

. The IQR is useful for identifying outliers.

Standard deviation

A measure of the average distance of each data value from the mean. A smaller standard deviation means the values are closer to the mean, while a larger one means the data are more spread out.

Extreme values (outliers)

Extremely large or extremely small numbers that are far away from the majority of the data.

Later, you’ll learn formal rules for identifying outliers. For now, treat an outlier as a value that looks unusually far from the rest of the data.

We’ll compare two data sets: one without an extreme value and one with an extreme value.

Example: Dataset without extreme values $4, 8, 6, 10, 7$

First, order the data: $4, 6, 7, 8, 10$ .

Next, compute the mean by adding the values and dividing by the number of values.

$\overset{x}{ˉ} = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$ .

The median is the middle value in the ordered list. Here, the third value is $7$ .

Check whether any value occurs most often (the mode). Since all values are unique, there is no mode.

Answer: Mean $= 7$ , median $= 7$ , there is no mode

Example: Dataset with an extreme value $4, 6, 7, 8, 100$

Order the data: $4, 6, 7, 8, 100$ .

Compute the measures of central tendency: mean, median, and mode.

$\frac{4 + 6 + 7 + 8 + 100}{5} = \frac{125}{5} = 25$ .

Since there are 5 values (an odd number), the median is the 3rd value in the ordered list, which is $7$ .

Since every value is unique, there is no mode.

Answer: Mean $= 25$ , median $= 7$ , there is no mode.

Sidenote

Not all measures of central tendency are affected equally

Notice that the mean shifts from 7 to 25 because of the extreme value, while the median stays at 7. This shows the median’s resistance to extreme values (outliers).

Measures of spread

Measures of spread describe how much the values in a data set vary around the center.

Common measures include:

Range (difference between the highest and lowest values)
Interquartile range (IQR) (spread of the middle 50% of the data)
Variance (average squared deviation from the mean)
Standard deviation (square root of the variance, describing typical distance from the mean)

These measures help you describe consistency versus variability and can hint at outliers or clustering. For the Praxis, focus on range, interquartile range, and standard deviation. Introductory statistics courses go deeper into variance and its role.

Range

Formula $max (x_{i}) - min (x_{i})$
Total span of the data. Sensitive to extreme values.

Interquartile range IQR

Formula $IQR = Q_{3} - Q_{1}$
Spread of the middle fifty percent of data.
Ignores extremes.

Standard deviation (population)

Formula $σ = \frac{\sum _{i = 1}^{n} ( x _{i} - x ˉ ) ^{2}}{n}$
Average distance of values from the mean.

Example: Computing range and IQR $4, 6, 7, 8, 10$

To find the interquartile range (IQR), start by ordering the data and identifying the median, which is 7. Then split the data into a lower half and an upper half around the median.

The lower half is $4, 6$ , so the first quartile $Q_{1}$ is the average of those two values: $\frac{4 + 6}{2} = \frac{10}{2} = 5$ .

The upper half is $8, 10$ , so the third quartile $Q_{3}$ is $\frac{8 + 10}{2} = \frac{18}{2} = 9$ .

The IQR is then calculated as $Q_{3} - Q_{1} = 9 - 5 = 4$ .

Answer: $4$

Example: Computing standard deviation $4, 6, 7, 8, 10$

Compute the mean, $\frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$

To calculate the standard deviation, start by finding the deviations of each data point from the mean. If the mean is 7, then the deviations are: $4 - 7 = - 3$ , $6 - 7 = - 1$ , $7 - 7 = 0$ , $8 - 7 = 1$ , $10 - 7 = 3$ .

Next, square each deviation to eliminate negatives: $(- 3)^{2} = 9$ , $(- 1)^{2} = 1$ , $0^{2} = 0$ , $1^{2} = 1$ , and $3^{2} = 9$ .

These squared deviations are $9, 1, 0, 1, 9$ , which sum to 20.

Divide this sum by the number of data points (5) to find the variance: $\frac{20}{5} = 4$ .

Finally, take the square root of the variance to get the standard deviation: $4 = 2$ .

Answer: $2$

Choosing the right measure

Situation	Best measure of center	Best measure of spread	Why
Categorical data (e.g., favorite color)	Mode	Not applicable	Mean and median cannot be calculated, only mode makes sense.
Numerical data, no extreme values, symmetric	Mean	Standard deviation	Uses all values, accurate for well-behaved data.
Numerical data, no extreme values, skewed	Median	IQR	Median resists skew, IQR ignores extremes.
Numerical data with extreme values (outliers)	Median	IQR	Both are resistant to outliers, mean and standard deviation would be distorted.
Small data set	Median	Range	Easy to compute, IQR and standard deviation less meaningful with tiny samples.

Quick rule:

Mean and standard deviation: only when data is numeric, roughly symmetric, and outlier free.
Median and IQR: use when data is numeric but skewed or has outliers.
Mode: use for categorical data or when identifying the most common value matters.

Effects of transformations

When a constant $c$ is added to each value in a data set, the mean, median, and mode all increase by $c$ . Measures of spread such as the range, interquartile range (IQR), and standard deviation stay the same, because the distances between values don’t change.

In contrast, when each data value is multiplied by a positive constant $k$ , the mean, median, and mode are all multiplied by $k$ . The range, IQR, and standard deviation are also multiplied by $k$ , because all distances are scaled by the same factor.

Example: Applying transformations $4, 6, 7, 8, 10$

The original data set has the following statistics (as computed previously):

Mean = $7$

Median = $7$

Range = $6$

Interquartile range (IQR) = $4$

Standard deviation (SD) = $2$

After adding $3$ to each value, the new data set is: $7, 9, 10, 11, 13$

Mean = $10$

Median = $10$

Range = $6$ (unchanged)

IQR = $4$ (unchanged)

SD = $2$ (unchanged)

After multiplying each value by $2$ , the new data set is: $8, 12, 14, 16, 20$

Mean = $14$

Median = $14$

Range = $12$

IQR = $8$

SD = $4$

Answer:

Adding a constant shifts the mean and median but does not change range, IQR, or SD.

Multiplying by a constant scales the mean, median, range, IQR, and SD by that factor.

Example $5, 7, 12, 10, 7, 15, 13, 12$

For the data set above, find the following:

Mean

Median

Mode

Range

Interquartile range (IQR)

Standard deviation

(spoiler)

The mean is calculated as $\overset{x}{ˉ} = \frac{5 + 7 + 7 + 10 + 12 + 12 + 13 + 15}{8} = \frac{81}{8} = 10.125$
The ordered data is: $5, 7, 7, 10, 12, 12, 13, 15$ . Since there are 8 values, the median is the average of the 4th and 5th values the $Median = \frac{10 + 12}{2} = \frac{22}{2} = 11$
Mode is the value that occurs most often is 7 and 12, which appears twice.
$Range = max - min = 15 - 5 = 10$
The lower half of the data is: $5, 7, 7, 10$ .
So, $Q_{1}$ is the average of the 2nd and 3rd values: $Q_{1} = \frac{7 + 7}{2} = \frac{14}{2} = 7$
The upper half is: $12, 12, 13, 15$ . So, $Q_{3}$ is the average of the 2nd and 3rd values: $Q_{3} = \frac{12 + 13}{2} = \frac{25}{2} = 12.5$
$IQR = Q_{3} - Q_{1} = 12.5 - 7 = 5.5$
To find the standard deviation we use the mean $\overset{x}{ˉ} = 10.125$ , compute the deviations: $5 - 10.125 = - 5.125$ $7 - 10.125 = - 3.125$ $7 - 10.125 = - 3.125$ $10 - 10.125 = - 0.125$ $12 - 10.125 = 1.875$ $12 - 10.125 = 1.875$ $13 - 10.125 = 2.875$ $15 - 10.125 = 4.875$
Now square each deviation: $26.2656, 9.7656, 9.7656, 0.0156, 3.5156, 3.5156, 8.2656, 23.7656$
The sum of these squared deviations is approximately $84.8752$ .
Divide by 8 to get the variance: $σ^{2} = \frac{84.8752}{8} = 10.6094$
Then take the square root to calculate the standard deviation: $σ = 10.6094 \approx 3.257$

Answer:

Mean: $10.125$
Median: $11$
Mode: $7$ and $12$
Range: $10$
IQR: $5.5$
Standard deviation: $\approx 3.257$

Sidenote

How to choose numbers for quartile calculations

When your data set has an odd number of terms (e.g., $5$ or $7$ ), the median is the $\frac{n + 1}{2}$ th term.

Example: $2, 5, 8, 9, 10, 10, 11$

Here $n = 7$ , so the median is the 4th term ( $9$ ).

Lower quartile ( $Q_{1}$ ) uses the numbers below the median: $2, 5, 8$
Upper quartile ( $Q_{3}$ ) uses the numbers above the median: $10, 10, 11$

When your data set has an even number of terms (e.g., $6$ or $8$ ), the median is the average of the $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th terms.

Example: $1, 3, 3, 5, 7, 8, 9, 10$

Here $n = 8$ , so the median is $\frac{5 + 7}{2} = 6$ .

Lower quartile ( $Q_{1}$ ) uses the lower half: $1, 3, 3, 5$
Upper quartile ( $Q_{3}$ ) uses the upper half: $7, 8, 9, 10$

Special case, middle numbers are the same: $1, 4, 5, 6, 6, 8, 9, 12$

Here $n = 8$ , so the $\frac{n}{2}$ th term and the $\frac{n}{2} + 1$ th term are both $6$ . The median is still considered between them, so:

Lower quartile ( $Q_{1}$ ) includes all numbers before the median: $1, 4, 5, 6$
Upper quartile ( $Q_{3}$ ) includes all numbers after the median: $6, 8, 9, 12$

This keeps the quartiles balanced, even if the middle values are identical.

Always order data before computing median or quartiles.
Use mean for balanced average but exercise caution with extreme values.
Use median when distribution is skewed or contains extreme values.
Mode is useful for categorical or discrete data.
Range gives quick sense of total spread but is sensitive to extremes.
IQR focuses on central spread ignoring extremes.
Standard deviation quantifies average deviation from the mean.
Remember how transformations affect each measure.

Understanding central tendencies

Definitions

Mean (arithmetic average)

Formula

\overset{x}{ˉ} = \frac{\sum _{i = 1}^{n} x _{i}}{n}

. The balance point of the data. Sensitive to extreme values.

Median (middle value)

Order data from smallest to largest.

If $n$ is odd, the median is the $\frac{n + 1}{2}$ th value.
If $n$ is even, the median is the average of the $\frac{n}{2}$ th and $(\frac{n}{2} + 1)$ th values. Resistant to extreme values.

Mode (most frequent value)

The value or values that occur most often. A data set can be unimodal, bimodal, multimodal, or have no mode.

Range

The difference between the highest and lowest values in a data set. It gives a quick sense of how spread out the data are.

Interquartile range (IQR)

The spread of the middle 50% of the data, found by subtracting the first quartile (

Q_{1}

) from the third quartile (

Q_{3}

IQR = Q_{3} - Q_{1}

. The IQR is useful for identifying outliers.

Standard deviation

A measure of the average distance of each data value from the mean. A smaller standard deviation means the values are closer to the mean, while a larger one means the data are more spread out.

Extreme values (outliers)

Extremely large or extremely small numbers that are far away from the majority of the data.

Later, you’ll learn formal rules for identifying outliers. For now, treat an outlier as a value that looks unusually far from the rest of the data.

We’ll compare two data sets: one without an extreme value and one with an extreme value.

Example: Dataset without extreme values $4, 8, 6, 10, 7$

First, order the data: $4, 6, 7, 8, 10$ .

Next, compute the mean by adding the values and dividing by the number of values.

$\overset{x}{ˉ} = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$ .

The median is the middle value in the ordered list. Here, the third value is $7$ .

Check whether any value occurs most often (the mode). Since all values are unique, there is no mode.

Answer: Mean $= 7$ , median $= 7$ , there is no mode

Example: Dataset with an extreme value $4, 6, 7, 8, 100$

Order the data: $4, 6, 7, 8, 100$ .

Compute the measures of central tendency: mean, median, and mode.

$\frac{4 + 6 + 7 + 8 + 100}{5} = \frac{125}{5} = 25$ .

Since there are 5 values (an odd number), the median is the 3rd value in the ordered list, which is $7$ .

Since every value is unique, there is no mode.

Answer: Mean $= 25$ , median $= 7$ , there is no mode.

Sidenote

Not all measures of central tendency are affected equally

Notice that the mean shifts from 7 to 25 because of the extreme value, while the median stays at 7. This shows the median’s resistance to extreme values (outliers).

Measures of spread

Measures of spread describe how much the values in a data set vary around the center.

Common measures include:

Range (difference between the highest and lowest values)
Interquartile range (IQR) (spread of the middle 50% of the data)
Variance (average squared deviation from the mean)
Standard deviation (square root of the variance, describing typical distance from the mean)

Range

Formula $max (x_{i}) - min (x_{i})$
Total span of the data. Sensitive to extreme values.

Interquartile range IQR

Formula $IQR = Q_{3} - Q_{1}$
Spread of the middle fifty percent of data.
Ignores extremes.

Standard deviation (population)

Formula $σ = \frac{\sum _{i = 1}^{n} ( x _{i} - x ˉ ) ^{2}}{n}$
Average distance of values from the mean.

Example: Computing range and IQR $4, 6, 7, 8, 10$

To find the interquartile range (IQR), start by ordering the data and identifying the median, which is 7. Then split the data into a lower half and an upper half around the median.

The lower half is $4, 6$ , so the first quartile $Q_{1}$ is the average of those two values: $\frac{4 + 6}{2} = \frac{10}{2} = 5$ .

The upper half is $8, 10$ , so the third quartile $Q_{3}$ is $\frac{8 + 10}{2} = \frac{18}{2} = 9$ .

The IQR is then calculated as $Q_{3} - Q_{1} = 9 - 5 = 4$ .

Answer: $4$

Example: Computing standard deviation $4, 6, 7, 8, 10$

Compute the mean, $\frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$

To calculate the standard deviation, start by finding the deviations of each data point from the mean. If the mean is 7, then the deviations are: $4 - 7 = - 3$ , $6 - 7 = - 1$ , $7 - 7 = 0$ , $8 - 7 = 1$ , $10 - 7 = 3$ .

Next, square each deviation to eliminate negatives: $(- 3)^{2} = 9$ , $(- 1)^{2} = 1$ , $0^{2} = 0$ , $1^{2} = 1$ , and $3^{2} = 9$ .

These squared deviations are $9, 1, 0, 1, 9$ , which sum to 20.

Divide this sum by the number of data points (5) to find the variance: $\frac{20}{5} = 4$ .

Finally, take the square root of the variance to get the standard deviation: $4 = 2$ .

Answer: $2$

Choosing the right measure

Situation	Best measure of center	Best measure of spread	Why
Categorical data (e.g., favorite color)	Mode	Not applicable	Mean and median cannot be calculated, only mode makes sense.
Numerical data, no extreme values, symmetric	Mean	Standard deviation	Uses all values, accurate for well-behaved data.
Numerical data, no extreme values, skewed	Median	IQR	Median resists skew, IQR ignores extremes.
Numerical data with extreme values (outliers)	Median	IQR	Both are resistant to outliers, mean and standard deviation would be distorted.
Small data set	Median	Range	Easy to compute, IQR and standard deviation less meaningful with tiny samples.

Quick rule:

Mean and standard deviation: only when data is numeric, roughly symmetric, and outlier free.
Median and IQR: use when data is numeric but skewed or has outliers.
Mode: use for categorical data or when identifying the most common value matters.

Effects of transformations

Example: Applying transformations $4, 6, 7, 8, 10$

The original data set has the following statistics (as computed previously):

Mean = $7$

Median = $7$

Range = $6$

Interquartile range (IQR) = $4$

Standard deviation (SD) = $2$

After adding $3$ to each value, the new data set is: $7, 9, 10, 11, 13$

Mean = $10$

Median = $10$

Range = $6$ (unchanged)

IQR = $4$ (unchanged)

SD = $2$ (unchanged)

After multiplying each value by $2$ , the new data set is: $8, 12, 14, 16, 20$

Mean = $14$

Median = $14$

Range = $12$

IQR = $8$

SD = $4$

Answer:

Adding a constant shifts the mean and median but does not change range, IQR, or SD.

Multiplying by a constant scales the mean, median, range, IQR, and SD by that factor.

Example $5, 7, 12, 10, 7, 15, 13, 12$

For the data set above, find the following:

Mean

Median

Mode

Range

Interquartile range (IQR)

Standard deviation

(spoiler)

The mean is calculated as $\overset{x}{ˉ} = \frac{5 + 7 + 7 + 10 + 12 + 12 + 13 + 15}{8} = \frac{81}{8} = 10.125$
The ordered data is: $5, 7, 7, 10, 12, 12, 13, 15$ . Since there are 8 values, the median is the average of the 4th and 5th values the $Median = \frac{10 + 12}{2} = \frac{22}{2} = 11$
Mode is the value that occurs most often is 7 and 12, which appears twice.
$Range = max - min = 15 - 5 = 10$
The lower half of the data is: $5, 7, 7, 10$ .
So, $Q_{1}$ is the average of the 2nd and 3rd values: $Q_{1} = \frac{7 + 7}{2} = \frac{14}{2} = 7$
The upper half is: $12, 12, 13, 15$ . So, $Q_{3}$ is the average of the 2nd and 3rd values: $Q_{3} = \frac{12 + 13}{2} = \frac{25}{2} = 12.5$
$IQR = Q_{3} - Q_{1} = 12.5 - 7 = 5.5$
To find the standard deviation we use the mean $\overset{x}{ˉ} = 10.125$ , compute the deviations: $5 - 10.125 = - 5.125$ $7 - 10.125 = - 3.125$ $7 - 10.125 = - 3.125$ $10 - 10.125 = - 0.125$ $12 - 10.125 = 1.875$ $12 - 10.125 = 1.875$ $13 - 10.125 = 2.875$ $15 - 10.125 = 4.875$
Now square each deviation: $26.2656, 9.7656, 9.7656, 0.0156, 3.5156, 3.5156, 8.2656, 23.7656$
The sum of these squared deviations is approximately $84.8752$ .
Divide by 8 to get the variance: $σ^{2} = \frac{84.8752}{8} = 10.6094$
Then take the square root to calculate the standard deviation: $σ = 10.6094 \approx 3.257$

Answer:

Mean: $10.125$
Median: $11$
Mode: $7$ and $12$
Range: $10$
IQR: $5.5$
Standard deviation: $\approx 3.257$

Sidenote

How to choose numbers for quartile calculations

When your data set has an odd number of terms (e.g., $5$ or $7$ ), the median is the $\frac{n + 1}{2}$ th term.

Example: $2, 5, 8, 9, 10, 10, 11$

Here $n = 7$ , so the median is the 4th term ( $9$ ).

Lower quartile ( $Q_{1}$ ) uses the numbers below the median: $2, 5, 8$
Upper quartile ( $Q_{3}$ ) uses the numbers above the median: $10, 10, 11$

When your data set has an even number of terms (e.g., $6$ or $8$ ), the median is the average of the $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th terms.

Example: $1, 3, 3, 5, 7, 8, 9, 10$

Here $n = 8$ , so the median is $\frac{5 + 7}{2} = 6$ .

Lower quartile ( $Q_{1}$ ) uses the lower half: $1, 3, 3, 5$
Upper quartile ( $Q_{3}$ ) uses the upper half: $7, 8, 9, 10$

Special case, middle numbers are the same: $1, 4, 5, 6, 6, 8, 9, 12$

Here $n = 8$ , so the $\frac{n}{2}$ th term and the $\frac{n}{2} + 1$ th term are both $6$ . The median is still considered between them, so:

Lower quartile ( $Q_{1}$ ) includes all numbers before the median: $1, 4, 5, 6$
Upper quartile ( $Q_{3}$ ) includes all numbers after the median: $6, 8, 9, 12$

This keeps the quartiles balanced, even if the middle values are identical.

Achievable Praxis Core: Math (5733)

2. Data analysis, statistics, and probability

Our Praxis Core: Math course is currently in development and is a work-in-progress.

Understanding central tendencies

10 min read

Font

Discuss

Feedback

Definitions

Mean (arithmetic average)

Formula

\overset{x}{ˉ} = \frac{\sum _{i = 1}^{n} x _{i}}{n}

. The balance point of the data. Sensitive to extreme values.

Median (middle value)

Order data from smallest to largest.

If $n$ is odd, the median is the $\frac{n + 1}{2}$ th value.
If $n$ is even, the median is the average of the $\frac{n}{2}$ th and $(\frac{n}{2} + 1)$ th values. Resistant to extreme values.

Mode (most frequent value)

The value or values that occur most often. A data set can be unimodal, bimodal, multimodal, or have no mode.

Range

The difference between the highest and lowest values in a data set. It gives a quick sense of how spread out the data are.

Interquartile range (IQR)

The spread of the middle 50% of the data, found by subtracting the first quartile (

Q_{1}

) from the third quartile (

Q_{3}

IQR = Q_{3} - Q_{1}

. The IQR is useful for identifying outliers.

Standard deviation

A measure of the average distance of each data value from the mean. A smaller standard deviation means the values are closer to the mean, while a larger one means the data are more spread out.

Extreme values (outliers)

Extremely large or extremely small numbers that are far away from the majority of the data.

Later, you’ll learn formal rules for identifying outliers. For now, treat an outlier as a value that looks unusually far from the rest of the data.

We’ll compare two data sets: one without an extreme value and one with an extreme value.

Example: Dataset without extreme values $4, 8, 6, 10, 7$

First, order the data: $4, 6, 7, 8, 10$ .

Next, compute the mean by adding the values and dividing by the number of values.

$\overset{x}{ˉ} = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$ .

The median is the middle value in the ordered list. Here, the third value is $7$ .

Check whether any value occurs most often (the mode). Since all values are unique, there is no mode.

Answer: Mean $= 7$ , median $= 7$ , there is no mode

Example: Dataset with an extreme value $4, 6, 7, 8, 100$

Order the data: $4, 6, 7, 8, 100$ .

Compute the measures of central tendency: mean, median, and mode.

$\frac{4 + 6 + 7 + 8 + 100}{5} = \frac{125}{5} = 25$ .

Since there are 5 values (an odd number), the median is the 3rd value in the ordered list, which is $7$ .

Since every value is unique, there is no mode.

Answer: Mean $= 25$ , median $= 7$ , there is no mode.

Sidenote

Not all measures of central tendency are affected equally

Notice that the mean shifts from 7 to 25 because of the extreme value, while the median stays at 7. This shows the median’s resistance to extreme values (outliers).

Measures of spread

Measures of spread describe how much the values in a data set vary around the center.

Common measures include:

Range (difference between the highest and lowest values)
Interquartile range (IQR) (spread of the middle 50% of the data)
Variance (average squared deviation from the mean)
Standard deviation (square root of the variance, describing typical distance from the mean)

Range

Formula $max (x_{i}) - min (x_{i})$
Total span of the data. Sensitive to extreme values.

Interquartile range IQR

Formula $IQR = Q_{3} - Q_{1}$
Spread of the middle fifty percent of data.
Ignores extremes.

Standard deviation (population)

Formula $σ = \frac{\sum _{i = 1}^{n} ( x _{i} - x ˉ ) ^{2}}{n}$
Average distance of values from the mean.

Example: Computing range and IQR $4, 6, 7, 8, 10$

To find the interquartile range (IQR), start by ordering the data and identifying the median, which is 7. Then split the data into a lower half and an upper half around the median.

The lower half is $4, 6$ , so the first quartile $Q_{1}$ is the average of those two values: $\frac{4 + 6}{2} = \frac{10}{2} = 5$ .

The upper half is $8, 10$ , so the third quartile $Q_{3}$ is $\frac{8 + 10}{2} = \frac{18}{2} = 9$ .

The IQR is then calculated as $Q_{3} - Q_{1} = 9 - 5 = 4$ .

Answer: $4$

Example: Computing standard deviation $4, 6, 7, 8, 10$

Compute the mean, $\frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$

To calculate the standard deviation, start by finding the deviations of each data point from the mean. If the mean is 7, then the deviations are: $4 - 7 = - 3$ , $6 - 7 = - 1$ , $7 - 7 = 0$ , $8 - 7 = 1$ , $10 - 7 = 3$ .

Next, square each deviation to eliminate negatives: $(- 3)^{2} = 9$ , $(- 1)^{2} = 1$ , $0^{2} = 0$ , $1^{2} = 1$ , and $3^{2} = 9$ .

These squared deviations are $9, 1, 0, 1, 9$ , which sum to 20.

Divide this sum by the number of data points (5) to find the variance: $\frac{20}{5} = 4$ .

Finally, take the square root of the variance to get the standard deviation: $4 = 2$ .

Answer: $2$

Choosing the right measure

Situation	Best measure of center	Best measure of spread	Why
Categorical data (e.g., favorite color)	Mode	Not applicable	Mean and median cannot be calculated, only mode makes sense.
Numerical data, no extreme values, symmetric	Mean	Standard deviation	Uses all values, accurate for well-behaved data.
Numerical data, no extreme values, skewed	Median	IQR	Median resists skew, IQR ignores extremes.
Numerical data with extreme values (outliers)	Median	IQR	Both are resistant to outliers, mean and standard deviation would be distorted.
Small data set	Median	Range	Easy to compute, IQR and standard deviation less meaningful with tiny samples.

Quick rule:

Mean and standard deviation: only when data is numeric, roughly symmetric, and outlier free.
Median and IQR: use when data is numeric but skewed or has outliers.
Mode: use for categorical data or when identifying the most common value matters.

Effects of transformations

Example: Applying transformations $4, 6, 7, 8, 10$

The original data set has the following statistics (as computed previously):

Mean = $7$

Median = $7$

Range = $6$

Interquartile range (IQR) = $4$

Standard deviation (SD) = $2$

After adding $3$ to each value, the new data set is: $7, 9, 10, 11, 13$

Mean = $10$

Median = $10$

Range = $6$ (unchanged)

IQR = $4$ (unchanged)

SD = $2$ (unchanged)

After multiplying each value by $2$ , the new data set is: $8, 12, 14, 16, 20$

Mean = $14$

Median = $14$

Range = $12$

IQR = $8$

SD = $4$

Answer:

Adding a constant shifts the mean and median but does not change range, IQR, or SD.

Multiplying by a constant scales the mean, median, range, IQR, and SD by that factor.

Example $5, 7, 12, 10, 7, 15, 13, 12$

For the data set above, find the following:

Mean

Median

Mode

Range

Interquartile range (IQR)

Standard deviation

(spoiler)

The mean is calculated as $\overset{x}{ˉ} = \frac{5 + 7 + 7 + 10 + 12 + 12 + 13 + 15}{8} = \frac{81}{8} = 10.125$
The ordered data is: $5, 7, 7, 10, 12, 12, 13, 15$ . Since there are 8 values, the median is the average of the 4th and 5th values the $Median = \frac{10 + 12}{2} = \frac{22}{2} = 11$
Mode is the value that occurs most often is 7 and 12, which appears twice.
$Range = max - min = 15 - 5 = 10$
The lower half of the data is: $5, 7, 7, 10$ .
So, $Q_{1}$ is the average of the 2nd and 3rd values: $Q_{1} = \frac{7 + 7}{2} = \frac{14}{2} = 7$
The upper half is: $12, 12, 13, 15$ . So, $Q_{3}$ is the average of the 2nd and 3rd values: $Q_{3} = \frac{12 + 13}{2} = \frac{25}{2} = 12.5$
$IQR = Q_{3} - Q_{1} = 12.5 - 7 = 5.5$
To find the standard deviation we use the mean $\overset{x}{ˉ} = 10.125$ , compute the deviations: $5 - 10.125 = - 5.125$ $7 - 10.125 = - 3.125$ $7 - 10.125 = - 3.125$ $10 - 10.125 = - 0.125$ $12 - 10.125 = 1.875$ $12 - 10.125 = 1.875$ $13 - 10.125 = 2.875$ $15 - 10.125 = 4.875$
Now square each deviation: $26.2656, 9.7656, 9.7656, 0.0156, 3.5156, 3.5156, 8.2656, 23.7656$
The sum of these squared deviations is approximately $84.8752$ .
Divide by 8 to get the variance: $σ^{2} = \frac{84.8752}{8} = 10.6094$
Then take the square root to calculate the standard deviation: $σ = 10.6094 \approx 3.257$

Answer:

Mean: $10.125$
Median: $11$
Mode: $7$ and $12$
Range: $10$
IQR: $5.5$
Standard deviation: $\approx 3.257$

Sidenote

How to choose numbers for quartile calculations

When your data set has an odd number of terms (e.g., $5$ or $7$ ), the median is the $\frac{n + 1}{2}$ th term.

Example: $2, 5, 8, 9, 10, 10, 11$

Here $n = 7$ , so the median is the 4th term ( $9$ ).

Lower quartile ( $Q_{1}$ ) uses the numbers below the median: $2, 5, 8$
Upper quartile ( $Q_{3}$ ) uses the numbers above the median: $10, 10, 11$

When your data set has an even number of terms (e.g., $6$ or $8$ ), the median is the average of the $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th terms.

Example: $1, 3, 3, 5, 7, 8, 9, 10$

Here $n = 8$ , so the median is $\frac{5 + 7}{2} = 6$ .

Lower quartile ( $Q_{1}$ ) uses the lower half: $1, 3, 3, 5$
Upper quartile ( $Q_{3}$ ) uses the upper half: $7, 8, 9, 10$

Special case, middle numbers are the same: $1, 4, 5, 6, 6, 8, 9, 12$

Here $n = 8$ , so the $\frac{n}{2}$ th term and the $\frac{n}{2} + 1$ th term are both $6$ . The median is still considered between them, so:

Lower quartile ( $Q_{1}$ ) includes all numbers before the median: $1, 4, 5, 6$
Upper quartile ( $Q_{3}$ ) includes all numbers after the median: $6, 8, 9, 12$

This keeps the quartiles balanced, even if the middle values are identical.

Always order data before computing median or quartiles.
Use mean for balanced average but exercise caution with extreme values.
Use median when distribution is skewed or contains extreme values.
Mode is useful for categorical or discrete data.
Range gives quick sense of total spread but is sensitive to extremes.
IQR focuses on central spread ignoring extremes.
Standard deviation quantifies average deviation from the mean.
Remember how transformations affect each measure.

Understanding central tendencies

Definitions

Mean (arithmetic average)

Formula

\overset{x}{ˉ} = \frac{\sum _{i = 1}^{n} x _{i}}{n}

. The balance point of the data. Sensitive to extreme values.

Median (middle value)

Order data from smallest to largest.

If $n$ is odd, the median is the $\frac{n + 1}{2}$ th value.
If $n$ is even, the median is the average of the $\frac{n}{2}$ th and $(\frac{n}{2} + 1)$ th values. Resistant to extreme values.

Mode (most frequent value)

The value or values that occur most often. A data set can be unimodal, bimodal, multimodal, or have no mode.

Range

The difference between the highest and lowest values in a data set. It gives a quick sense of how spread out the data are.

Interquartile range (IQR)

The spread of the middle 50% of the data, found by subtracting the first quartile (

Q_{1}

) from the third quartile (

Q_{3}

IQR = Q_{3} - Q_{1}

. The IQR is useful for identifying outliers.

Standard deviation

A measure of the average distance of each data value from the mean. A smaller standard deviation means the values are closer to the mean, while a larger one means the data are more spread out.

Extreme values (outliers)

Extremely large or extremely small numbers that are far away from the majority of the data.

Later, you’ll learn formal rules for identifying outliers. For now, treat an outlier as a value that looks unusually far from the rest of the data.

We’ll compare two data sets: one without an extreme value and one with an extreme value.

Example: Dataset without extreme values $4, 8, 6, 10, 7$

First, order the data: $4, 6, 7, 8, 10$ .

Next, compute the mean by adding the values and dividing by the number of values.

$\overset{x}{ˉ} = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$ .

The median is the middle value in the ordered list. Here, the third value is $7$ .

Check whether any value occurs most often (the mode). Since all values are unique, there is no mode.

Answer: Mean $= 7$ , median $= 7$ , there is no mode

Example: Dataset with an extreme value $4, 6, 7, 8, 100$

Order the data: $4, 6, 7, 8, 100$ .

Compute the measures of central tendency: mean, median, and mode.

$\frac{4 + 6 + 7 + 8 + 100}{5} = \frac{125}{5} = 25$ .

Since there are 5 values (an odd number), the median is the 3rd value in the ordered list, which is $7$ .

Since every value is unique, there is no mode.

Answer: Mean $= 25$ , median $= 7$ , there is no mode.

Sidenote

Not all measures of central tendency are affected equally

Notice that the mean shifts from 7 to 25 because of the extreme value, while the median stays at 7. This shows the median’s resistance to extreme values (outliers).

Measures of spread

Measures of spread describe how much the values in a data set vary around the center.

Common measures include:

Range (difference between the highest and lowest values)
Interquartile range (IQR) (spread of the middle 50% of the data)
Variance (average squared deviation from the mean)
Standard deviation (square root of the variance, describing typical distance from the mean)

Range

Formula $max (x_{i}) - min (x_{i})$
Total span of the data. Sensitive to extreme values.

Interquartile range IQR

Formula $IQR = Q_{3} - Q_{1}$
Spread of the middle fifty percent of data.
Ignores extremes.

Standard deviation (population)

Formula $σ = \frac{\sum _{i = 1}^{n} ( x _{i} - x ˉ ) ^{2}}{n}$
Average distance of values from the mean.

Example: Computing range and IQR $4, 6, 7, 8, 10$

To find the interquartile range (IQR), start by ordering the data and identifying the median, which is 7. Then split the data into a lower half and an upper half around the median.

The lower half is $4, 6$ , so the first quartile $Q_{1}$ is the average of those two values: $\frac{4 + 6}{2} = \frac{10}{2} = 5$ .

The upper half is $8, 10$ , so the third quartile $Q_{3}$ is $\frac{8 + 10}{2} = \frac{18}{2} = 9$ .

The IQR is then calculated as $Q_{3} - Q_{1} = 9 - 5 = 4$ .

Answer: $4$

Example: Computing standard deviation $4, 6, 7, 8, 10$

Compute the mean, $\frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7$

To calculate the standard deviation, start by finding the deviations of each data point from the mean. If the mean is 7, then the deviations are: $4 - 7 = - 3$ , $6 - 7 = - 1$ , $7 - 7 = 0$ , $8 - 7 = 1$ , $10 - 7 = 3$ .

Next, square each deviation to eliminate negatives: $(- 3)^{2} = 9$ , $(- 1)^{2} = 1$ , $0^{2} = 0$ , $1^{2} = 1$ , and $3^{2} = 9$ .

These squared deviations are $9, 1, 0, 1, 9$ , which sum to 20.

Divide this sum by the number of data points (5) to find the variance: $\frac{20}{5} = 4$ .

Finally, take the square root of the variance to get the standard deviation: $4 = 2$ .

Answer: $2$

Choosing the right measure

Situation	Best measure of center	Best measure of spread	Why
Categorical data (e.g., favorite color)	Mode	Not applicable	Mean and median cannot be calculated, only mode makes sense.
Numerical data, no extreme values, symmetric	Mean	Standard deviation	Uses all values, accurate for well-behaved data.
Numerical data, no extreme values, skewed	Median	IQR	Median resists skew, IQR ignores extremes.
Numerical data with extreme values (outliers)	Median	IQR	Both are resistant to outliers, mean and standard deviation would be distorted.
Small data set	Median	Range	Easy to compute, IQR and standard deviation less meaningful with tiny samples.

Quick rule:

Mean and standard deviation: only when data is numeric, roughly symmetric, and outlier free.
Median and IQR: use when data is numeric but skewed or has outliers.
Mode: use for categorical data or when identifying the most common value matters.

Effects of transformations

Example: Applying transformations $4, 6, 7, 8, 10$

The original data set has the following statistics (as computed previously):

Mean = $7$

Median = $7$

Range = $6$

Interquartile range (IQR) = $4$

Standard deviation (SD) = $2$

After adding $3$ to each value, the new data set is: $7, 9, 10, 11, 13$

Mean = $10$

Median = $10$

Range = $6$ (unchanged)

IQR = $4$ (unchanged)

SD = $2$ (unchanged)

After multiplying each value by $2$ , the new data set is: $8, 12, 14, 16, 20$

Mean = $14$

Median = $14$

Range = $12$

IQR = $8$

SD = $4$

Answer:

Adding a constant shifts the mean and median but does not change range, IQR, or SD.

Multiplying by a constant scales the mean, median, range, IQR, and SD by that factor.

Example $5, 7, 12, 10, 7, 15, 13, 12$

For the data set above, find the following:

Mean

Median

Mode

Range

Interquartile range (IQR)

Standard deviation

(spoiler)

The mean is calculated as $\overset{x}{ˉ} = \frac{5 + 7 + 7 + 10 + 12 + 12 + 13 + 15}{8} = \frac{81}{8} = 10.125$
The ordered data is: $5, 7, 7, 10, 12, 12, 13, 15$ . Since there are 8 values, the median is the average of the 4th and 5th values the $Median = \frac{10 + 12}{2} = \frac{22}{2} = 11$
Mode is the value that occurs most often is 7 and 12, which appears twice.
$Range = max - min = 15 - 5 = 10$
The lower half of the data is: $5, 7, 7, 10$ .
So, $Q_{1}$ is the average of the 2nd and 3rd values: $Q_{1} = \frac{7 + 7}{2} = \frac{14}{2} = 7$
The upper half is: $12, 12, 13, 15$ . So, $Q_{3}$ is the average of the 2nd and 3rd values: $Q_{3} = \frac{12 + 13}{2} = \frac{25}{2} = 12.5$
$IQR = Q_{3} - Q_{1} = 12.5 - 7 = 5.5$
To find the standard deviation we use the mean $\overset{x}{ˉ} = 10.125$ , compute the deviations: $5 - 10.125 = - 5.125$ $7 - 10.125 = - 3.125$ $7 - 10.125 = - 3.125$ $10 - 10.125 = - 0.125$ $12 - 10.125 = 1.875$ $12 - 10.125 = 1.875$ $13 - 10.125 = 2.875$ $15 - 10.125 = 4.875$
Now square each deviation: $26.2656, 9.7656, 9.7656, 0.0156, 3.5156, 3.5156, 8.2656, 23.7656$
The sum of these squared deviations is approximately $84.8752$ .
Divide by 8 to get the variance: $σ^{2} = \frac{84.8752}{8} = 10.6094$
Then take the square root to calculate the standard deviation: $σ = 10.6094 \approx 3.257$

Answer:

Mean: $10.125$
Median: $11$
Mode: $7$ and $12$
Range: $10$
IQR: $5.5$
Standard deviation: $\approx 3.257$

Sidenote

How to choose numbers for quartile calculations

When your data set has an odd number of terms (e.g., $5$ or $7$ ), the median is the $\frac{n + 1}{2}$ th term.

Example: $2, 5, 8, 9, 10, 10, 11$

Here $n = 7$ , so the median is the 4th term ( $9$ ).

Lower quartile ( $Q_{1}$ ) uses the numbers below the median: $2, 5, 8$
Upper quartile ( $Q_{3}$ ) uses the numbers above the median: $10, 10, 11$

When your data set has an even number of terms (e.g., $6$ or $8$ ), the median is the average of the $\frac{n}{2}$ th and $\frac{n}{2} + 1$ th terms.

Example: $1, 3, 3, 5, 7, 8, 9, 10$

Here $n = 8$ , so the median is $\frac{5 + 7}{2} = 6$ .

Lower quartile ( $Q_{1}$ ) uses the lower half: $1, 3, 3, 5$
Upper quartile ( $Q_{3}$ ) uses the upper half: $7, 8, 9, 10$

Special case, middle numbers are the same: $1, 4, 5, 6, 6, 8, 9, 12$

Here $n = 8$ , so the $\frac{n}{2}$ th term and the $\frac{n}{2} + 1$ th term are both $6$ . The median is still considered between them, so:

Lower quartile ( $Q_{1}$ ) includes all numbers before the median: $1, 4, 5, 6$
Upper quartile ( $Q_{3}$ ) includes all numbers after the median: $6, 8, 9, 12$

This keeps the quartiles balanced, even if the middle values are identical.