Transcript Properties of the Normal Curve

Revised 2002
Statistics
Show #3 of 3
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The NORMAL Curve
Properties
&
Example
2
Properties of the Normal Curve
It
is important
to (symmetric)
remember thatdistribution
this rule applies only to
•Bell
Shaped
NORMALLY DISTRIBUTED DATA.
•Mean = Median ( is the mathematical symbol used
Using the mean and standard deviation to describe the spread
for
this measure of center)
of the data can be done for any set of data.
•The
standard
) is used
for thedoes
measure
of
Just because
you deviation(
choose thismeasure
of spread
not
spread.
mean
that the rule applies!
THE 68-95-99.7 RULE
•Approximately 68% of the data lies within 1 standard
deviation of the mean
•Approximately 95% of the data lies within 2 standard
deviations of the mean
•Approximately 99.7% of the data lies within 3
standard deviations of the mean.
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A brief explanation...Percentile ranks…
• A percentile rank indicates a place in any set of data where a certain
percentage of the data falls AT or BELOW.
• For example, the MEDIAN is the 50th percentile,
• which means that 50% of the data values are EQUAL TO or LESS THAN
that median.
• It can also be said that the MEDIAN is greater than or equal to 50% of the
data values in the set.
M
95%
the that
dataisthat
the same
50% of
theofdata
theissame
as or as or less than some data value, D.
...SET
OF
ranked
list95th
of all
the values in the
This
value
(D) (M)
is also
called the
%ile
less
than
M.
ThisDATA…
value
is(a
also
called the 50th %ile
D
set)
5% of
the data
values
• Another example…
• The 95th percentile (95%ile) for a set of data would be...
• that data value that is greater than or equal to 95% of all of the data
values in the set.
• Notice, also that it is less than or equal to 5% of the values in the set4
Percentile ranks and the Normal Distribution…
• So, if you know that one property of a Normal Distribution is that 68% of
the data lies within 1 standard deviation from the mean...
• And you know that the mean = median for this type of distribution...
• You can say something about percentile ranks for this set of data...
50%ile
...SET OF NormallyM
distributed DATA…
M
Measure one s.d. to the left
34%
34%
data
and 68%
one toof
thethe
right
of M.
This is where you find 68%84%ile
• You know that M =16%ile
50%ile
of the data.
• and you know that a NORMAL
DISTRIBUTION means that the data is
SYMMETRIC about that data value…
• So, 34% of the data is in that part of the set of values that is one standard
deviation to the left of M
• and 34% is in the set to the right of M
• To find the percentile rank for the value to the LEFT of the mean, subtract:
50% -34% = 16%ile
Watch!
• Similarly, on the other side, 50% + 34% = 84%ile
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Using the properties of the Normal Curve...
• Bell
Shaped
distribution rule to determine
You
can
use(symmetric)
the 68-95-99.7
• Mean = Median () = 50th percentile
percentile
ranks
for other
points
on the
Normal
• The standard
deviation()
is used for
the measure
of spread.
• 68-95-99.7 rule...
Curve...
DISTRIBUTION OF VALUES:
.15% 2.5%
16%
50%
84% 97.5%
99.85%
68%
95%
99.7%
-3 -2 - 

+  + 2 + 3
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Normal Curve… common notation
Another common notation that is used when describing normally
DISTRIBUTION OF VALUES:
distributed data is called a STANDARD VALUE (standard score, z-score)
using STANDARD SCORES (Z-SCORES)
This notation simply uses the standard deviation as a “standard unit, “
• describes
(ex.) a standard
score(z-score)
of
and
how far away
(in standard deviations)
each actual data
+1 = an actual score of +
value is from the mean.
The
sign ofathe
standard
its direction
from
the mean.
• (ex.)
z-score
ofvalue
-3 =indicates
an actual
score of
-3
.15% 2.5% 16% 50% 84% 97.5% 99.85%
-3
-2
-1
-3 -2 - 
0

+1
+2
+3
+  + 2 + 3
Each actual data value corresponds to a standard value.
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Application of the Normal Curve
DISTRIBUTION OF VALUES
for weights of adult women whose height is approximately 5ft. 5 in.:
 = 134 pounds
.15% 2.5% 16%
-3
116
 = 6 pounds
50%
-2
-1
0
122
128
134
84% 97.5% 99.85%
+1
140
+2
146
+3
152
Weights in pounds
Put the actual values on the graph and complete the application
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Application of the Normal Curve
DISTRIBUTION OF VALUES for weights of adult women whose height is
approximately 5ft. 5 in.:
 = 134 pounds
 = 6 pounds
.15% 2.5% 16%
50% 84% 97.5% 99.85%
75%
116
122
128
134
140
146
152
(a) A woman weighing 140 lb. will be in which percentile?
84th %ile
(b) A woman weighing 122 lb. will be in which percentile?
2.5th %ile
(c) A woman the 75th percentile will weigh
About 137 pounds
approximately…?
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Application Analysis
• DISTRIBUTION OF VALUES for weights of adult women whose
height is approximately 5ft. 5 in.:
 = 134 pounds
=6
pounds
• (a) A woman weighing 140 lb. will be in the 84th percentile:
84 percent of women in this height category
weigh 140 lb or less. (also, 16% weigh more than 140)
• (b) A woman weighing 122 lb. will be in the 2.5th percentile:
2.5 percent of women
weigh 122 lb or less.(and 97.5% weigh more)
• (c) A woman in the 75th percentile will weigh approximately…?
The 75th percentile is between the 50th and 84th;
There are charts that can help you estimate more
therefore, the weight is between 134 and 140 lb
accurately between the standard percentiles; but we
won’t discuss them here.
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Analysis Using the Normal Distribution...
• Many distributions that occur naturally exhibit a normal
(or nearly normal) tendency…
–
–
–
–
Weights of children
Weights of adults in certain height categories
Scores on state-wide or national tests
etc…
• The properties exhibited by normally distributed data
allow us to make predictions about similar sets of data,
• and to draw conclusions about these sets without having
to handle each piece of data.
• See your text for more examples of normal distributions
and the uses of their properties!
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End of show #3
This is the final show in the
Statistics Unit
REVISED 2002
Prepared by Kimberly Conti, SUNY College @ Fredonia
Suggestions and comments to: [email protected]
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