1. Section 7.4 Approximating the Binomial Distribution Using the Normal Distribution HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.

2. 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. HAWKES LEARNING SYSTEMS math courseware specialists Review of Binomial Distribution: Sampling Distributions 7.4 Approx. the Binomial Dist. Using the Normal Dist.

3. 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: HAWKES LEARNING SYSTEMS math courseware specialists Normal Distribution Approximation of a Binomial Distribution: Sampling Distributions 7.4 Approx. the Binomial Dist. Using the Normal Dist.

4. A continuity correction is a correction factor employed when using a continuous distribution to approximate a discrete distribution. HAWKES LEARNING SYSTEMS math courseware specialists Continuity Correction: Sampling Distributions 7.4 Approx. the Binomial Dist. Using the Normal Dist. Examples of the Continuity Correction StatementSymbolicallyArea At least 45, or no less than 45≥ 45Area to the right of 44.5 More than 45, or greater than 45> 45Area to the right of 45.5 At most 45, or no more than 45≤ 45Area to the left of 45.5 Less than 45, or fewer than 45< 45Area to the left of 44.5 Exactly 45, or equal to 45= 45Area between 44.5 and 45.5

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: HAWKES LEARNING SYSTEMS math courseware specialists 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.

6. 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. HAWKES LEARNING SYSTEMS math courseware specialists Process for Using the Normal Curve to Approximate the Binomial Distribution: Sampling Distributions 7.4 Approx. the Binomial Dist. Using the Normal Dist.

7. 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: HAWKES LEARNING SYSTEMS math courseware specialists 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 5  2.121

8. HAWKES LEARNING SYSTEMS math courseware specialists 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

9. 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: HAWKES LEARNING SYSTEMS math courseware specialists 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

10. HAWKES LEARNING SYSTEMS math courseware specialists 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