2.3 Interpreting data

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.

Interpreting data

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Spotting patterns, trends, and outliers

You’ll often use data displays to spot overall shapes and patterns and to notice values that stand out. You’ll also use the five-number summary - minimum, $Q_{1}$ , median ( $Q_{2}$ ), $Q_{3}$ , and maximum - to describe a data set and create a boxplot. These tools help you summarize data, identify clusters or gaps, and support conclusions with specific evidence.

Definitions

Cluster: A concentration of data values within a specific range, indicating a subgroup or phase in the data. Clusters show where values tend to group together and can highlight common behaviors or conditions.
Outlier: A data value that is substantially higher or lower than the rest of the data set. You can often spot an outlier by eye as a value lying far from the main grouping, and you can verify it mathematically by computing $Q_{1}$ and $Q_{3}$ , then $IQR = Q_{3} - Q_{1}$ , and checking if the value is greater than $Q_{3} + 1.5 \times IQR$ or less than $Q_{1} - 1.5 \times IQR$ .
Trend: A general direction in which data values are moving over time or across categories. Trends can be increasing, decreasing, or constant, and they help reveal long-term patterns in a data set.

Patterns and trends describe how data values behave as a group - for example, steadily rising, steadily falling, or clustering around certain values. Outliers matter because they can distort summary measures like the mean or the range.

A box-and-whisker plot (or boxplot) summarizes a data set using five key numbers: minimum, first quartile $Q_{1}$ , median $Q_{2}$ , third quartile $Q_{3}$ , and maximum values. It shows the center and spread of the distribution.

In a boxplot, outliers are data values that lie far from the rest of the dataset. Formally:

$Outlier if: x > Q_{3} + 1.5 \times IQR or x < Q_{1} - 1.5 \times IQR$

where $IQR = Q_{3} - Q_{1}$ .

Values beyond this range are often plotted individually as points or small circles.
Outliers can indicate unusual cases, errors in data collection, or natural variability.

Finding the five-number summary

Take the following data set:

$65, 68, 71, 74, 77, 82, 82, 89$

The values below show the five-number summary for this data set. (If you need a refresher on how each value is found, refer back to section $1$ of this unit.)

Minimum (Min) = $65$

Maximum (Max) = $89$

Median ( $Q_{2}$ ) = $75.5$

First quartile ( $Q_{1}$ ) = $69.5$

Third quartile ( $Q_{3}$ ) = $82$

Visual interpretation

Some problems ask you to choose the correct box-and-whisker plot (also called a box plot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.

A box plot can be drawn horizontally or vertically.
In a horizontal box plot, the minimum is on the left and the maximum is on the right.
In a vertical box plot, the minimum is at the bottom and the maximum is at the top.

Let’s sketch one using the five-number summary from the previous example.

Example: Drawing a five-number summary

Suppose you want to manually construct the boxplot.

To create a box plot, start by ordering the data set:

$65, 68, 71, 74, 77, 82, 82, 89$

To draw the box plot:

Draw a box from $Q_{1} = 69.5$ to $Q_{3} = 82$

Place a vertical line inside the box at the median $Q_{2} = 75.5$

Extend whiskers from the box out to the minimum $(65)$ and maximum $(89)$

Answer: Draw the box from $69.5$ to $82$ , place the median line at $75.5$ , and draw whiskers to $65$ and $89$ .

Example: Identify the outlier and compare means

$10, 12, 15, 14, 13, 100, 16, 14$

Find the following:

The outlier

The mean of all eight values

The mean of the seven typical values excluding the outlier

(spoiler)

Order the data: $10, 12, 13, 14, 14, 15, 16, 100$

First we need to compute the IQR by finding the corresponding $Q_{1}$ and $Q_{3}$ values.
$Q_{1} = \frac{12 + 13}{2} = 12.5$
$Q_{3} = \frac{15 + 16}{2} = 15.5$
Recall we find the IQR by taking the $Q_{1}$ and subtracting it from the $Q_{3}$ value. $IQR = 15.5 - 12.5 = 3$

By the definition of outliers, if the number in the data set exceeds the sum of $Q_{3}$ and $1.5 \times IQR$ then it is an outlier.

Outlier cutoff: $Q_{3} + 1.5 \times IQR = 15.5 + (1.5 \times 3) = 15.5 + 4.5 = 20$ Since $100 > 20$ , it is an outlier.
Mean of all values: $\frac{10 + 12 + 13 + 14 + 14 + 15 + 16 + 100}{8} = \frac{194}{8} = 24.25$
Mean without outlier: $\frac{94}{7} \approx 13.43$

Answer: The outlier is $100$ . The mean of all values is $24.25$ , and the mean without the outlier is approximately $13.43$ .

Why this matters: The outlier inflates the mean by over 80%, which can hide what’s typical in the data. For example, in home prices, a few multi million dollar mansions can pull the mean upward, which is one reason real estate often reports the middle value instead.

Example: Detect trend, cluster, and outlier Given the following data set describe cluster(s), trend, or outlier(s) if applicable.

$5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

(spoiler)

Two clusters: $5$ - $10$ and $15$ - $21$
- Could also argue three clusters $5$ - $10$ , $15$ - $17$ , and $20$ - $21$

Given the data set: $5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

The data is already ordered from least to greatest.
Median ( $Q_{2}$ ): There are $10$ values in the data set. Since $10$ is even, the median is the average of the $5^{th}$ and $6^{th}$ values: $Q_{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5$
Lower quartile ( $Q_{1}$ ): The lower half of the data includes the values below the median: $5, 7, 8, 9, 10$ This set has $5$ values, so $Q_{1}$ is the middle (third) value: $Q_{1} = 8$
Upper quartile ( $Q_{3}$ ): The upper half of the data includes the values above the median: $15, 16, 17, 20, 21$ This set also has $5$ values, so $Q_{3}$ is the middle (third) value: $Q_{3} = 17$
Interquartile range (IQR): $IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

To determine if any values in the data set are outliers, we use the $1.5 \times IQR$ rule.

We already found:

$Q_{1} = 8$
$Q_{3} = 17$
$IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

Now calculate the lower and upper bounds for outliers:

Lower bound = $Q_{1} - 1.5 \times IQR = 8 - 1.5 \times 9 = 8 - 13.5 = - 5.5$
Upper bound = $Q_{3} + 1.5 \times IQR = 17 + 1.5 \times 9 = 17 + 13.5 = 30.5$

Any value below $- 5.5$ or above $30.5$ is considered an outlier. Since all values in the data set fall between $5$ and $21$ , there are no outliers.

Answer: The data show two clusters ( $5$ - $10$ and $15$ - $21$ ) and no outliers.

Why this matters: Recognizing clusters can reveal meaningful subgroups, and confirming that there are no outliers helps you trust that your summaries represent the full data set.

Justifying conclusions with data

Strong conclusions point to specific numbers, trends, or features in the display, and they avoid claims the data can’t support.

Example: Trend description and justification Monthly website visits are shown below:

Month	Visits
Jan	1200
Feb	1500
Mar	1800

Write a sentence describing the trend and justify it with data.

(spoiler)

Visits rose by exactly $300$ each month from $1200$ in January to $1800$ in March, indicating a consistent upward trend likely driven by marketing or seasonal factors.

Answer: Visits increased by $300$ each month from January to March.

Why this matters: Using exact changes (like “up $300$ each month”) keeps your conclusion precise and tied directly to the data.

Look for overall patterns, trends, clusters, or cycles before focusing on individual values
Identify any values that stand apart and confirm them as outliers using the $1.5 \times IQR$ rule
Use the five-number summary to describe the spread and center of a data set, and represent it visually with a boxplot
Anchor conclusions in specific numeric changes or clear features from the display
Avoid claims that extend beyond what the data actually show

Interpreting data

Spotting patterns, trends, and outliers

Definitions

Cluster: A concentration of data values within a specific range, indicating a subgroup or phase in the data. Clusters show where values tend to group together and can highlight common behaviors or conditions.
Outlier: A data value that is substantially higher or lower than the rest of the data set. You can often spot an outlier by eye as a value lying far from the main grouping, and you can verify it mathematically by computing $Q_{1}$ and $Q_{3}$ , then $IQR = Q_{3} - Q_{1}$ , and checking if the value is greater than $Q_{3} + 1.5 \times IQR$ or less than $Q_{1} - 1.5 \times IQR$ .
Trend: A general direction in which data values are moving over time or across categories. Trends can be increasing, decreasing, or constant, and they help reveal long-term patterns in a data set.

In a boxplot, outliers are data values that lie far from the rest of the dataset. Formally:

$Outlier if: x > Q_{3} + 1.5 \times IQR or x < Q_{1} - 1.5 \times IQR$

where $IQR = Q_{3} - Q_{1}$ .

Values beyond this range are often plotted individually as points or small circles.
Outliers can indicate unusual cases, errors in data collection, or natural variability.

Finding the five-number summary

Take the following data set:

$65, 68, 71, 74, 77, 82, 82, 89$

The values below show the five-number summary for this data set. (If you need a refresher on how each value is found, refer back to section $1$ of this unit.)

Minimum (Min) = $65$

Maximum (Max) = $89$

Median ( $Q_{2}$ ) = $75.5$

First quartile ( $Q_{1}$ ) = $69.5$

Third quartile ( $Q_{3}$ ) = $82$

Visual interpretation

Some problems ask you to choose the correct box-and-whisker plot (also called a box plot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.

A box plot can be drawn horizontally or vertically.
In a horizontal box plot, the minimum is on the left and the maximum is on the right.
In a vertical box plot, the minimum is at the bottom and the maximum is at the top.

Let’s sketch one using the five-number summary from the previous example.

Example: Drawing a five-number summary

Suppose you want to manually construct the boxplot.

To create a box plot, start by ordering the data set:

$65, 68, 71, 74, 77, 82, 82, 89$

To draw the box plot:

Draw a box from $Q_{1} = 69.5$ to $Q_{3} = 82$

Place a vertical line inside the box at the median $Q_{2} = 75.5$

Extend whiskers from the box out to the minimum $(65)$ and maximum $(89)$

Answer: Draw the box from $69.5$ to $82$ , place the median line at $75.5$ , and draw whiskers to $65$ and $89$ .

Example: Identify the outlier and compare means

$10, 12, 15, 14, 13, 100, 16, 14$

Find the following:

The outlier

The mean of all eight values

The mean of the seven typical values excluding the outlier

(spoiler)

Order the data: $10, 12, 13, 14, 14, 15, 16, 100$

First we need to compute the IQR by finding the corresponding $Q_{1}$ and $Q_{3}$ values.
$Q_{1} = \frac{12 + 13}{2} = 12.5$
$Q_{3} = \frac{15 + 16}{2} = 15.5$
Recall we find the IQR by taking the $Q_{1}$ and subtracting it from the $Q_{3}$ value. $IQR = 15.5 - 12.5 = 3$

By the definition of outliers, if the number in the data set exceeds the sum of $Q_{3}$ and $1.5 \times IQR$ then it is an outlier.

Outlier cutoff: $Q_{3} + 1.5 \times IQR = 15.5 + (1.5 \times 3) = 15.5 + 4.5 = 20$ Since $100 > 20$ , it is an outlier.
Mean of all values: $\frac{10 + 12 + 13 + 14 + 14 + 15 + 16 + 100}{8} = \frac{194}{8} = 24.25$
Mean without outlier: $\frac{94}{7} \approx 13.43$

Answer: The outlier is $100$ . The mean of all values is $24.25$ , and the mean without the outlier is approximately $13.43$ .

Why this matters: The outlier inflates the mean by over 80%, which can hide what’s typical in the data. For example, in home prices, a few multi million dollar mansions can pull the mean upward, which is one reason real estate often reports the middle value instead.

Example: Detect trend, cluster, and outlier Given the following data set describe cluster(s), trend, or outlier(s) if applicable.

$5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

(spoiler)

Two clusters: $5$ - $10$ and $15$ - $21$
- Could also argue three clusters $5$ - $10$ , $15$ - $17$ , and $20$ - $21$

Given the data set: $5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

The data is already ordered from least to greatest.
Median ( $Q_{2}$ ): There are $10$ values in the data set. Since $10$ is even, the median is the average of the $5^{th}$ and $6^{th}$ values: $Q_{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5$
Lower quartile ( $Q_{1}$ ): The lower half of the data includes the values below the median: $5, 7, 8, 9, 10$ This set has $5$ values, so $Q_{1}$ is the middle (third) value: $Q_{1} = 8$
Upper quartile ( $Q_{3}$ ): The upper half of the data includes the values above the median: $15, 16, 17, 20, 21$ This set also has $5$ values, so $Q_{3}$ is the middle (third) value: $Q_{3} = 17$
Interquartile range (IQR): $IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

To determine if any values in the data set are outliers, we use the $1.5 \times IQR$ rule.

We already found:

$Q_{1} = 8$
$Q_{3} = 17$
$IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

Now calculate the lower and upper bounds for outliers:

Lower bound = $Q_{1} - 1.5 \times IQR = 8 - 1.5 \times 9 = 8 - 13.5 = - 5.5$
Upper bound = $Q_{3} + 1.5 \times IQR = 17 + 1.5 \times 9 = 17 + 13.5 = 30.5$

Any value below $- 5.5$ or above $30.5$ is considered an outlier. Since all values in the data set fall between $5$ and $21$ , there are no outliers.

Answer: The data show two clusters ( $5$ - $10$ and $15$ - $21$ ) and no outliers.

Why this matters: Recognizing clusters can reveal meaningful subgroups, and confirming that there are no outliers helps you trust that your summaries represent the full data set.

Justifying conclusions with data

Strong conclusions point to specific numbers, trends, or features in the display, and they avoid claims the data can’t support.

Example: Trend description and justification Monthly website visits are shown below:

Month	Visits
Jan	1200
Feb	1500
Mar	1800

Write a sentence describing the trend and justify it with data.

(spoiler)

Visits rose by exactly $300$ each month from $1200$ in January to $1800$ in March, indicating a consistent upward trend likely driven by marketing or seasonal factors.

Answer: Visits increased by $300$ each month from January to March.

Why this matters: Using exact changes (like “up $300$ each month”) keeps your conclusion precise and tied directly to the data.

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.

Interpreting data

8 min read

Font

Discuss

Feedback

Spotting patterns, trends, and outliers

Definitions

Cluster: A concentration of data values within a specific range, indicating a subgroup or phase in the data. Clusters show where values tend to group together and can highlight common behaviors or conditions.
Outlier: A data value that is substantially higher or lower than the rest of the data set. You can often spot an outlier by eye as a value lying far from the main grouping, and you can verify it mathematically by computing $Q_{1}$ and $Q_{3}$ , then $IQR = Q_{3} - Q_{1}$ , and checking if the value is greater than $Q_{3} + 1.5 \times IQR$ or less than $Q_{1} - 1.5 \times IQR$ .
Trend: A general direction in which data values are moving over time or across categories. Trends can be increasing, decreasing, or constant, and they help reveal long-term patterns in a data set.

In a boxplot, outliers are data values that lie far from the rest of the dataset. Formally:

$Outlier if: x > Q_{3} + 1.5 \times IQR or x < Q_{1} - 1.5 \times IQR$

where $IQR = Q_{3} - Q_{1}$ .

Values beyond this range are often plotted individually as points or small circles.
Outliers can indicate unusual cases, errors in data collection, or natural variability.

Finding the five-number summary

Take the following data set:

$65, 68, 71, 74, 77, 82, 82, 89$

The values below show the five-number summary for this data set. (If you need a refresher on how each value is found, refer back to section $1$ of this unit.)

Minimum (Min) = $65$

Maximum (Max) = $89$

Median ( $Q_{2}$ ) = $75.5$

First quartile ( $Q_{1}$ ) = $69.5$

Third quartile ( $Q_{3}$ ) = $82$

Visual interpretation

Some problems ask you to choose the correct box-and-whisker plot (also called a box plot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.

A box plot can be drawn horizontally or vertically.
In a horizontal box plot, the minimum is on the left and the maximum is on the right.
In a vertical box plot, the minimum is at the bottom and the maximum is at the top.

Let’s sketch one using the five-number summary from the previous example.

Example: Drawing a five-number summary

Suppose you want to manually construct the boxplot.

To create a box plot, start by ordering the data set:

$65, 68, 71, 74, 77, 82, 82, 89$

To draw the box plot:

Draw a box from $Q_{1} = 69.5$ to $Q_{3} = 82$

Place a vertical line inside the box at the median $Q_{2} = 75.5$

Extend whiskers from the box out to the minimum $(65)$ and maximum $(89)$

Answer: Draw the box from $69.5$ to $82$ , place the median line at $75.5$ , and draw whiskers to $65$ and $89$ .

Example: Identify the outlier and compare means

$10, 12, 15, 14, 13, 100, 16, 14$

Find the following:

The outlier

The mean of all eight values

The mean of the seven typical values excluding the outlier

(spoiler)

Order the data: $10, 12, 13, 14, 14, 15, 16, 100$

First we need to compute the IQR by finding the corresponding $Q_{1}$ and $Q_{3}$ values.
$Q_{1} = \frac{12 + 13}{2} = 12.5$
$Q_{3} = \frac{15 + 16}{2} = 15.5$
Recall we find the IQR by taking the $Q_{1}$ and subtracting it from the $Q_{3}$ value. $IQR = 15.5 - 12.5 = 3$

By the definition of outliers, if the number in the data set exceeds the sum of $Q_{3}$ and $1.5 \times IQR$ then it is an outlier.

Outlier cutoff: $Q_{3} + 1.5 \times IQR = 15.5 + (1.5 \times 3) = 15.5 + 4.5 = 20$ Since $100 > 20$ , it is an outlier.
Mean of all values: $\frac{10 + 12 + 13 + 14 + 14 + 15 + 16 + 100}{8} = \frac{194}{8} = 24.25$
Mean without outlier: $\frac{94}{7} \approx 13.43$

Answer: The outlier is $100$ . The mean of all values is $24.25$ , and the mean without the outlier is approximately $13.43$ .

Why this matters: The outlier inflates the mean by over 80%, which can hide what’s typical in the data. For example, in home prices, a few multi million dollar mansions can pull the mean upward, which is one reason real estate often reports the middle value instead.

Example: Detect trend, cluster, and outlier Given the following data set describe cluster(s), trend, or outlier(s) if applicable.

$5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

(spoiler)

Two clusters: $5$ - $10$ and $15$ - $21$
- Could also argue three clusters $5$ - $10$ , $15$ - $17$ , and $20$ - $21$

Given the data set: $5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

The data is already ordered from least to greatest.
Median ( $Q_{2}$ ): There are $10$ values in the data set. Since $10$ is even, the median is the average of the $5^{th}$ and $6^{th}$ values: $Q_{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5$
Lower quartile ( $Q_{1}$ ): The lower half of the data includes the values below the median: $5, 7, 8, 9, 10$ This set has $5$ values, so $Q_{1}$ is the middle (third) value: $Q_{1} = 8$
Upper quartile ( $Q_{3}$ ): The upper half of the data includes the values above the median: $15, 16, 17, 20, 21$ This set also has $5$ values, so $Q_{3}$ is the middle (third) value: $Q_{3} = 17$
Interquartile range (IQR): $IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

To determine if any values in the data set are outliers, we use the $1.5 \times IQR$ rule.

We already found:

$Q_{1} = 8$
$Q_{3} = 17$
$IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

Now calculate the lower and upper bounds for outliers:

Lower bound = $Q_{1} - 1.5 \times IQR = 8 - 1.5 \times 9 = 8 - 13.5 = - 5.5$
Upper bound = $Q_{3} + 1.5 \times IQR = 17 + 1.5 \times 9 = 17 + 13.5 = 30.5$

Any value below $- 5.5$ or above $30.5$ is considered an outlier. Since all values in the data set fall between $5$ and $21$ , there are no outliers.

Answer: The data show two clusters ( $5$ - $10$ and $15$ - $21$ ) and no outliers.

Why this matters: Recognizing clusters can reveal meaningful subgroups, and confirming that there are no outliers helps you trust that your summaries represent the full data set.

Justifying conclusions with data

Strong conclusions point to specific numbers, trends, or features in the display, and they avoid claims the data can’t support.

Example: Trend description and justification Monthly website visits are shown below:

Month	Visits
Jan	1200
Feb	1500
Mar	1800

Write a sentence describing the trend and justify it with data.

(spoiler)

Visits rose by exactly $300$ each month from $1200$ in January to $1800$ in March, indicating a consistent upward trend likely driven by marketing or seasonal factors.

Answer: Visits increased by $300$ each month from January to March.

Why this matters: Using exact changes (like “up $300$ each month”) keeps your conclusion precise and tied directly to the data.

Look for overall patterns, trends, clusters, or cycles before focusing on individual values
Identify any values that stand apart and confirm them as outliers using the $1.5 \times IQR$ rule
Use the five-number summary to describe the spread and center of a data set, and represent it visually with a boxplot
Anchor conclusions in specific numeric changes or clear features from the display
Avoid claims that extend beyond what the data actually show

Interpreting data

Spotting patterns, trends, and outliers

Definitions

Cluster: A concentration of data values within a specific range, indicating a subgroup or phase in the data. Clusters show where values tend to group together and can highlight common behaviors or conditions.
Outlier: A data value that is substantially higher or lower than the rest of the data set. You can often spot an outlier by eye as a value lying far from the main grouping, and you can verify it mathematically by computing $Q_{1}$ and $Q_{3}$ , then $IQR = Q_{3} - Q_{1}$ , and checking if the value is greater than $Q_{3} + 1.5 \times IQR$ or less than $Q_{1} - 1.5 \times IQR$ .
Trend: A general direction in which data values are moving over time or across categories. Trends can be increasing, decreasing, or constant, and they help reveal long-term patterns in a data set.

In a boxplot, outliers are data values that lie far from the rest of the dataset. Formally:

$Outlier if: x > Q_{3} + 1.5 \times IQR or x < Q_{1} - 1.5 \times IQR$

where $IQR = Q_{3} - Q_{1}$ .

Values beyond this range are often plotted individually as points or small circles.
Outliers can indicate unusual cases, errors in data collection, or natural variability.

Finding the five-number summary

Take the following data set:

$65, 68, 71, 74, 77, 82, 82, 89$

The values below show the five-number summary for this data set. (If you need a refresher on how each value is found, refer back to section $1$ of this unit.)

Minimum (Min) = $65$

Maximum (Max) = $89$

Median ( $Q_{2}$ ) = $75.5$

First quartile ( $Q_{1}$ ) = $69.5$

Third quartile ( $Q_{3}$ ) = $82$

Visual interpretation

Some problems ask you to choose the correct box-and-whisker plot (also called a box plot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.

A box plot can be drawn horizontally or vertically.
In a horizontal box plot, the minimum is on the left and the maximum is on the right.
In a vertical box plot, the minimum is at the bottom and the maximum is at the top.

Let’s sketch one using the five-number summary from the previous example.

Example: Drawing a five-number summary

Suppose you want to manually construct the boxplot.

To create a box plot, start by ordering the data set:

$65, 68, 71, 74, 77, 82, 82, 89$

To draw the box plot:

Draw a box from $Q_{1} = 69.5$ to $Q_{3} = 82$

Place a vertical line inside the box at the median $Q_{2} = 75.5$

Extend whiskers from the box out to the minimum $(65)$ and maximum $(89)$

Answer: Draw the box from $69.5$ to $82$ , place the median line at $75.5$ , and draw whiskers to $65$ and $89$ .

Example: Identify the outlier and compare means

$10, 12, 15, 14, 13, 100, 16, 14$

Find the following:

The outlier

The mean of all eight values

The mean of the seven typical values excluding the outlier

(spoiler)

Order the data: $10, 12, 13, 14, 14, 15, 16, 100$

First we need to compute the IQR by finding the corresponding $Q_{1}$ and $Q_{3}$ values.
$Q_{1} = \frac{12 + 13}{2} = 12.5$
$Q_{3} = \frac{15 + 16}{2} = 15.5$
Recall we find the IQR by taking the $Q_{1}$ and subtracting it from the $Q_{3}$ value. $IQR = 15.5 - 12.5 = 3$

By the definition of outliers, if the number in the data set exceeds the sum of $Q_{3}$ and $1.5 \times IQR$ then it is an outlier.

Outlier cutoff: $Q_{3} + 1.5 \times IQR = 15.5 + (1.5 \times 3) = 15.5 + 4.5 = 20$ Since $100 > 20$ , it is an outlier.
Mean of all values: $\frac{10 + 12 + 13 + 14 + 14 + 15 + 16 + 100}{8} = \frac{194}{8} = 24.25$
Mean without outlier: $\frac{94}{7} \approx 13.43$

Answer: The outlier is $100$ . The mean of all values is $24.25$ , and the mean without the outlier is approximately $13.43$ .

Why this matters: The outlier inflates the mean by over 80%, which can hide what’s typical in the data. For example, in home prices, a few multi million dollar mansions can pull the mean upward, which is one reason real estate often reports the middle value instead.

Example: Detect trend, cluster, and outlier Given the following data set describe cluster(s), trend, or outlier(s) if applicable.

$5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

(spoiler)

Two clusters: $5$ - $10$ and $15$ - $21$
- Could also argue three clusters $5$ - $10$ , $15$ - $17$ , and $20$ - $21$

Given the data set: $5, 7, 8, 9, 10, 15, 16, 17, 20, 21$

The data is already ordered from least to greatest.
Median ( $Q_{2}$ ): There are $10$ values in the data set. Since $10$ is even, the median is the average of the $5^{th}$ and $6^{th}$ values: $Q_{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5$
Lower quartile ( $Q_{1}$ ): The lower half of the data includes the values below the median: $5, 7, 8, 9, 10$ This set has $5$ values, so $Q_{1}$ is the middle (third) value: $Q_{1} = 8$
Upper quartile ( $Q_{3}$ ): The upper half of the data includes the values above the median: $15, 16, 17, 20, 21$ This set also has $5$ values, so $Q_{3}$ is the middle (third) value: $Q_{3} = 17$
Interquartile range (IQR): $IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

To determine if any values in the data set are outliers, we use the $1.5 \times IQR$ rule.

We already found:

$Q_{1} = 8$
$Q_{3} = 17$
$IQR = Q_{3} - Q_{1} = 17 - 8 = 9$

Now calculate the lower and upper bounds for outliers:

Lower bound = $Q_{1} - 1.5 \times IQR = 8 - 1.5 \times 9 = 8 - 13.5 = - 5.5$
Upper bound = $Q_{3} + 1.5 \times IQR = 17 + 1.5 \times 9 = 17 + 13.5 = 30.5$

Any value below $- 5.5$ or above $30.5$ is considered an outlier. Since all values in the data set fall between $5$ and $21$ , there are no outliers.

Answer: The data show two clusters ( $5$ - $10$ and $15$ - $21$ ) and no outliers.

Why this matters: Recognizing clusters can reveal meaningful subgroups, and confirming that there are no outliers helps you trust that your summaries represent the full data set.

Justifying conclusions with data

Strong conclusions point to specific numbers, trends, or features in the display, and they avoid claims the data can’t support.

Example: Trend description and justification Monthly website visits are shown below:

Month	Visits
Jan	1200
Feb	1500
Mar	1800

Write a sentence describing the trend and justify it with data.

(spoiler)

Visits rose by exactly $300$ each month from $1200$ in January to $1800$ in March, indicating a consistent upward trend likely driven by marketing or seasonal factors.

Answer: Visits increased by $300$ each month from January to March.

Why this matters: Using exact changes (like “up $300$ each month”) keeps your conclusion precise and tied directly to the data.