9. INTERPRETING DENSITY CURVES USING THE EMPIRICAL
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Transcript 9. INTERPRETING DENSITY CURVES USING THE EMPIRICAL
NORMAL Distribution
The Empirical Rule
Normal Distribution?
These density curves are symmetric,
single-peaked, and bell-shaped.
We capitalize Normal to remind you
that these curves are special.
Normal distribution is described by giving
its mean μ and its standard deviation σ
Shape of the Normal
curve
The standard deviation σ controls the
spread of a Normal curve
The Empirical Rule
68-95-99.7 rule
In the Normal distribution with mean μ and standard
deviation σ:
• Approximately 68% of the observations fall within σ
of the mean μ.
• Approximately 95% of the observations fall within
2σ of μ.
• Approximately 99.7% of the observations fall
within 3σ of μ.
The Normal Distribution
and Empirical Rule
Example: YOUNG
WOMEN’s HEIGHT
The distribution of heights of young women aged 18 to 24
is approximately Normal with mean μ = 64.5 inches and
standard deviation σ = 2.5 inches.
Importance of Normal
Curve
• scores on tests taken by many even though many
people (such as SAT exams and sets of data follow a
many psychological tests),
Normal distribution,
• repeated careful measurements
of the same quantity, and
• characteristics of biological
populations (such as yields of
corn and lengths of animal
pregnancies).
many do not.
Most income
distributions, for
example, are skewed to
the right and so are not
Normal
Standard Normal
distribution
Standard Normal Distribution
The standard Normal distribution is the
Normal distribution N(0, 1) with mean 0 and
standard deviation
Standard Normal
Calculation
The Standard Normal Table
Table A is a table of areas under the standard Normal
curve. The table entry for each value z is the area
under the curve to the left of z.
How to use the
table of values
Area to the
LEFT
Using the standard Normal table
Problem: Find the proportion of observations from the
standard Normal distribution that are less than 2.22.
illustrates the relationship between the
value z = 2.22 and the area 0.9868.
illustrates the relationship
between the value z = 2.22 and
the area 0.9868.
Example
Area to the
RIGHT
Using the standard Normal table
Problem: Find the proportion of observations from the
standard Normal distribution that are greater than −2.15
z = −2.15
Area = 0.0158
Area = 1-0.0158
Area = .9842
Practice
(a) z < 2.85
(a) 0.9978.
(b) z > 2.85
(b)0.0022.
(c) z > −1.66
(c) 0.9515.
(d) −1.66 < z < 2.85
(d) 0.9493.
CODY’S quiz score
relative to his classmates
79 81 80 77 73 83 74 93 78 80 75 67 73
77 83 86 90 79 85 89 84 77 72 83 82 x
z = 0.99
Area = .8389
Cody’s actual score
relative to the other
students who took the
same test is 84%