SME
Grade 9 Math · Problem type
Histogram from a data list
Use this for “Minutes of homework: 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40. Make a relative frequency histogram.”
Chart
12 values in 6 bins of width 10, from 10 to 70.
Heights
are
relative
frequencies:
count
n
.
Mean x̄ = 38.
Median = 39.
Shape: roughly symmetric.
This is an example. Type your own values to start a new problem.
n
Number of values
example
a
First bin starts at
example
w
Bin width
example
x₁
Value 1
example
x₂
Value 2
example
x₃
Value 3
example
x₄
Value 4
example
x₅
Value 5
example
x₆
Value 6
example
x₇
Value 7
example
x₈
Value 8
example
x₉
Value 9
example
x₁₀
Value 10
example
x₁₁
Value 11
example
x₁₂
Value 12
example
min
Least
worked out (not typed)
max
Greatest
worked out (not typed)
k
Number of bins
worked out (not typed)
e
Last bin ends at
worked out (not typed)
M
Median
worked out (not typed)
x̄
Mean
worked out (not typed)
Formulas
min
=
least
of
the
n
values
18 = least of the 12 values
max
=
greatest
of
the
n
values
60 = greatest of the 12 values
k
=
⌊
max −
a
w
⌋
+
1
6
=
⌊
60 − 10
10
⌋
+
1
e
=
a
+
k
×
w
70 = 10 + 6 × 10
M
=
median
of
the
n
values
39 = median of the 12 values
x̄
=
sum
of
the
n
values
n
38
=
sum
of
the
12
values
12
Assumptions
•
Pick a start at or below the least value and a bin width; each bin takes its left end and leaves out its right.
•
Add bins until the greatest value is in one: k = ⌊(max − a) ÷ w⌋ + 1.
•
Each height is a relative frequency, the bin’s count ÷ n, so the heights add to 1.
Step-by-step
WE KNOW
n = 12, a = 10, w = 10, x₁ = 35, x₂ = 42, x₃ = 18, x₄ = 50, x₅ = 27, x₆ = 44, x₇ = 38, x₈ = 60, x₉ = 30, x₁₀ = 47, x₁₁ = 25, x₁₂ = 40
FIND
least (min), greatest (max), number of bins (k), last bin ends at (e), median (M), mean (x̄)
Step 1 · Find least (min)
Use min = least of the n values
The smallest value, first in order.
min
=
least
of
x₁
,
x₂
,
x₃
,
x₄
,
x₅
,
x₆
,
x₇
,
x₈
,
x₉
,
x₁₀
,
x₁₁
,
x₁₂
min = least of 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40
In order: 18, 25, 27, 30, 35, 38, 40, 42, 44, 47, 50, 60
min = 18
Step 2 · Find greatest (max)
Use max = greatest of the n values
The largest value, last in order.
max
=
greatest
of
x₁
,
x₂
,
x₃
,
x₄
,
x₅
,
x₆
,
x₇
,
x₈
,
x₉
,
x₁₀
,
x₁₁
,
x₁₂
max = greatest of 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40
max = 60
Step 3 · Find number of bins (k)
Use k = ⌊(max − a) ÷ w⌋ + 1
Count the whole widths from the start to the greatest value, then one more bin to hold it.
k
=
⌊
60 − 10
10
⌋
+
1
[10,
20):
1
→
1
12
=
0.08333
[20,
30):
2
→
2
12
=
0.1667
[30,
40):
3
→
3
12
=
0.25
[40,
50):
4
→
4
12
=
0.3333
[50,
60):
1
→
1
12
=
0.08333
[60,
70):
1
→
1
12
=
0.08333
The
heights
add
to
1:
1
12
+
2
12
+
3
12
+
4
12
+
1
12
+
1
12
=
12
12
=
1
k
=
6
Step 4 · Find last bin ends at (e)
Use e = a + k × w
Start
at
a
and
add
k
widths:
the
right
end
of
the
last
bin.
e
=
10
+
6
×
10
e
=
10
+
60
e
=
70
Step 5 · Find median (M)
Use M = median of the n values
The middle value in order (halfway between the middle two); half the values lie on each side.
M
=
median
of
x₁
,
x₂
,
x₃
,
x₄
,
x₅
,
x₆
,
x₇
,
x₈
,
x₉
,
x₁₀
,
x₁₁
,
x₁₂
M
=
median
of
35,
42,
18,
50,
27,
44,
38,
60,
30,
47,
25,
40
The middle two are 38 and 40.
38 + 40
2
=
39
M
=
39
Step 6 · Find mean (x̄)
Use x̄ = sum of the n values ÷ n
Add the values, then divide by how many there are.
x̄
=
(
x₁
+
x₂
+
x₃
+
x₄
+
x₅
+
x₆
+
x₇
+
x₈
+
x₉
+
x₁₀
+
x₁₁
+
x₁₂
)
÷
n
x̄
=
(35
+
42
+
18
+
50
+
27
+
44
+
38
+
60
+
30
+
47
+
25
+
40)
÷
12
35 + 42 + 18 + 50 + 27 + 44 + 38 + 60 + 30 + 47 + 25 + 40 = 456
456
12
=
38
x̄
=
38
Check
Put the numbers back in. Both sides should match.
18 = least of 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40 ✓
60 = greatest of 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40 ✓
6
=
⌊
60 − 10
10
⌋
+
1
✓
70 = 10 + 6 × 10 ✓
39 = median of 35, 42, 18, 50, 27, 44, 38, 60, 30, 47, 25, 40 ✓
(35 + 42 + 18 + 50 + 27 + 44 + 38 + 60 + 30 + 47 + 25 + 40) ÷ 12 = 38 ✓
Topic check
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