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Section 7.3
Areas Under Any Normal Curve
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To Work with Any Normal Distributions
• Convert x values to z values using the
formula:
z
x

Then use Table 3 of the Appendix to find
corresponding areas and probabilities.
Rounding
• Round z values to the hundredths positions
before using Table 3.
• Leave area results with four digits to the right
of the decimal point.
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Example
Let x be a normal distribution with μ = 10 and σ = 2.
Find the probability that an x value selected at random
from this distribution is between 11 and 14. P(11  x  14)
Solution
Since probabilities correspond to areas under the distribution curve,
we want to find the area under the x curve above the interval
x = 11 and x = 14.
To do so, we will convert the x values to standard z values and then
use table 3 of the Appendix to find the corresponding area under
the standard curve.
x   14  10
x   11  10
z


2
For x = 11 z    2  0.5
for x = 14

2
The interval (11  x  14) Corresponds to interval (0.5  z  2)
P(11  x  14)  P(0.50  z  2.00)  P( z  2.00)  P( z  0.50)  0.9772  0.6915  0.2857
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Application of the Normal Curve
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. If you order
a pizza, find the probability that the delivery time will be:
.3413
a. between 25 and 27 minutes. a. __________
b. less than 30 minutes.
.9938
b. __________
c. less than 22.7 minutes.
.1251
c. __________
5
Inverse
Normal Probability Distribution
• Finding z or x values that correspond to a
given area under the normal curve
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Inverse Normal Left Tail Case
Look up area A in body of Table 3 and use
corresponding z value.
7
Inverse Normal Right Tail Case:
Look up the number 1 – A in body of Table 3
and use corresponding z value.
8
Inverse Normal Center Case:
Look up the number (1 – A)/2 in body of Table
3 and use corresponding ± z value.
9
Using Table 3 for
Inverse Normal Distribution
• Use the nearest area value rather than interpolating.
• When the area is exactly halfway between two area
values, use the z value exactly halfway between the z
values of the corresponding table areas.
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When the area is exactly
halfway between two area values
• When the z value corresponding to an area is
smaller than 2, use the z value corresponding
to the smaller area.
• When the z value corresponding to an area is
larger than 2, use the z value corresponding to
the larger area.
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Find the indicated z score:
– 2.57
z = _______
12
Find the indicated z score:
2.33
z = _______
13
Find the indicated z scores:
Solution:
Look up the number (1 – A)/2 in body of Table 3
and use corresponding ± z value.
(1-.7814)/2 = .1093
–1.23
–z = _____
1.23
z = ____
14
Find the indicated z scores:
Solution:
99% = .9900 and (1-.9900)/2 = .0100/2 = .0050
± 2.58
± z =__________
15
Find the indicated z scores:
Solution:
95% = .9500 and (1-.9500)/2 = .0500/2 = .0250
± 1.96
± z = ________
16
Application of Determining z Scores
The Verbal SAT test has a mean score of 500
and a standard deviation of 100. Scores are
normally distributed. A major university
determines that it will accept only students
whose Verbal SAT scores are in the top 4%.
What is the minimum score that a student
must earn to be accepted?
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Application of Determining z Scores
Mean = 500
standard
deviation = 100
The cut-off score is 1.75 standard deviations above the
mean.
The cut-off score is 500 + 1.75(100) = 675.
Assignment 15, 16 and 17
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Example
• Graph and examine a situation where the mean score is 46
and the standard deviation is 8.5 for a normally distributed set
of data.
• Go to Y= normalpdf(X,46,8.5)
• Adjust the window.
• [20.5 , 71.5] x [0 , 0.0588] Xscl=8.5 Yscl=0 Xres=1
• GRAPH.
Examine: What is the probability of a value falling between
the mean and the first standard deviation to the right?
• Answer: approximately 34%
• Notice how this percentage supports the information found in
the chart at the top of this page for the percentage of
information falling within one standard deviation above the
mean.
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Example
• The lifetime of a battery is normally distributed with
a mean life of 40 hours and a standard deviation of
1.2 hours. Find the probability that a randomly
selected battery lasts longer than 42 hours.
• The most accurate answer to this problem cannot be obtained
by using the chart of the standard normal distribution. One
standard deviation above the mean would be located at 41.2
hours, 2 standard deviations would be at 42.4, and one and
one-half standard deviations would be at 41.8 standard
deviations. None of these locations corresponds to the
needed 42 hours. We need more power than we have in the
chart. Calculator to the rescue!!
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Example cont.
• Let's get a visual look at the situation by examining the graph. The
location of 42 hours indicates that our answer is going to be quite small.
• Go to Y= normalpdf(X,40,1.2).
• GRAPH.
• NOTE: If you do not graph first, you may not see the normal curve
displayed in the answer due to the previously set window on the
calculator.
• Now: What is the probability of a value falling to the right of 42 hours
(between 42 hours and infinity)? Answer: approximately 4.8%
•
ShadeNorm( go to DISTR and
right arrow to DRAW.
Choose #1:ShadeNorm(42,1E99,40,1.2).
•
ENTER. The percentage is
read from the Area =.
• ShadeNorm (lower bound, upperbound, mean, standard deviation)
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