Download Exercise: Empirical Rule

Exercise: Empirical Rule
Use the empirical rule to answer the following:
!  Monthly
maintenance costs are distributed
normally with a µ=$250 and σ=$50
" 1)
What percent of months have maintenance
costs in the range of $200 to $300?
" 2)
What is the chance (i.e. probability) that a
randomly chosen month has a maintenance
cost of $150 or less?
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Answer:
" 1)
" 2)
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5.2 Properties of the
Normal Distribution
part 2
!  Standard
Scores
!  Finding
percentiles when you can’t use the
empirical rule (when the data value is something
other than 1, 2 or 3 standard deviations from the
mean)
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Normal Percentiles
!  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
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
(continued): 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 IQ of 85: z =
85 – 100 = -0.94
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standard score for IQ of 100: z = 100 – 100 = 0.00
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standard score for IQ of 125: z = 125 – 100 = 1.56
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An IQ of 85 is
0.94 standard
deviations
below the
mean.
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Standard Scores
!  What
percentile are the following IQ scores?
85 (standard score: z = -0.94)
" 100 (standard score: z = 0)
" 125 (standard score: z = 1.56)
" 
!  We
can’t use the empirical rule here.
!  We’ll
have to use a table to find the
percentages (Appendix A in our book).
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A subset of Appendix A
(shown here) is provided
in Section 5.2 in the book.
This table shows the
percentage of
observations below any
given standard score.
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What percent of
observations are
below a standard
score of z = -0.94?
The closest
standard score is
-0.95
and 17.11% of the
observations are
below a standard
score of -0.95.
17%
shaded
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What percent of
observations are
below a standard
score of z = 0?
Ans: 50%
50%
shaded
A standard score of
0 is at the 50th
percentile.
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What percent of
observations are
below a standard
score of z = 1.56?
This standard
score lies between
1.5 and 1.6 on the
table. We can
approximate the
percentile for this
standard score as
(93.32% + 94.52%)
2
Or 93.92%, which is the 93.92nd percentile.
94%
shaded
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Standard Scores
!  What
percentile are the following IQ scores?
" IQ
of 85 (z=-0.94) is at the 17th percentile.
" IQ of 100 (z=0) is at the 50th percentile
" IQ of 125 (z=1.56) is at the 94th percentile.
"  Recall
that negative z-scores are below the mean and
positive z-scores are above the mean.
!  Thus,
we can get the percentiles even though
we’re not exactly 1, 2 or 3 standard
deviations from the mean.
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Standard Scores
!  We
will use Appendix A from the book (a
subset of that table was shown above) to
compute percentiles and probabilities
because it has finer resolution (more
decimals).
!  THIS
TAKES PRACTICE!
!  Active
work: See worksheet on normal curve
scores.
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More Exercises: z-scores
!  Assume
you have a normal distribution.
Use the z-score table in Appendix A to
answer:
" 1)
What percent of observations lie below a zscore of 0?
" 2)
What percent of observations lie below a zscore of 1.72?
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More Exercises: z-scores
" 3)
What percent of observations fall
BETWEEN z-scores of 0 and 1.72?
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5.2 Properties of the
Normal Distribution
part 3
!  Connecting
z-scores to probabilities.
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!  Example:
The Stanford-Binet IQ test is
normally distributed and scaled so that
scores have a mean of 100 and a standard
deviation of 16.
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!  Example:
The Stanford-Binet IQ test is
normally distributed and scaled so that
scores have a mean of 100 and a standard
deviation of 16.
" If
you draw someone at random, what is the
probability that they have an IQ score of 90 or
less?
" To
answer this, we just need to know what
percent of IQ scores are at 90 or lower.
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!  Example:
The Stanford-Binet IQ test is
normally distributed with a mean of 100
and standard deviation of 16.
Let X be an IQ score of a person.
Short-hand notation:
The 2 parameters needed to
define a normal distribution.
X ~ N(µ=100,σ=16)
“is distributed”
Normal
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" If
you draw someone at random, what is the
probability that they have an IQ score of 90 or less?
" We
need to answer:
When X ~ N(µ=100,σ=16), what is P(X ≤ 90)?
X is a data value (or IQ score in this case).
We will convert it to a z-score…
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z = standard score =
" P(X
data value – mean
standard deviation
≤ 90) = P( X –µ ≤ 90 –100
σ
16
)
= P(Z ≤ – 10/16)
= P(Z ≤ – 0.63)
= 0.2643
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z = standard score =
" P(X
data value – mean
standard deviation
≤ 90) = P(Z ≤ – 0.63) = 0.2643
Looked up
on z-table
An IQ score of 90 has a z-score of - 0.63
" The
probability of randomly drawing someone
with an IQ score of 90 or lower is 0.2643.
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QUICK-CHECK:
The Empirical Rule tells
me the percent that is
below an IQ of 90 has
to be between16% (to
the left of 1σ below the
mean) and 50% (to the
left of the mean itself).
IQ 90
So, 26.43% is totally in-line with my
Empirical Rule information because
being 0.63 standard deviations is
between 1 and 0 standard
deviations down from the mean.
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Using new notation, exercise 1:
!  Let
X ~ N(µ=40,σ=5).
Find P(X < 51):
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Using new notation, exercise 2:
!  Suppose
bowling scores are normally
distributed with a mean of 186 and a
standard deviation of 30. Find the
percentage of games with a score of 120
or HIGHER.
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