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1
WEEK 6, Monday, Oct 4
Today: Cover Sections 45, 47
.
Finding percentiles of Normal Distributions (Chap 4, Sec 45).
TI83
.
Use invNorm(.90,
μ
,
σ
) and get 90
th
percentile of the Normal distribution N(
μ
,
σ
).
Excel
.
Use = norminv(.90,
μ
,
σ
) and get 90
th
percentile of the Normal distribution N(
μ
,
σ
).
Percentiles using zTable:
Calculating the position
x
where
P
(
X
<
x
) = 0.90 is a twostep
process
.
First step
is to use the zTable to find the position z where
P
(
Z
<
z
) = 0.90, that is,
to find the 90
th
percentile of the Standard Normal distribution N(0, 1).
It happens to be the
value
z
= 1.28.
Second step
is to calculate the position
x
by the formula
.
28
.
1
z
x
(The textbook calls this equation the inverse transformation.)
The position
x
is the 90
th
percentile of the Normal distribution N(
,
).
Example 1.
Finding a 90
th
Percentile of N(165, 10)
.
What is the 90
th
percentile of the
distribution of weights of healthy, male, St. Bernard, age 6, dogs assuming that distribution
has mean 165 lbs, standard deviation 10 lbs and is a Normal Distribution?
Draw the density function and then the arrow that cuts the area to 90% below and 10% above.
It gives the 90
th
percentile of the weight distribution.
Make a guess for the
x
position of the
arrow.
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Mean = 165 lbs, StDev = 10 lbs
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0.040
135
140
145
150
155
160
165
170
175
180
185
190
195
x (lbs)
f(y)
TI83
.
Use invNorm(.9, 165, 10) and get 177.8155 lbs
.
Excel
.
Use = norminv(.9, 165, 10) and you get 177.8155 lbs
.
zTable
.
From the table the position has zscore 1.28.
The inverse of the standardization function
z
x
is
z
x
.
Hence, in this example the raw score
x
corresponding to
z
= 1.28 is
x
= 165
lbs
+ (1.28) 10
lbs = 177.8
lbs
Using zTables, 120 + (.15)(44)
= 126.6.
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 Summer '10
 DennisGilliland
 Normal Distribution

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