SME
Grade 11 Math · Problem type
The value at a percentile
Use this for “Scores have μ = 500 and σ = 100. What score is at the 90th percentile?”
Chart
X is normal with μ = 500 and σ = 100.
P(X ≤ 628.1552) = 0.9000.
x = 628.1552 is 1.2816 standard deviations above the mean: z = 1.2816.
This is an example. Type your own values to start a new problem.
x
=
μ
+
z
×
σ
This is z = (x − μ) ÷ σ solved for x. The z comes from the percentile: invNorm(P).
P
Share below x (the percentile)
example
Formulas
P
=
Φ(
z
)
0.9 = Φ(1.2816)
z
=
x
−
μ
σ
1.2816
=
628.16 − 500
100
Assumptions
•
The data are roughly normal, with mean μ and standard deviation σ.
•
At the 90th percentile, 90% of the data are below: the area left of x is 0.90.
•
invNorm(P) is the z with area P to its left.
Step-by-step
WE KNOW
μ = 500, σ = 100, P = 0.9
FIND
z-score (z), value at the percentile (x)
Step 1 · Find z-score (z)
Use P = Φ(z)
invNorm
undoes
Φ:
the
z
with
that
area
to
its
left.
z
=
invNorm(
P
)
z
=
invNorm(0.9)
z
=
1.2816
Step 2 · Find value at the percentile (x)
Use z = (x − μ) ÷ σ
Start
at
the
mean
and
go
z
standard
deviations.
x
=
μ
+
z
×
σ
x
=
500
+
1.2816
×
100
x
=
500
+
128.16
x
=
628.16
Check
Put the numbers back in. Both sides should match.
0.9 = Φ(1.2816) ✓
1.2816
=
628.16 − 500
100
✓
Topic check
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