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STAT1010 – Empirical rule
Review Exercises: Recall…
(mosaic plot)
Given these 3 things:
P(R)=0.10
1.00
0.75
low
P(H)=0.20
0.50
P(H|R)=0.20
0.25
0.00
high
Find: 1) P(R) = P(not R)=
2) P(H|R)=P(not H | R)=
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Review Exercises:
3) Suppose P(A)=0.5, P(B)=0.75, P(A and B)=0.4
Find: P(A or B)
4) Suppose P(A)=0.2, P(B)=0.65, P(A or B)=0.75
Find: P(A and B)
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5.2 Properties of the
Normal Distribution
!  READ
THIS SECTION IN THE BOOK!!!
" Understand
the examples.
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STAT1010 – Empirical rule
5.2 Properties of the
Normal Distribution
!  Important
!  Important
!  Important
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Normal Distribution
!  Most
widely used distribution.
!  Arises naturally in physical phenomena
!  Two parameters COMPLETELY define a
normal distribution, µ and σ.
!  µ is the center (mean) of the distribution.
!  σ is the standard deviation of the distribution
(quantifies spread).
!  Symmetrical distribution.
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Normal Distribution
!  Can
occur anywhere along the real number
line. µ specifies the center (position) and σ
specifies the spread.
Different normal distributions for selected values of
the parameters µ and σ.
Recall that the area under the curve is 1 (or 100%)
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STAT1010 – Empirical rule
Empirical Rule (68-95-99.7 Rule)
!  Special
result of the normal distribution:
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Empirical Rule (68-95-99.7 Rule)
Consider the normal model
with µ = 0 and σ =1.
100% of the data falls in (- ∞ , ∞ )
µ
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Empirical Rule (68-95-99.7 Rule)
! 
68% of the values fall within 1
standard deviation of the mean.
68%
1σ
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STAT1010 – Empirical rule
Empirical Rule (68-95-99.7 Rule)
! 
95% of the values fall within 2
standard deviations of the mean.
95%
2σ
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Empirical Rule (68-95-99.7 Rule)
! 
99.7% of the values fall within 3
standard deviations of the mean.
0.15% left
in this tail
99.7%
3σ
0.15% left in this tail
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Empirical Rule (68-95-99.7 Rule)
! 
Very useful for
determining what %
of the observations
fall between two
x-values
! 
Very useful for
determining what %
of the observations
fall in the tail
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STAT1010 – Empirical rule
Finding a Percentile using the
Empirical Rule
!  Example
1: The amount of cereal in a box
varies a little from box to box. Suppose the
amount in a box has a normal distribution
with µ=15 oz. and σ=0.2 oz.
What percentage of boxes have between
14.6 and 15.4 oz. of cereal?
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Finding a Percentile using the
Empirical Rule
14.6
14.8
15 15.2 15.4
ß Specific to cereal
What percent is within 2 standard deviations of the mean?
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Finding a Percentile using the
Empirical Rule
Example 1: What percentage of boxes have
between 14.6 and 15.4 oz. of cereal?
Two standard deviations down from µ:
15 – 2(0.2) = 14.6
Two standard deviations up from µ:
15 + 2(0.2) = 15.4
Answer: 95%
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STAT1010 – Empirical rule
Finding a Percentile using the
Empirical Rule
!  Example
2: Octane ratings are normally
distributed with µ=91 and σ=1.5
What percentage of octane ratings
fall below 92.5?
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Finding a Percentile using the
Empirical Rule
68% of the values fall between 89.5 and 92.5
(µ - 1σ, µ + 1σ)
95% of the values fall between 88.0 and 94.0
(µ - 2σ, µ + 2σ)
99.7% of the values fall between 86.5 and 95.5
(µ - 3σ, µ + 3σ)
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Finding a Percentile using the
Empirical Rule
!  What
percentage of octane ratings fall
below 92.5?
!  Draw a picture.
!  How far away from the mean is 92.5 in
terms of number of standard deviations?
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STAT1010 – Empirical rule
Draw a picture
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Finding a Percentile using the
Empirical Rule
!  What
percentage of octane ratings fall
below 95.5?
!  Draw a picture.
!  How far away from the mean is 95.5 in
terms of number of standard deviations?
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Draw a picture
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STAT1010 – Empirical rule
Finding a Percentile using the
Empirical Rule
!  Example
3: On a visit to the doctor’s
office, your fourth-grade daughter is told
that her height is 1 standard deviation
above the mean for her age and sex. What
is her percentile for height? Assume that
heights of fourth-grade girls are normally
distributed.
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Finding a Percentile using the
Empirical Rule
The total percent in green is 84% (this represents
percentage of heights below 1 standard deviation above
the mean). The girl’s height is at the 84th percentile.
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Identifying Unusual Results
!  For
a normal distribution, 95% of all
observations lie within 2 standard
deviations of the mean.
!  As
a rule of thumb, “unusual values” are
values that are more than 2 standard
deviations away from the mean.
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STAT1010 – Empirical rule
Normal Percentiles
!  What
about percentages that are not
exactly 68% or 95% or 99.7%?
!  What
if we’re 1.5 standard deviations
up from the mean? How do we
compute such a percentile?
!  Solution:
Standard Scores
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Standard Scores
!  Language
like “one standard deviation
from the mean” is very generic and can
be applied to ANY normal distribution (i.e.
any µ and any σ).
!  This
makes using the empirical rule very
useful whether you’re dealing with bowling
scores, weights, heights, etc.
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Standard Scores
!  When
we can’t use the empirical rule, we
will instead relate our specific normal
distribution (i.e. a particular µ and σ) to a
very special normal distribution called the
‘Standard Normal’ distribution which has
µ=0 and σ=1.
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STAT1010 – Empirical rule
Standard Scores
!  Every
normal distribution can be related to
the standard normal (µ=0,σ=1).
!  To
do this, we quantify “how many
standard deviations from the mean” any
particular value is.
!  The
number of standard deviations a data
value is above or below the mean is called
it’s standard score (or z-score).
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Standard Scores
The number of standard deviations a data
value lies above or below the mean is called
its standard score (or z-score), defined by
z = standard score =
data value – mean
standard deviation
The standard score is positive for data
values above the mean and negative for
data values below the mean.
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Standard Scores
!  Example
1: For our cereal, the amount in
a box has a normal distribution with µ=15
oz. and σ=0.2 oz.
!  How
many standard deviations away from
the mean is a box with 15.25 oz?
!  What
is the standard score of a box with
15.25 oz?
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STAT1010 – Empirical rule
Standard Scores
z = standard score = data value – mean = 15.25 - 15 = 1.25
standard deviation
0.20
!  15.25
is 0.25 oz. up from the mean.
many standard deviations up from the
mean?
!  How
" Recall,
1 standard deviation is 0.2 oz
Compute: 0.25/0.2=1.25
The distance 0.25 is 1.25 standard deviations.
So, 15.25 is 1.25 standard deviations up from
15 + (1.25 x 0.2)=15.25
the mean.
mean
1.25 standard
deviations
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Standard Scores
!  The
standard score for a box with 15.25 oz
is
15.25 - 15
0.2
z = standard score =
= 1.25
(this z-score is positive because it is above the mean)
‘Standard Normal’
distribution
(µ=0,σ=1)
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Standard Scores
!  Example
2: The Stanford-Binet IQ test is
scaled so that scores have a mean of 100
and a standard deviation of 16. Find the
standard scores for IQs of 85, 100, and
125.
standard score for 85: z =
85 – 100
16
= -0.94
standard score for 100: z = 100 – 100 = 0.00
16
standard score for 125: z =
125 – 100
= 1.56
16
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STAT1010 – Empirical rule
Standard Scores
!  We
can interpret these standard scores as
follows:
" 85
is 0.94 standard deviation below the mean,
100 is equal to the mean, and 125 is 1.56
standard deviations above the mean.
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Standard Scores
!  The
Stanford-Binet IQ test.
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