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Chapter 3
Chapter
Chapter Outline
5
Normal Probability
Distributions
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Section 5.5 Objectives
• How to determine when the normal distribution can
approximate the binomial distribution
• How to find the correction for continuity
• How to use the normal distribution to approximate
binomial probabilities
Section 5.5
Normal Approximations to Binomial
Distributions
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Normal Approximation to a Binomial
Normal Approximation to a Binomial Distribution
• If np ³ 5 and nq ³ 5, then the binomial random
variable x is approximately normally distributed with
§ mean μ = np
§ standard deviation σ = npq
• Let n represent the number of independent trials, p is
the probability of success in a single trial, and q is the
probability of failure in a single trial.
Larson/Farber 5th ed
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Normal Approximation to a Binomial
• The normal distribution is used to approximate the
binomial distribution when it would be impractical to
use the binomial distribution to find a probability.
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Chapter 3
Normal Approximation to a Binomial
Example: Approximating the Binomial
• Binomial distribution: p = 0.25
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If
you can, find the mean and standard deviation.
1. Sixty-two percent of adults in the U.S. have an
HDTV in their homes. You randomly select 45
adults in the U.S. and ask them if they have an
HDTV in their homes.
• As n increases the histogram approaches a normal
curve.
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Solution: Approximating the Binomial
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If
you can find, find the mean and standard deviation.
2. Twelve percent of adults in the U.S. who do not
have an HDTV in their home are planning to
purchase one in the next two years. You randomly
select 30 adults in the U.S. who do not have an
HDTV and ask them if they are planning to
purchase one in the next two years.
• Mean: μ = np = 27.9
• Standard Deviation: σ = npq = 45 × 0.62 × 0.38 » 3.26
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Solution: Approximating the Binomial
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• The binomial distribution is discrete and can be
represented by a probability histogram.
• To calculate exact binomial probabilities, the
binomial formula is used for each value of x and the
results are added.
• Geometrically this corresponds to adding the areas of
bars in the probability histogram.
• Because np < 5, you cannot use the normal
distribution to approximate the distribution of x.
Larson/Farber 5th ed
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Correction for Continuity
• You cannot use the normal approximation
n = 30, p = 0.12, q = 0.88
np = (30)(0.12) = 3.6
nq = (30)(0.88) = 26.4
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Example: Approximating the Binomial
• You can use the normal approximation
n = 45, p = 0.62, q = 0.38
np = (45)(0.62) = 27.9
nq = (45)(0.38) = 17.1
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Chapter 3
Example: Using a Correction for
Continuity
Correction for Continuity
• When you use a continuous normal distribution to
approximate a binomial probability, you need to
move 0.5 unit to the left and right of the midpoint to
include all possible x-values in the interval
(correction for continuity).
Exact binomial probability
Use a correction for continuity to convert the binomial
intervals to a normal distribution interval.
1. The probability of getting between 270 and 310
successes, inclusive.
Solution:
• The discrete midpoint values are 270, 271, …, 310.
• The corresponding interval for the continuous normal
distribution is 269.5 < x < 310.5. The normal
distribution probability is P(269.5 < x < 310.5).
Normal approximation
P(x = c)
c
P(c – 0.5 < x < c + 0.5)
c– 0.5 c c+ 0.5
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Example: Using a Correction for
Continuity
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Example: Using a Correction for
Continuity
Use a correction for continuity to convert the binomial
intervals to a normal distribution interval.
Use a correction for continuity to convert the binomial
intervals to a normal distribution interval.
2. The probability of getting at least 158 successes.
3. The probability of getting less than 63 successes.
Solution:
• The discrete midpoint values are 158, 159, 160, ….
• The corresponding interval for the continuous normal
distribution is x > 157.5. The normal distribution
probability is P(x > 157.5).
Solution:
• The discrete midpoint values are …,60, 61, 62.
• The corresponding interval for the continuous normal
distribution is x < 62.5. The normal distribution
probability is P(x < 62.5).
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Using the Normal Distribution to
Approximate Binomial Probabilities
In Words
1. Verify that the binomial
distribution applies.
2. Determine if you can use
the normal distribution to
approximate x, the binomial
variable.
3. Find the mean m and
standard deviations for the
distribution.
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Larson/Farber 5th ed
Using the Normal Distribution to
Approximate Binomial Probabilities
In Symbols
Specify n, p, and q.
In Words
4. Apply the appropriate
continuity correction.
Shade the corresponding
area under the normal
curve.
5. Find the corresponding zscore(s).
6. Find the probability.
Is np ³ 5?
Is nq ³ 5?
m = np
s =
16
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npq
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In Symbols
Add or subtract 0.5
from endpoints.
z =
x-m
s
Use the Standard
Normal Table.
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Chapter 3
Example: Approximating a Binomial
Probability
Solution: Approximating a Binomial
Probability
• Apply the continuity correction:
Fewer than 20 (…17, 18, 19) corresponds to the
continuous normal distribution interval x = 19.5
Sixty-two percent of adults in the U.S. have an HDTV
in their home. You randomly select 45 adults in the U.S.
and ask them if they have an HDTV in their home.
What is the probability that fewer than 20 of them
respond yes? (Source: Opinion Research Corporation)
Normal Distribution
μ = 27.9 σ = 3.26
Standard Normal
μ=0 σ=1
z=
Solution:
• Can use the normal approximation
μ = 45 (0.62) = 27.9 σ = 45 × 0.62 × 0.38 » 3.26
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x-m
s
=
19.5 - 27.9
» -2.58
3. 26
0.9429
μ =0
20
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Solution: Approximating a Binomial
Probability
• Apply the continuity correction:
Exactly 96 corresponds to the continuous normal
distribution interval 95.5 < x < 96.5
A survey reports that 62% of Internet users use
Windows® Internet Explorer ® as their browser. You
randomly select 150 Internet users and ask each whether
they use Internet Explorer as their browser. What is the
probability that exactly 96 will say yes? (Source: Net
Normal Distribution
μ = 93 σ = 5.94
Applications)
Solution:
• Can use the normal approximation
x-m
Standard Normal
μ=0 σ=1
95.5 - 93
z1 =
=
» 0.42
s
5.94
x - m 96.5 - 93
z2 =
=
» 0.59
s
5. 94
P(0.42 < z < 0.59)
0.8212
0.7611
np = (150)(0.62) = 93 ≥ 5 nq = (150)(0.38) = 57 ≥ 5
z
μ =0 0.59
0.42
σ = 150 × 0.62 × 0.38 » 5.94
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z
1.58
P(z < =2.58) = 0.0049
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Example: Approximating a Binomial
Probability
μ = 150∙0.62 = 93
P(z < 1.58)
P(0.42 < z < 0.59) = 0.7224 – 0.6628 = 0.0596
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Section 5.5 Summary
• Determined when the normal distribution can
approximate the binomial distribution
• Found the correction for continuity
• Used the normal distribution to approximate binomial
probabilities
Copyright © 2015, 2012, and 2009 Pearson Education, Inc.
Larson/Farber 5th ed
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