http wwwyoutubecomwatchvxDIyAOBayU The Standard Normal Distribution 1 Notice that the x axis is standard scores also called z scores This means that the distribution has a ID: 136642
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Slide1
http://www.youtube.com/watch?v=xDIyAOBa_yU
The Standard Normal Distribution
Note
:
This file is on my webpage for this course. See the syllabus for the web address.Slide2
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.Slide3
The most significant thing about the normal distribution is that
predictable proportions of cases occur in specific regions of the curve.
50%
50%Slide4
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% below.Slide5
IQ Percentile Problem 1
:
IQ:
m=100, s=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 = x
i
-
m
/
s
= (115-100)/15 = 15/15; z = +1.0
(1
sd above the mean)
IQ=115, z = 1.0Slide6
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.
IQ=115, z = +1.0
The part with the
slanty
lines
repre-sents
the portion of the distribution we’re looking for.Slide7
IQ Percentile Problem 2
:
IQ:
m=100, s=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 = x
i
-
m
/
s
= (85-100)/15 = -15/15; z = -1.0
(1
sd below
the mean)
IQ=85, z = -1.0Slide8
Step 2:
Calculate the area under the curve for all scores below z=-1.
Area under the curve values below z=-1.0: 50.00-34.13=
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.
IQ=85, z = -1.0Slide9
z = x
i-m/
s =
85-100/15 = -15/15z = -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.
IQ=85, z=-1.0