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.
Finding the five-number summary
To find Q1 and Q3: order the data, locate the median, then split into a lower half and an upper half. If n is odd, exclude the median from both halves; if n is even, split evenly. Q1 is the median of the lower half and Q3 is the median of the upper half.
Take the following data set:
65,68,71,74,77,82,82,89
There are n=8 values, already ordered. The median is the average of the 4th and 5th values:
Q2=274+77=75.5
The lower half is 65,68,71,74, so:
Q1=268+71=69.5
The upper half is 77,82,82,89, so:
Q3=282+82=82
This gives the five-number summary:
Minimum (Min) = 65
First quartile (Q1) = 69.5
Median (Q2) = 75.5
Third quartile (Q3) = 82
Maximum (Max) = 89
Visual interpretation
Some problems ask you to choose the correct box-and-whisker plot (also called a boxplot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.
The box spans from Q1 to Q3, the median Q2 is marked by a line inside the box, and whiskers extend from the box out to the minimum and maximum. Boxplots can be drawn horizontally or vertically; the box still spans from Q1 to Q3 with the median line inside - only the axis orientation changes.
Here’s what the five-number summary from the previous example looks like as a boxplot. Since no values fall outside the 1.5×IQR fences, the whiskers extend all the way to the actual minimum and maximum.
Boxplot of quiz scores
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, compute Q1 and Q3 to find the IQR.
Q1=212+13=12.5
Q3=215+16=15.5
IQR=Q3−Q1=15.5−12.5=3
Now apply the 1.5×IQR rule. A value is an outlier if it exceeds Q3+1.5×IQR:
Mean of all values: 810+12+13+14+14+15+16+100=8194=24.25
Mean without outlier: 794≈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.
Example: Detect trend, clusters, and outliers
Given the following data set, describe any clusters, trends, or outliers.
5,7,8,9,10,15,16,17,20,21
(spoiler)
The data is already ordered from least to greatest.
Clusters: There are two visible clusters - one from 5 to 10 and another from 15 to 21, with a gap in between.
Outliers: All values fall between 5 and 21 with no extreme gaps from the group, so there are no outliers.
Answer: The data show two clusters (5-10 and 15-21) and no outliers.
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. When describing a trend, name the direction and support it with at least one specific value - for example, “visits increased by 300 each month from January (1,200) to March (1,800).”
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×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
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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.
Finding the five-number summary
To find Q1 and Q3: order the data, locate the median, then split into a lower half and an upper half. If n is odd, exclude the median from both halves; if n is even, split evenly. Q1 is the median of the lower half and Q3 is the median of the upper half.
Take the following data set:
65,68,71,74,77,82,82,89
There are n=8 values, already ordered. The median is the average of the 4th and 5th values:
Q2=274+77=75.5
The lower half is 65,68,71,74, so:
Q1=268+71=69.5
The upper half is 77,82,82,89, so:
Q3=282+82=82
This gives the five-number summary:
Minimum (Min) = 65
First quartile (Q1) = 69.5
Median (Q2) = 75.5
Third quartile (Q3) = 82
Maximum (Max) = 89
Visual interpretation
Some problems ask you to choose the correct box-and-whisker plot (also called a boxplot) for a given data set. To do that, you need to know how the five-number summary maps onto the picture.
The box spans from Q1 to Q3, the median Q2 is marked by a line inside the box, and whiskers extend from the box out to the minimum and maximum. Boxplots can be drawn horizontally or vertically; the box still spans from Q1 to Q3 with the median line inside - only the axis orientation changes.
Here’s what the five-number summary from the previous example looks like as a boxplot. Since no values fall outside the 1.5×IQR fences, the whiskers extend all the way to the actual minimum and maximum.
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, compute Q1 and Q3 to find the IQR.
Q1=212+13=12.5
Q3=215+16=15.5
IQR=Q3−Q1=15.5−12.5=3
Now apply the 1.5×IQR rule. A value is an outlier if it exceeds Q3+1.5×IQR:
Mean of all values: 810+12+13+14+14+15+16+100=8194=24.25
Mean without outlier: 794≈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.
Example: Detect trend, clusters, and outliers
Given the following data set, describe any clusters, trends, or outliers.
5,7,8,9,10,15,16,17,20,21
(spoiler)
The data is already ordered from least to greatest.
Clusters: There are two visible clusters - one from 5 to 10 and another from 15 to 21, with a gap in between.
Outliers: All values fall between 5 and 21 with no extreme gaps from the group, so there are no outliers.
Answer: The data show two clusters (5-10 and 15-21) and no outliers.
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. When describing a trend, name the direction and support it with at least one specific value - for example, “visits increased by 300 each month from January (1,200) to March (1,800).”