SME
Grade 11 Math · Algebra 2
Normal distributions, z-scores and margin of error
Use this for “Heights have a mean of 170 cm and a standard deviation of 8 cm. What is the z-score of 182 cm, and what share is shorter?”
Chart
X is normal with μ = 170 cm and σ = 8 cm.
P(X ≤ 182) = 0.9332.
x = 182 is 1.5 standard deviations above the mean: z = 1.5.
This is an example. Type your own values to start a new problem.
z
=
x (cm)
−
μ (cm)
σ (cm)
P
Share below x
calculated
Formulas
z
=
x
−
μ
σ
1.5
=
182 − 170
8
P
=
Φ(
z
)
0.9332 = Φ(1.5)
Assumptions
•
The data are roughly bell-shaped and symmetric about the mean μ.
•
A z-score counts standard deviations from the mean: above the mean it is positive.
•
The area under the curve left of x is the share of the data below x.
Step-by-step
WE KNOW
μ = 170 cm, σ = 8 cm, x = 182 cm
FIND
z-score (z), share below x (P)
Step 1 · Find z-score (z)
Use z = (x − μ) ÷ σ
How far the value is from the mean, in standard deviations.
z
=
182 − 170
8
z
=
12
8
z
=
1.5
Step 2 · Find share below x (P)
Use P = Φ(z)
Φ(
z
),
the
z
-table
value
or
normalcdf(−∞,
z
):
the
area
left
of
z
.
P
=
Φ(1.5)
P
=
0.9332
Check
Put the numbers back in. Both sides should match.
1.5
=
182 − 170
8
✓
0.9332 = Φ(1.5) ✓
Topic check
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