Year 11 General Univariate Data Analysis

Dot Plots And Stem-And-Leaf Plots

20 practice questions 1 video lesson Theory + worked examples

Theory

Use dot plots and stem-and-leaf plots to calculate summary statistics (mean, median, quartiles) and to identify outliers. Mean from a dot plot uses $\bar{x} = \frac{\sum f x}{n}$ . Back-to-back stem-and-leaf plots compare two groups (left side reads right-to-left). An outlier is any value outside the 1.5 × IQR fences: $Q_{1} - 1.5 IQR$ or $Q_{3} + 1.5 IQR$ .

This subtopic builds on the basic dot plot and stem-and-leaf plot. Now we use them to calculate summary statistics, identify outliers, and compare two datasets using a back-to-back display.

For a dot plot, each dot is one data value. To find the mean, multiply each $x$ -value by the number of dots above it (the frequency), sum the products, and divide by the total count.

For a stem-and-leaf plot, read off all the values; the plot is already sorted, so the median, quartiles, minimum and maximum are easy to spot.

A back-to-back stem-and-leaf plot compares two groups by sharing one stem column, with one group's leaves on the left (read right-to-left) and the other on the right.

The 1.5 × IQR rule identifies outliers: a value $x$ is an outlier if

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

These two boundaries are the lower fence and upper fence.

The first diagram shows how to compute the mean from a dot plot using $\sum f x \div n$ . The second is a back-to-back stem-and-leaf plot comparing two classes.

12 households, 14 pets in total — mean

\approx 1.17

pets per household.

Shared stems in the middle. Class A's leaves read right-to-left; Class B's read left-to-right.

The key formulas are for the mean from frequencies and the outlier fences.

Mean from a dot plot or frequency table

$\bar{x} = \frac{\sum f x}{n}$

where each $x$ -value is multiplied by its frequency $f$ (number of dots), summed, and divided by the total number of data points $n$ .

\bar{x} = \frac{\sum f x}{n}

Outlier fences (1.5 × IQR rule)

$lower fence = Q_{1} - 1.5 \times IQR$

$upper fence = Q_{3} + 1.5 \times IQR$

upper fence = Q_{3} + 1.5 \times IQR

Outlier criterion

A value $x$ is an outlier if

$x < lower fence or x > upper fence$

Reading direction reminder. On a back-to-back stem-and-leaf plot, the leaf nearest the stem on either side is the smallest. The left side reads right-to-left; the right side reads left-to-right.

Mean from a dot plot

Count the dots above each value — this is the frequency $f$ .
Multiply each $x$ -value by its frequency to get $f x$ .
Sum the $f x$ values and the total number of dots $n = \sum f$ .
Divide: $\bar{x} = \frac{\sum f x}{n}$ .

Summary statistics from a stem-and-leaf plot

The plot is already sorted, so the minimum is the first leaf of the first stem and the maximum is the last leaf of the last stem.
Find the position $(n + 1) / 2$ of the median and count along the leaves.
For $Q_{1}$ and $Q_{3}$ , find the medians of the lower and upper halves.

Checking for outliers (1.5 × IQR rule)

Find $Q_{1}$ , $Q_{3}$ , and $IQR = Q_{3} - Q_{1}$ .
Compute the lower fence = $Q_{1} - 1.5 \times IQR$ and the upper fence = $Q_{3} + 1.5 \times IQR$ .
Any value below the lower fence or above the upper fence is an outlier.

EXAMPLE 1 — MEAN FROM A DOT PLOT

For the pets dot plot (

0 \to 3

1 \to 5

2 \to 3

3 \to 1

; total

12

), calculate the mean number of pets.

SOLUTION

Multiply each $x$ by its frequency, sum, and divide by $n = 12$ .

$\sum f x$	$=$	$0 (3) + 1 (5) + 2 (3) + 3 (1)$
	$=$	$0 + 5 + 6 + 3 = 14$
$\bar{x}$	$=$	$\frac{14}{12} \approx 1.17$

Answer: the mean is approximately $1.17$ pets.

\bar{x} \approx 1.17

EXAMPLE 2 — Q₁ FROM A STEM-AND-LEAF PLOT

A stem-and-leaf plot of

12

ages: stem

2 | 3 7

; stem

3 | 1 4 5 8

; stem

4 | 0 2 6 9

; stem

5 | 3 7

. Find

Q_{1}

SOLUTION

$n = 12$ . The lower half contains the smallest $6$ values: $23, 27, 31, 34, 35, 38$ . $Q_{1}$ is the median of these — the average of the $3$ rd and $4$ th values.

$Q_{1}$	$=$	$\frac{31 + 34}{2}$
	$=$	$32.5$

Answer: $Q_{1} = 32.5$ .

Q_{1} = 32.5

EXAMPLE 3 — THE IQR

For the same dataset,

Q_{3} = 47.5

. Calculate the IQR.

SOLUTION

The IQR is $Q_{3} - Q_{1}$ .

IQR	$=$	$Q_{3} - Q_{1}$
	$=$	$47.5 - 32.5$
	$=$	$15$

Answer: the IQR is $15$ .

IQR = 15

EXAMPLE 4 — OUTLIER CHECK

A test has

Q_{1} = 15

Q_{3} = 25

, so

IQR = 10

. Is a score of

45

an outlier?

SOLUTION

Compute the upper fence and compare to $45$ .

Upper fence	$=$	$Q_{3} + 1.5 \times IQR$
	$=$	$25 + 1.5 (10)$
	$=$	$40$

Since $45 > 40$ , the value $45$ is above the upper fence.

Answer: yes — $45$ is an outlier.

45 > 40 \Rightarrow outlier

Common pitfalls

On the left of a back-to-back plot, read right-to-left. The leaf closest to the stem is the smallest digit. Don't reverse this when picking out the minimum or computing the median.

Range = max − min, not (highest stem) − (lowest stem). For a stem-and-leaf plot, look at the actual data values. For stems

2

and

6

with leaves giving min

= 23

and max

= 61

, the range is

61 - 23 = 38

— not just

6 - 2 = 4

The IQR is multiplied by 1.5, not added.

1.5 \times 10 = 15

, not

11.5

. Get this wrong and the fences will be far off.

An outlier is past the fences, not at them. A value exactly equal to the upper fence (e.g.

40

when the fence is

40

) is not typically counted as an outlier — only values strictly above the upper fence (or strictly below the lower fence).

Mean and frequency. When computing the mean from a dot plot, divide by the total number of dots, not the number of distinct

x

-values. Three dots above

0

means three contributions of

0

, not one.

Frequently asked questions

How do I calculate the mean from a dot plot?

Each dot is one data value. Multiply each x-value by its frequency (the number of dots above it), add the products, and divide by the total number of dots. So for 3 zeros, 5 ones, and 4 twos, the mean is (0 times 3 plus 1 times 5 plus 2 times 4) divided by 12.

How do I read a back-to-back stem-and-leaf plot?

Both groups share a single stem column in the middle. The left group's leaves are read RIGHT-to-LEFT (the leaf closest to the stem is the smallest digit). The right group reads left-to-right as normal.

What is the 1.5 times IQR rule for outliers?

A value is an outlier if it falls below Q1 minus 1.5 times the IQR, or above Q3 plus 1.5 times the IQR. These two boundaries are called the lower and upper fences.

What is the upper fence and lower fence?

The upper fence equals Q3 plus 1.5 times the IQR. The lower fence equals Q1 minus 1.5 times the IQR. Any data value outside these fences is considered an outlier.

Why do we multiply the IQR by 1.5?

The factor 1.5 is a standard convention chosen so that, in a roughly normal distribution, about 99 percent of values fall within the fences. Values outside are unusual enough to investigate.

Can a dataset have more than one outlier?

Yes. Every value below the lower fence and every value above the upper fence is an outlier. Sometimes a dataset has several outliers at one or both ends.

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Summarising Data

Dot Plots And Stem-And-Leaf Plots

📖 Theory

Mean from a dot plot or frequency table

Outlier fences (1.5 × IQR rule)

Outlier criterion

Mean from a dot plot

Summary statistics from a stem-and-leaf plot

Checking for outliers (1.5 × IQR rule)

Common pitfalls

Frequently asked questions

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✏️ Practice Questions

Theory

Video Lesson

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