SME
Cartography
Thematic mapping
Use this for “County values run from 12 to 92. With 5 equal-interval classes, what is the class width, and which class holds 50?”
Chart
20 values in 5 bins of width 16, from 12 to 92.
Bin 44 to 60: 4 of 20 = 0.2.
Shape: skewed right (a long tail to the right).
This is an example. Type your own values to start a new problem.
w
=
−
k
x
A value
example
c
Its class
worked out (not typed)
Formulas
w
=
xₘₐₓ − xₘᵢₙ
k
16
=
92 − 12
5
c
=
min(
k
,
⌊
x
− xₘᵢₙ
w
⌋
+
1)
3
=
min(5,
⌊
50 − 12
16
⌋
+
1)
Assumptions
•
Each class takes in its lower break and stops short of the next; the top class also takes the maximum.
•
Equal intervals suit data spread evenly over its range; skewed data leaves most places in one class, which quantile breaks fix.
•
The bars show 20 sample places binned at the breaks, so the classes can be seen filling.
Step-by-step
WE KNOW
xₘᵢₙ = 12, xₘₐₓ = 92, k = 5, x = 50
FIND
class width (w), its class (c)
Step 1 · Find class width (w)
Use w = (xₘₐₓ − xₘᵢₙ) ÷ k
Share the range, highest minus lowest, equally among the classes.
w
=
92 − 12
5
w
=
80
5
w
=
16
Step 2 · Find its class (c)
Use c = min(k, ⌊(x − xₘᵢₙ) ÷ w⌋ + 1)
Count
the
whole
class
widths
between
the
lowest
value
and
x
;
the
class
is
one
more,
and
the
maximum
stays
in
the
top
class.
c
=
min(5,
⌊
50 − 12
16
⌋
+
1)
Breaks: 12, 28, 44, 60, 76, 92
50 − 12
16
=
2.375
c
=
3
Check
Put the numbers back in. Both sides should match.
16
=
92 − 12
5
✓
3
=
min(5,
⌊
50 − 12
16
⌋
+
1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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