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Normal Approximations to Binomial Distributions Larson/Farber 4th ed1

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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 Larson/Farber 4th ed2

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Normal Approximation to a Binomial 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 Larson/Farber 4th ed3 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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Normal Approximation to a Binomial Binomial distribution: p = 0.25 Larson/Farber 4th ed4 As n increases the histogram approaches a normal curve.

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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. 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. Larson/Farber 4th ed5

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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: Larson/Farber 4th ed6

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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. 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. Larson/Farber 4th ed7

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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. Larson/Farber 4th ed8

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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. Larson/Farber 4th ed9

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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). Larson/Farber 4th ed10 Exact binomial probability P(x =c) c Normal approximation P(c – 0.5 < x < c + 0.5) c c+ 0.5c– 0.5

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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. Larson/Farber 4th ed11 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. Larson/Farber 4th ed12 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. Larson/Farber 4th ed13 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 14Larson/Farber 4th ed 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. Is np 5? Is nq 5? Specify n, p, and q. In Words In Symbols

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Using the Normal Distribution to Approximate Binomial Probabilities 15Larson/Farber 4th ed 4.Apply the appropriate continuity correction. Shade the corresponding area under the normal curve. 5.Find the corresponding z- score(s). 6.Find the probability. Add or subtract 0.5 from endpoints. Use the Standard Normal Table. In Words In Symbols

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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) Larson/Farber 4th ed16 Solution: Can use the normal approximation (see slide 89) μ = 65∙0.51 = 33.15

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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 Larson/Farber 4th ed17 39.5μ =33.15 P(x < 39.5) Normal Distribution μ = 33.15 σ = 4.03 x 1.58μ =0 P(z < 1.58) Standard Normal μ = 0 σ = 1 z 0.942 9 P(z < 1.58) = 0.9429

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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) Larson/Farber 4th ed18 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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Solution: Approximating a Binomial Probability Apply the continuity correction: Exactly 176 corresponds to the continuous normal distribution interval 175.5 < x < 176.5 Larson/Farber 4th ed19 0.761 1 0.92μ =0 P(0.71 < z < 0.92) Standard Normal μ = 0 σ = 1 z 0.71 0.821 2 P(0.71 < z < 0.92) = 0.8212 – 0.7611 = 0.0601 176.5 μ =172 P(175.5 < x < 176.5) Normal Distribution μ = 172 σ = 4.91 x 175.5

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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 Larson/Farber 4th ed20

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