SME
Grade 11 Math · Problem type
The 68–95–99.7 rule
Use this for “IQ scores have μ = 100 and σ = 15. About what percent are between 70 and 130?”
Chart
X is normal with μ = 100 and σ = 15.
P(70 ≤ X ≤ 130) = 0.9545.
Within 1σ: 0.6827; within 2σ: 0.9545; within 3σ: 0.9973.
The 68–95–99.7 rule rounds these.
This is an example. Type your own values to start a new problem.
a
=
μ
−
k
×
σ
b
=
μ
+
k
×
σ
P
Share within k standard deviations
worked out (not typed) · %
Formulas
a
=
μ
−
k
×
σ
70 = 100 − 2 × 15
b
=
μ
+
k
×
σ
130 = 100 + 2 × 15
P
=
share
within
k
standard
deviations
95 = share within 2 standard deviations
Assumptions
•
The data are roughly normal, with mean μ and standard deviation σ.
•
About 68%, 95% and 99.7% of the data are within 1, 2 and 3 standard deviations of the mean.
•
The rule gives rounded shares; a z-table gives more digits.
Step-by-step
WE KNOW
μ = 100, σ = 15, k = 2
FIND
lower end (a), upper end (b), share within k standard deviations (P)
Step 1 · Find lower end (a)
Use a = μ − k × σ
Go
down
k
standard
deviations
from
the
mean.
a
=
100
−
2
×
15
a
=
100
−
30
a
=
70
Step 2 · Find upper end (b)
Use b = μ + k × σ
Go
up
k
standard
deviations
from
the
mean.
b
=
100
+
2
×
15
b
=
100
+
30
b
=
130
Step 3 · Find share within k standard deviations (P)
Use P = share within k standard deviations
The 68–95–99.7 rule: 1, 2 or 3 standard deviations each way.
P
=
share
within
2
standard
deviations
P
=
95%
Check
Put the numbers back in. Both sides should match.
70 = 100 − 2 × 15 ✓
130 = 100 + 2 × 15 ✓
95 = share within 2 standard deviations ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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