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ELEMENTARY
STATISTICS
Chapter 7
Normal Distributions: Finding Probabilities
C.M. Pascual
C.M.Pascual
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Properties of the Normal Distribution
1. The normal distribution curve is bell-shaped;
2. The mean, median, and mode are equal and
located at the center of the distribution;
3. The normal distribution curve is unimodal;
4. The curve is symmetrical about the mean;
5. The curve is continuous;
6. The curve never touches the x-axis;
7. The total area under the normal distribution
curve is equal to 1 or 100%.
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Other Normal Distributions


If
 0 or  1 (or both), we will
convert values to standard scores using
Formula 5-2, then procedures for working
with all normal distributions are the same
as those for the standard normal
distribution.
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Other Normal Distributions


If
 0 or  1 (or both), we will
convert values to standard scores using
Formula 5-2, then procedures for working
with all normal distributions are the same
as those for the standard normal
distribution.
Formula 7-2
C.M.Pascual
z=
x-µ

4
Converting to Standard Normal
Distribution
P
(a)

x
Figure 7-13
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Converting to Standard Normal
Distribution
x-
z=

P
P
(a)

x
(b)
0
z
Figure 5-13
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Probability of Weight between 143
pounds and 201 pounds
z=
x = 143
s = 29
143
201
201 - 143
29
= 2.00
Weight
z
Figure 7-14
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0
2.00
7
Probability of Weight between 143
pounds and 201 pounds
Value found
in Appendix B
x = 143
s = 29
143
201
Weight
z
Figure 7-14
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0
2.00
8
Probability of Weight between 143
pounds and 201 pounds
0.4772
x = 143
s = 29
143
201
Weight
z
Figure 7-14
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0
2.00
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Probability of Weight between 143
pounds and 201 pounds
There is a 0.4772 probability
of randomly selecting a
woman with a weight between
143 and 201 lbs.
x = 143
s = 29
143
201
Weight
z
Figure 7-14
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0
2.00
10
Probability of Weight between 143
pounds and 201 pounds
OR - 47.72% of women have
weights between 143 lb and
201 lb.
x = 143
s = 29
143
201
Weight
z
Figure 7-14
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0
2.00
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Example 1
• Determine the area under the standard normal
curve between 0 to 1.96
• Solution:
0
z=1.96
• Using Appendix B, the area = 0.4750
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Example 2
• Determine the area under the standard normal
curve between z= - 1.53
• Solution:
By symmetry
z= -1.53
0
z=1.53
• Using Appendix B, the area = 0.4370
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Example 3
• Determine the area under the standard normal curve
right of z= 0.71
• Solution: Area to the
Right of 0 is 0.5
Area = 0.2389
0
0.71
• Using Appendix B, at z=0.71, the area = 0.2611;
• At right of z = 0.5 – 0,2611 = 0.2389
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Example 4
• Find the area under the standard normal curve left of
z= -2.12
• Solution: By symmetry Area to the left of 0 is 0.5
Area = 0.4830
Area = 0.2389
z = - 2.12
0
• Using Appendix B, at z=0.71, the area = 0.4830;
• At right of z = 0.5 – 0.4830 = 0.0170
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Example 5
• Find the area under the standard normal curve between z= 1.08
and z = 2.96
• Solution:
• Area between 0 and 1.08 = 0.3599; Area between 0 and 2.96
= 0.4985
Area = 0.1386
0
z-=1.08
z=2.96
• Area between 1.08 and 2.96 = 0.4985-0.3599 = 0.1386
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Example 6
• Find the area under the standard normal curve between z=-1.3
and z = 0.99
• Solution:
• Area between -1.3 and 0 = 0.4032; Area between 0 and 0.99
= 0.3389
Area = 0.7421
z=-1.3
0
z=0.99
• Area between -1.3 and 0.99 = 0.4032+0.3389 = 0.7421
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Example 7
• Find the value of z for the normal curve with area of 0.029
• Solution:
• Area between 0 and z = 0.5 – 0.029 = 0.471; From Appendix
B search for area 0f 0.471 with z value of 1.90
Area = 0.029
• Therefore the value of z = 1.90
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0
z =?
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Example 8
• Find the probability P(0 < z < 1.65) using the standard normal
distribution.
• Solution:
• Area between 0 and z = 1.65 = 0.4505 or 45.05%;
Area = 0.4505
• Therefore the value of z = 1.90
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0
z =1.65
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Example 9
• Determine the value of z when x=12, µ=16 and σ = 3
• Solution:
• Using the formula z = (x - µ)/ σ
•
= (12 – 16)/3
•
= - 1.33
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Example 10
• For a continuous random variable that has a normal
distribution with mean of 20 and a standard deviation of 4,
find the area under the normal curve from x=20 and x=27.
• Solution:
• Find standardize the normal distribution by converting the x
values to z values:
• Using the formula z = (x - µ)/ σ
•
z= (20 – 20)/4 = 0 and z = (27 – 20)/4 = 1.75
• Then using Appendix B; at z= 1.75; Area = 0.4608
• The area between 0 and 1.75 is 0.4608
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