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Section 5.5
Normal Approximations to Binomial Distributions
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Section 5.5 Objectives
 Determine when the normal distribution can approximate
the binomial distribution
 Find the correction for continuity
 Use the normal distribution to approximate binomial
probabilities
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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.
Normal Approximation to a Binomial Distribution
 If np  5 and nq  5, then the binomial random variable x is
approximately normally distributed with
 mean μ = np
 standard deviation
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σ  npq
Normal Approximation to a Binomial
 Binomial distribution: p = 0.25
• As n increases the histogram approaches a normal
curve.
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Example: 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 the mean and standard deviation.
1. Fifty-one percent of adults in the U.S. whose New Year’s
resolution was to exercise more achieved their
resolution.You randomly select 65 adults in the U.S.
whose resolution was to exercise more and ask each if he
or she achieved that resolution.
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Solution: Approximating the Binomial
 You can use the normal approximation
n = 65, p = 0.51,
q = 0.49
np = (65)(0.51) = 33.15 ≥ 5
nq = (65)(0.49) = 31.85 ≥ 5
 Mean: μ = np = 33.15
 Standard Deviation:
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σ  npq  65  0.51 0.49  4.03
Example: 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. Fifteen percent of adults in the U.S. do not make New
Year’s resolutions.You randomly select 15 adults in the
U.S. and ask each if he or she made a New Year’s
resolution.
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Solution: Approximating the Binomial
 You cannot use the normal approximation
n = 15, p = 0.15,
q = 0.85
np = (15)(0.15) = 2.25 < 5
nq = (15)(0.85) = 12.75 ≥ 5
 Because np < 5, you cannot use the normal distribution to
approximate the distribution of x.
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Correction for Continuity
 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.
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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
P(x = c)
c
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Normal approximation
P(c – 0.5 < x < c + 0.5)
c– 0.5 c c+ 0.5
Example: Using a Correction for
Continuity
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
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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.
2. The probability of getting at least 158 successes.
Solution:
• The discrete midpoint values are 158, 159, 160, ….
• The corresponding interval for the continuous normal
distribution is
x > 157.5
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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.
3. The probability of getting less than 63 successes.
Solution:
• The discrete midpoint values are …,60, 61, 62.
• The corresponding interval for the continuous normal
distribution is
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  and
standard deviation for the
distribution.
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In Symbols
Specify n, p, and q.
Is np  5?
Is nq  5?
  np
  npq
Using the Normal Distribution to
Approximate Binomial Probabilities
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.
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In Symbols
Add or subtract 0.5
from endpoints.
z
x-

Use the Standard
Normal Table.
Example: Approximating a Binomial
Probability
Fifty-one percent of adults in the U. S. whose New Year’s resolution
was to exercise more achieved their resolution.You randomly select
65 adults in the U. S. whose resolution was to exercise more and
ask each if he or she achieved that resolution. What is the
probability that fewer than forty of them respond yes? (Source:
Opinion Research Corporation)
Solution:
• Can use the normal approximation (see slide 89)
μ = 65∙0.51 = 33.15 σ  65  0.51  0.49  4.03
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Solution: Approximating a Binomial
Probability
 Apply the continuity correction:
Fewer than 40 (…37, 38, 39) corresponds to the continuous
normal distribution interval x < 39.5
Normal Distribution
μ = 33.15 σ = 4.03
P(x < 39.5)
z
x-

Standard Normal
μ=0 σ=1

39.5 - 33.15
 1.58
4.03
P(z < 1.58)
0.9429
x
μ =33.15
39.5
P(z < 1.58) = 0.9429
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z
μ =0
1.58
Example: Approximating a Binomial
Probability
A survey reports that 86% of Internet users use Windows® Internet
Explorer ® as their browser.You randomly select 200 Internet users
and ask each whether he or she uses Internet Explorer as his or her
browser. What is the probability that exactly 176 will say yes?
(Source: 0neStat.com)
Solution:
• Can use the normal approximation
np = (200)(0.86) = 172 ≥ 5 nq = (200)(0.14) = 28 ≥ 5
μ = 200∙0.86 = 172
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σ  200  0.86  0.14  4.91
Solution: Approximating a Binomial
Probability
 Apply the continuity correction:
Exactly 176 corresponds to the continuous normal distribution
interval 175.5 < x < 176.5
Normal Distribution
Standard Normal
μ = 172 σ = 4.91 x -  175.5 - 172
μ=0 σ=1
z 

 0.71
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P(175.5 < x < 176.5)

4.91
x -  176.5 - 172
z2 

 0.92

4.91
P(0.71 < z < 0.92)
0.8212
0.7611
z
x
μ =172 176.5
175.5
μ =0 0.92
0.71
P(0.71 < z < 0.92) = 0.8212 – 0.7611 = 0.0601
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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
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