Section 7.4

Approximating the Binomial Distribution Using the Normal

Distribution

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• The experiment consists of n identical trials.• Each trial is independent of the others.• For each trial, there are only two possible

outcomes. For counting purposes, one outcome is labeled a success, the other a failure.

• For every trial, the probability of getting a success is called p. The probability of getting a failure is then 1 – p.

• The binomial random variable, X, is the number of successes in n trials.

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Review of Binomial Distribution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

If the conditions that np ≥ 5 and n(1 – p) ≥ 5 are met for a given binomial distribution, then a normal distribution can be used to approximate its probability distribution with the given mean and standard deviation:

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Normal Distribution Approximation of a Binomial Distribution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

A continuity correction is a correction factor employed when using a continuous distribution to approximate a discrete distribution.

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Continuity Correction:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

Examples of the Continuity Correction

Statement Symbolically Area

At least 45, or no less than 45 ≥ 45 Area to the right of 44.5

More than 45, or greater than 45 > 45 Area to the right of 45.5

At most 45, or no more than 45 ≤ 45 Area to the left of 45.5

Less than 45, or fewer than 45 < 45 Area to the left of 44.5

Exactly 45, or equal to 45 = 45 Area between 44.5 and 45.5

Use the continuity correction factor to describe the area under the normal

curve that approximates the probability that at least 2 people, in a

statistics class of 50, cheated on the last test. Assume that the number of

people who cheated is a binomial distribution with a mean of 5 and a

standard deviation of 2.12.

Calculate the probability:

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Begin by adding and subtracting 0.5 to and from 2.

Draw a normal curve indicating the interval 1.5 to 2.5 to represent 2.

Next, shade the area corresponding to the phrase at least 2.

Solution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

1. Determine the values of n and p.

2. Verify that the conditions np ≥ 5 and n(1 – p) ≥ 5.

3. Calculate the values of the mean and standard deviation using the formulas and .

4. Use a continuity correction to determine the interval corresponding to the value of x.

5. Draw a normal curve labeled with the information in the problem.

6. Convert the value of the random variable(s) to a z-value(s).

7. Use the normal curve table to find the appropriate area under the curve.

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Process for Using the Normal Curve to Approximate the Binomial Distribution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

After many hours of studying for your statistics test, you believe

that you have a 90% probability of answering any given question

correctly. Your test included 50 true/false questions. What is the

probability that you will miss no more than 4 questions?

Calculate the probability:

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n 50, p 0.10 since we are looking at questions missed.

np 5 and n(1 – p) 45, both which are greater than or

equal to 5.

Solution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

50(0.10) 5

2.121

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Use the continuity correction by adding and subtracting 0.5 to

and from 4.

Draw a normal curve indicating the interval 3.5 to 4.5 to

represent 4.

Solution (continued):

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

P(z ≤ 0.24) 0.4052

0.24

Many toothpaste commercials advertise that 3 out of 4 dentists

recommend their brand of toothpaste. What is the probability that

out of a random survey of 400 dentists, 300 will have

recommended Brand X toothpaste? Assume that the

commercials are correct, and therefore, there is a 75% chance

that any given dentist will recommend Brand X toothpaste.

Calculate the probability:

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n 400, p 0.75

np 300 and n(1 – p) 100, both which are greater than or

equal to 5.

Solution:

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

400(0.75) 300 8.660

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Use the continuity correction by adding and subtracting 0.5 to

and from 300.

Draw a normal curve indicating the interval 299.5 to 300.5 to

represent 300.

Solution (continued):

Sampling Distributions

7.4 Approx. the Binomial Dist. Using

the Normal Dist.

P(0.06 ≤ z ≤ 0.06) 0.0478

0.06 and 0.06