The Standard Normal Distribution
Normal distribution demo:
http://www.youtube.com/watch?v=xDIyAOBa_yU
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Notice that the x axis is standard scores, also
called z scores. This means that the distribution
has a population mean of zero, and a population
standard deviation of 1.
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The most significant thing about the normal
distribution is that predictable proportions of
cases occur in specific regions of the curve.
50%
50%
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Notice that: (1) 34.13% of the scores lie between the
mean and 1 sd above the mean, (2) 13.6% of the scores
lie between 1 sd above the mean and 2 sds above the
mean, (3) 2.14% of the scores lie between 2 sds above
the mean and 3 sds above the mean, and (4) 0.1% of
the scores in the entire region above 3 sds.
The curve is
symmetrical,
so the area
betw 0 and -1
= 34.13%, -1
to -2 = 13.6%,
etc. Also,
50% of the
cases are
above 0, 50%
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below.
IQ=115, z = 1.0
IQ Percentile Problem 1:
IQ: μ=100, σ=15. Convert an IQ score of 115 into a
percentile, using the standard normal distribution.
Step 1: Convert IQ=115 into a z score:
z = xi-μ/σ = (115-100)/15 = 15/15; z = +1.0
(1 sd above the mean)
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IQ=115, z = +1.0
The part with
the slanty
lines represents the
portion of the
distribution
we’re looking
for.
Step 2: Calculate the area under the curve for all
scores below z=1 (percentile=% of scores falling below a score).
Area under the curve below z=1.0: 34.13+50.00=
84.13. We get this by adding 34.13 (the area between the
mean and z=1) to 50.00 (50% is the area under the curve for values
less than zero; i.e., the entire left side of the bell curve). So, an IQ
score of 115 (z=1.0) has a percentile score of 84.13.
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IQ=85, z = -1.0
IQ Percentile Problem 2:
IQ: μ=100, σ=15. Convert an IQ score of 85 into a
percentile, using the standard normal distribution.
Step 1: Convert IQ=85 into a z score:
z = xi-μ/σ = (85-100)/15 = -15/15; z = -1.0
(1 sd below the mean)
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IQ=85, z = -1.0
Step 2: Calculate the area under the curve for all
scores below z=-1.
Area under the curve values below z=-1.0: 50.0034.13=15.87. We get this by subtracting 34.13 (the area
between the mean and z=-1) from 50.00 (50% is the total area under
the curve for values less than zero; i.e., the entire left side of the bell
curve).
So, an IQ score of 85 (z=-1.0) has a percentile
score of 15.87.
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z = xi-μ/σ = 85-100/15 = -15/15
z = -1.0
IQ=85, z=-1.0
Area under the curve for z=-1.0: 50.00-34.13=15.87.
We get this by subtracting 34.13 from 50.00 (50.00 is
the total area under the curve for values less than zero; i.e., the entire
left side of the bell curve.) We
therefore want the 50%
minus the area between zero and -1.0). So, an IQ
score of 85 has a percentile score of 15.87.
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