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5.5 Normal Approximations to Binomial Distributions Statistics Mrs. Spitz Fall 2008

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Objectives/Assignment How to decide 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 Assignment: pp. 241-243 #1-18 all

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Approximating a Binomial Distribution In Section 4.2, you learned how to find binomial probabilities. For instance, a surgical procedure has an 85% chance of success and a doctor performs the procedure on 10 patients, it is easy to find the probability of exactly two successful surgeries. But what if the doctor performs the surgical procedure on 150 patients and you want to find the probability of fewer than 100 successful surgeries? To do this using the techniques described in 4.2, you would have to use the binomial formula 100 times and find the sum of the resulting probabilities. This is not practical and a better approach is to use a normal distribution to approximate the binomial distribution.

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Normal Approximation to a Binomial Distribution To see why this result is valid, look at the following slide and binomial distributions for p = 0.25 and n = 4, 10, 25 and 50. Notice that as n increases, the histogram approaches a normal curve.

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Study Tip Properties of a binomial experiment –n–n independent trials –T–Two possible outcomes: success or failure –P–Probability of success is p; probability of a failure is 1 – p = q –p–p is constant for each trial

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Ex. 1: Approximating the Binomial Distribution Two binomial experiments are listed. Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If so, find the mean and standard deviation. If not, explain why. 1. Thirty-seven percent of Americans say they always fly an American flag on the Fourth of July. You randomly select 15 Americans and ask each if he or she always flies an American flag on the Fourth of July.

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Ex. 1: Approximating the Binomial Distribution - Solution 1. Thirty-seven percent of Americans say they always fly an American flag on the Fourth of July. You randomly select 15 Americans and ask each if he or she always flies an American flag on the Fourth of July. In this binomial experiment, n = 15, p = 0.37 and q = 0.63, so: np = 15(0.37) = 5.55 and nq = 15(0.63) = 9.45 Because np 5 and nq 5, you can use the normal distribution with = 5.55 and to approximate the distribution of x.

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Ex. 1b: Approximating the Binomial Distribution - Solution 2. Ninety-three percent of Americans want the national anthem to remain the same. You randomly select 65 Americans and ask each if he or she want the national anthem to remain the same. In this binomial experiment, n = 65, p = 0.93 and q = 0.07, so: np = 65(0.93) = 60.45 and nq = 65(0.07) = 4.55 Because nq 5, you cannot use the normal distribution to approximate the distribution of x.

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Note for TI users The TIs cannot calculate the cumulative binomial probability for n = 10,000, p = 0.4 and x = 9000. The probability can be calculated using a normal approximation. There are issues with memory limitation for your calculator for the binomial distribution.

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Correction for Continuity The binomial distribution is discrete and can be represented by a probability histogram. To calculate an exact binomial probability, you can use the binomial formula for each value of x and add the results. Geometrically, this corresponds to adding the areas of bars in the probability histogram. When you do this, remember that each bar has a width of one unit and x is the midpoint of the interval. When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 units to the left and right of the midpoint to include all possible x-values in the interval. When you do this, you are making a correction for continuity.

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Ex. 2: Using a Correction for Continuity Use the correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 1. The probability of getting between 270 and 310 successes, inclusive 2. The probability of getting more than 157 and less than 420 successes 3. The probability of getting less than 63 successes. SOLUTION: 1. The probability of getting between 270 and 310 successes, inclusive The midpoint values are 270, 271,... 310. The boundaries for the normal distribution are 269.5 < x < 310.5

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Ex. 2: Using a Correction for Continuity Use the correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 2. The probability of getting more than 157 and less than 420 successes SOLUTION: 2. The probability of getting more than 157 and less than 420 successes The midpoint values are 158, 159,... 419. The boundaries for the normal distribution are 157.5 < x < 419.5 (less than means but not equal to

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Ex. 2: Using a Correction for Continuity Use the correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 3. The probability of getting less than 63 successes. SOLUTION: 3. The probability of getting less than 63 successes. The midpoint values are... 60, 61, 62. The boundary for the normal distribution is x < 62.5 (less than means but not equal to

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Ex. 3: Approximating a Binomial Probability Thirty-seven percent of Americans say they always fly an American flag on the Fourth of July. You randomly select 15 Americans and ask each if he or she flies an American flag on the Fourth of July. What is the probability that fewer than eight of them reply yes? SOLUTION: From Example 1, you know that you can use a normal distribution with = 5.55 and 1.87 to approximate the binomial distribution. By applying the continuity correction, you can rewrite the discrete probability P(x < 8) as P (x < 7.5). The graph on the next slide shows a normal curve with = 5.55 and 1.87 and a shaded area to the left of 7.5. The z-score that corresponds to x = 7.5 is

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Continued... Using the Standard Normal Table, P (z<1.04) = 0.8508 So, the probability that fewer than eight people respond yes is 0.8508

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Ex. 4: Approximating a Binomial Probability Twenty-nine percent of Americans say they are confident that passenger trips to the moon will occur during their lifetime. You randomly select 200 Americans and ask if he or she thinks passenger trips to the moon will occur in his or her lifetime. What is the probability that at least 50 will say yes? SOLUTION: Because np = 200 0.29 = 58 and nq = 200 0.71 = 142, the binomial variable x is approximately normally distributed with and

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Ex. 4 Continued Using the correction for continuity, you can rewrite the discrete probability P (x 50) as the continuous probability P ( x 49.5). The graph shows a normal curve with = 58 and = 6.42, and a shaded area to the right of 49.5.

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Ex. 4 Continued The z-score that corresponds to 49.5 is So, the probability that at least 50 will say yes is: P(x 49.5) = 1 – P(z -1.32) = 1 – 0.0934 = 0.9066

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Study Tip In a discrete distribution, there is a difference between P (x c) and P( x > c). This is true because the probability that x is exactly c is not zero. IN a continuous distribution, however, there is no difference between P (x c) and P (x >c) because the probability that x is exactly c is zero.

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Ex. 5: Approximating a Binomial Probability A survey reports that 48% of Internet users use Netscape as their browser. You randomly select 125 Internet users and ask each whether he or she uses Netscape as his or her browser. What is the probability that exactly 63 will say yes? SOLUTION: Because np = 125 0.48 = 60 and nq = 125 0.52 = 65, the binomial variable x is approximately normally distributed with and

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Ex. 5 Continued Using the correction for continuity, you can rewrite the discrete probability P (x 63) as the continuous probability P ( 62.5 < x < 63.5). The graph shows a normal curve with = 60 and = 5.59, and a shaded area between 62.5 and 63.5.

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Ex. 5 Continued The z-scores that corresponds to 62.5 and 63.5 are:

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Ex. 5 Continued So, the probability that at least 50 will say yes is: P(62.5 < x < 63.5) = P (0.45 < z < 0.63) = P(z < 0.63) – P(z < 0.45) = 0.7357 – 0.6736 = 0.0621 There is a probability of about 0.06 that exactly 63 of the Internet users will say they use Netscape.

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Reminders for the next couple of weeks Friday and Monday– Work in Class on 5.5 and Review Chapter 5. Tuesday – Chapter 5 Test/Binder Grade Wednesday – Review for Semester Exam – old exams Thursday– Semester Exam Friday – 6.1Confidence Intervals for the Mean (Large Samples) Tuesday –6.2 Confidence Intervals for the Mean (Small Samples) Thursday – 6.3 Confidence Intervals for Population Proportions

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Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter The Normal Probability Distribution 7.

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