Click on the graph in the upper right and drag the cursor to move
the shaded area on the graph. Move the cursor until the shaded
region represents the probability of interest (the “Between
(RedShaded)” number shown in the Probabilities column on the
right side will show the probability).
The value on the horizontal axis that marks the right side of the
shaded area will be the percentile of interest.
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Finding Percentiles Using The Normal Table
Note: Confusingly, some people refer to the
q
from the
q
th
percentile as the percentile, rather than the value that corresponds
to that probability.
Finding percentiles from a Normal table
To find the
q
th percentile of a
N
(
μ, σ
)
distribution:
1
Use the table for the
N
(0
,
1)
distribution to find the
z
score of
the
q
th percentile.
2
Convert the
z
score to proper units.
Since the
z
score is defined as
z
=
value

mean
standard deviation
=
x

μ
σ
,
then solving for
x
means the
q
th percentile is
q
th percentile
=
x
=
μ
+
σz.
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Example: Children’s Heights III
Example
Suppose children’s heights follow a
N
(68
,
2)
distribution. How tall
does a child need to be in order to be in the 90th percentile?
z
.05
.06
.07
.08
.09
0.8
0.8023
0.8051
0.8078
0.8106
0.8133
0.9
0.8289
0.8315
0.8340
0.8365
0.8389
1
0.8531
0.8554
0.8577
0.8599
0.8621
1.1
0.8749
0.8770
0.8790
0.8810
0.8830
1.2
0.8944
0.8962
0.8980
0.8997
0.9015
1.3
0.9115
0.9131
0.9147
0.9162
0.9177
1.4
0.9265
0.9279
0.9292
0.9306
0.9319
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Example: Children’s Heights III
Example
Suppose children’s heights follow a
N
(68
,
2)
distribution. How tall
does a child need to be in order to be in the 90th percentile?
z
.05
.06
.07
.08
.09
0.8
0.8023
0.8051
0.8078
0.8106
0.8133
0.9
0.8289
0.8315
0.8340
0.8365
0.8389
1
0.8531
0.8554
0.8577
0.8599
0.8621
1.1
0.8749
0.8770
0.8790
0.8810
0.8830
1.2
0.8944
0.8962
0.8980
0.8997
0.9015
1.3
0.9115
0.9131
0.9147
0.9162
0.9177
1.4
0.9265
0.9279
0.9292
0.9306
0.9319
To find the 90th percentile, we want to find the
z
score that
corresponds to a probability of 0.90. From the Normal table, the
probability of 0.8997 is the closest to 0.90. The probability of
0.8997 corresponds to a
z
score of 1.28.
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Example: Children’s Heights III
Example (Continued)
Suppose children’s heights follow a
N
(68
,
2)
distribution. How tall
does a child need to be in order to be in the 90th percentile?
To find the 90th percentile, we want to find the
z
score that
corresponds to a probability of 0.90. From the Normal table, the
probability of 0.8997 is the closest to 0.90. The probability of
0.8997 corresponds to a
z
score of 1.28.
Converting the
z
score of 1.28 into inches, we have
z
= 1
.
28 =
x

μ
σ
=
x

68
2
,
so
x
=
μ
+
σz
= 68 + (2
×
1
.
28) = 70
.
56
. A child needs to be
70.56 inches tall to be in the 90th percentile of heights.
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The Normal Model and the Empirical Rule
Recall from Chapter 3:
The Empirical Rule
If the distribution is symmetric and unimodal, then
Approximately 68% of the observations (roughly twothirds)
will be within one standard deviation of the mean.
Approximately 95% of the observations will be within two
standard deviations of the mean.
Nearly all the observations will be within three standard
deviations of the mean.
The Empirical Rule applies for any arbitrary symmetric and
unimodal distribution as an approximate guideline. However, for the
Normal model, the Empirical Rule is (nearly) exact.