Transcript Section 7.2 - USC Upstate: Faculty

Section 7.2
• Standard Units and Areas
• Under the Standard Normal Distribution
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Standard Score
• The z value or z score tells the number of
standard deviations between the original
measurement x and the mean μ of the x
distribution.
• The z value is in standard units.
z 
x  

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x Values and Corresponding z Values
z
x

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Calculating z scores
The amount of time it takes for a pizza delivery is
approximately normally distributed with a mean
of 25 minutes and a standard deviation of 2
minutes.
a. Convert 21 minutes to a z score.
b. Convert 29.7 minutes to a z score.
Solution
a. z 
b.
z
x  

x


21  25
  2 . 00
2

29.7  25
2
 2.35
Raw Score
• A raw score is the result of converting from
standard units (z scores) back to original
measurements, x values.
• From
z
x

• We get the formula:
x=z+
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Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.60.
x  z     1 . 6 ( 2 )  25  28 . 2
The delivery time is 28.2 minutes.
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Standard Normal Distribution
is a normal distribution with μ = 0 and σ = 1

=0

=1
Any x values are converted to z
scores.
x=z+ z*1+0=z
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Importance of the
Standard Normal Distribution:
For μ = 0 and σ = 1
Any Normal Curve:
Areas will be equal.
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Areas of a Standard Normal
Distribution
We have seen how to convert any normal distribution
to the standard normal distribution. We can change
any x value to a z value and back again. But what is
the is the advantage of all that work?
The advantage is that there are extensive tables that
show the area under the standard normal curve for
almost any interval along the z axis.
The areas are important because each area is equal to
the probability that the measurement on an item
selected at random falls in that interval
Thus the standard normal distribution can be a
tremendously helpful tool.
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Areas of a Standard Normal
Distribution
• Appendix
• Table 3
• Pages A6 - A7
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Use of the Normal Probability Table
Appendix Table 3 is a left-tail style table.
Entries give the cumulative areas to the left
of a specified z.
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To Find the area to the
Left of a Given z score
• Find the row associated with the sign, units
and tenths portion of z in the left column of
Table 3.
• Move across the selected row to the column
headed by the hundredths digit of the given z.
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Find the area to the left of
z = – 2.84
13
To find the area to the left of
z = – 2.84
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To find the area to the left of
z = – 2.84
15
The area to the left of
z = – 2.84 is .0023
16
To Find the Area to the Left of a Given
Negative z Value:
Use Table 3 of the Appendix directly.
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To Find the Area to the Left
of a Given Positive z Value:
Use Table 3 of the Appendix directly.
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To Find the Area to the Right
of a Given z Value:
Subtract the area to the left of z from 1.0000.
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Alternate Way To Find the Area to the Right
of a Given Positive z Value:
Use the symmetry of the normal distribution.
Area to the right of z
= area to left of –z.
20
To Find the Area Between Two z Values
Subtract area to left of z1 from area to left of
z2 . (When z2 > z1.)
Convention for Using Table 3
• Treat any area to the left of a z value smaller
than 3.49 as 0.000
• Treat any area to the left of a z value greater
than 3.49 as 1.000
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Example
Use Table 3 of the Appendix to find the specified areas.
a. Find the area between z = 1.00 and z = 2.70.
b. Find the area to the right of z = 0.94.
Solution
a. (Area between 1.00 and 2.70) =
= (Area left to 2.70) – (Area left to 1.00)
= 0.9965 – 0.8413 = 0.1552
b. (Area to the right of 0.94) =
= (Area under entire curve) – (Area to the left of 0.94) =
1.000 – 0.8264 = 0.1736
Alternatively:
(Area to the right of 0.94) = (Area to the left of - 0.94)
= 0.1736.
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Use of the Normal Probability Table
a.
.9495
P( z < 1.64 ) = __________
.0034
b.
P( z < - 2.71 ) = __________
c.
P(0 < z < 1.24) =
.3925
______
d.
P(0 < z < 1.60) =
.4452
_______
e.
.4911
P( 2.37 < z < 0) = ______
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Use of the Normal Probability Table
f.
.9974
P( 3 < z < 3 ) = _______
g.
.9322
P( 2.34 < z < 1.57 ) = _____
i.
.0774
P( 1.24 < z < 1.88 ) = _______
.2254
P( 2.44 < z < 0.73 ) = _______
j.
.0084
P( z > 2.39 ) = ________
k.
.9236
P( z > 1.43 ) = __________
h.
Assignment 14
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Example:
Graph and investigate the normal distribution curve where the
mean is 0 and the standard deviation is 1.
For graphing the normal distribution, choose normalpdf.
The normalpdf (normal probability density function) is found under
DISTR (2nd VARS) #1normalpdf(.
Go to the Y = menu.
The parameters will be (variable, mean, standard deviation).
Adjust the WINDOW.
You will have to set your own window. Guideline is:
Xmin = mean - 3 SD
Xmax = mean + 3 SD
Xscl = SD
Ymin = 0
Ymax = 1/(2 SD)
Yscl = 0
GRAPH. Using TRACE, simply type the desired x value and the point
will be plotted.
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Example cont.
Investigate: What happens to the curve as the standard deviation
increases?
Double the standard deviation and see what happens to the graph.
When graphing 2 normal curves, the window will need to be
adjusted. Xmin = mean - 3 (largest SD)
Xmax = mean + 3 (largest SD)
Xscl = largest SD
Ymin = 0
Ymax = 1/(2 smallest SD)
Yscl = 0
Observe that as the standard deviation increases, the more
spread out the graph becomes.
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Example cont.
Now, the area under the curve between particular values represents
the probabilities of events occurring within that specific
range. This area can be seen using the command ShadeNorm(.
To find ShadeNorm(, go to DISTR and right arrow to DRAW. Choose
#1:ShadeNorm(.
ShadeNorm (lower bound, upperbound, mean, standard
deviation)
By entering parameters -1,1 you will see the area, indicating
approximately 68% probability of a score falling within 1 standard
deviation from the mean in a normally distributed set of
values. Since the calculator defaults to a mean of 0 and standard
deviation of 1, it was not necessary to enter these values in this
example, but is is a good idea to get in the habit of entering all 4
parameters.
Notice how this answer supports the percentage listed in the chart of
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the standard normal distribution
Example cont.
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