Download 6-4A Lecture

Section 6 – 4A: Finding Probabilities involving a Normal Distribution of x values
Finding P( z > 5) requires a Z Distribution and Z table
The last section developed the probabilities for data that came from a standard normal
distribution of Z values. These questions looked like P( z > 1 ) or P( –1.2 < z < 1.6). The Z curve
is the basis for answering questions about data that is represented by a standard normal distribution
of Z values. It is not very common to have a distribution of numbers where the mean is 0 and the
standard deviation is 1.
It is far more common for the data to come from a distribution of numbers where the mean is NOT 0
and the standard deviation is NOT 1. These distributions do not contain z values. These
distributions use x as the variable to represent any single number selected from the distribution of x
values. We will now develop the concept of finding probabilities that involve distributions of x values
that are not Standard Normal Z Distributions.
Finding P( x > 5 ) from a distribution of x values
The probability question P( x > 5) is a question about an x value. This question asks if I select one
x value from the distribution of x values what is the probability that the x value will be within
the stated range of x values. These x values are NOT from a distribution that is a Standard
Normal Z Distribution. The mean is NOT 0 and the standard deviation is NOT 1. You CANNOT
just use the x values and read the areas from the Z Table as we did in the past section with Z values.
Do we need to find an x Table for x values that works like the Z Table ?
No. This would never work. The area under the Z curve is based on the unique equation that
describes the single Z distribution. To find the area under a curve for a specific distribution of x
values you would need a table for that specific distribution. The impractical part of that plan is
that every normal distribution of x values has a different mean and standard deviation and is
represented by a different normal curve. You would need a different table for each of the infinite
normal distributions of x that exist. It is practical to have a table for the single z curve but it is not
practical to try and develop an infinite number of x tables for the different distributions of x.
So What's the Good News
If the distribution of x values is normal we can convert the x data to z data and then use the z
table to answer probabilities about the distribution of x values. This means that all the probability
questions about an x value from any normal distribution of x values can be answered with the use of a
single Z table.
Stat 300 6 – 4A Lecture
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© 2012 Eitel
The Relationship between x and Z values
Every x value in the Normal Distribution of x values has a unique z value
and
every z value in the Standard Normal Distribution has a unique x value
Any given x1 value can be converted into its corresponding z1 value
(x − µx)
by the formula Z =
σx
Convert each x value
into itʼs z value
x1
z1 =
x1 − µx
σx
Z1
x2
z2 =
x 2 − µx
σx
Z2
x3
.
.
.
xn
z3 =
x 3 − µx
σx
zn =
x n − µx
σx
a Normal Distribution
of x values
with
a mean of µ x
.
.
.
is converted into
and a SD of σ x
A Normal Distribution
of X Values
can be converted into
Z3
.
.
.
Zn
a Normal Distribution
of z values
with
a mean of 0
and a SD of 1
A Standard Normal Distribution
of Z Values
X -------> Z
by
(x − µx)
Z=
σx
µx
σx
Stat 300 6 – 4A Lecture
X
Mean =1
SD = 0
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Z
© 2012 Eitel
Converting an x value to its corresponding Z value
Any given x1 value can be converted into its corresponding z1 value
by the formula
(x − µ x )
Z=
σx
Example 1
A normal distribution has a mean of µ x = 15.6 and a standard deviation of σ x = 3.21 Convert each
listed x value into its corresponding z value. Round to 2 decimal places.
1a) Convert x = 20.8 to Z
1b) Convert x = 8.31 to Z
setup for finding z
z=
z=
setup for finding z
(x − µx)
z=
σx
(20.8 − 15.6)
3.21
z=
z = 1.62
(x − µx)
σx
(8.31− 15.6)
3.21
z = – 2.27
Example 2
A normal distribution has a mean of µ x = 152.98 and a standard deviation of σ x = 24.92 Convert
each x value into its corresponding z value. Round to 2 decimal places.
2a) Convert x = 192.15 to Z
2b) Convert x = 81.74 to Z
setup for finding z
setup for finding z
z=
z=
(x − µx)
z=
σx
(192.15 − 152.98)
24.92
z=
z = 1.57
Stat 300 6 – 4A Lecture
(x − µx)
σx
(81.74 − 152.98)
24.92
z = – 2.86
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© 2012 Eitel
Converting a Z value to its corresponding x value
Any given z1 value can be converted into its corresponding x1 value
by the formula
x = µ x + z (σ x )
Example 3
A normal distribution has µ x = 72.16 and σ x = 9.03 Convert each z value into its corresponding
x value. Round to 2 decimal places.
3a. Convert z = 2.41 to x
3b. Convert z = – 1.74 to x
setup for finding x
setup for finding x
x = µ x + z (σ x )
x = µ x + z (σ x )
x = 72.16 + 2.41 (9.03)
x = 72.16 − 1.74 (9.03)
x = 93.92
x = 56.45
Example 4
A normal distribution has µ x = 21.09 and σ x = 1.42 Convert each z value into its corresponding
x value. Round to 2 decimal places.
4A. Convert z = – .1.96 to x
4b.
Convert z = 1.645 to x
setup for finding x
setup for finding x
x = µ x + z (σ x )
x = µ x + z (σ x )
x = 21.09 − 1.96 (1.42)
x = 21.09 + 1.645 (1.42)
x = 18.31
x = 23.43
Stat 300 6 – 4A Lecture
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© 2012 Eitel