SECTION 5.5 Normal Approximations to Binomial Distributions 279 TRY IT YOURSELF 3 In a survey of teen drivers in the United States, 35% admit to texting while driving. You randomly select 100 teen drivers in the United States and ask them whether they admit to texting while driving. What is the probability that more than 30 respond yes? (Source: American Automobile Association) Answer: Page A39 Approximating a Binomial Probability A study found that 52% of U.S. drivers who use both alcohol and marijuana admit to driving aggressively. You randomly select 200 U.S. drivers who use both alcohol and marijuana and ask them whether they admit to driving aggressively. What is the probability that at least 100 drivers will say yes, they admit to driving aggressively? (Source: American Automobile Association) SOLUTION Because np = 20010.522 = 104 and nq = 20010.482 = 96, the binomial variable x is approximately normally distributed, with m = np = 104 and s = 1npq = 220010.52210.482 ≈ 7.07. Using the continuity correction, you can rewrite the discrete probability P1x Ú 1002 as the continuous probability P1x 7 99.52. The figure shows a normal curve with m = 104, s = 7.07, and the shaded area to the right of 99.5. x 99.5 μ= 104 Number responding yes 81 86 91 96 101 106 111 116 121 126 The z@score that corresponds to 99.5 is z = 99.5 - 104 2 20010.52210.482 ≈ -0.64. So, the probability that at least 100 drivers will say “yes” is approximately P1x 7 99.52 = P1z 7 -0.642 = 1 - P1z 6 -0.642 = 1 - 0.2611 = 0.7389. Interpretation The probability that at least 100 drivers will say “yes” is approximately 0.7389, or about 73.9%. TRY IT YOURSELF 4 In Example 4, what is the probability that at most 90 drivers will say yes, they admit to driving aggressively? Answer: Page A39 EXAMPLE 4 Tech Tip Recall that you can use technology to find a normal probability. For instance, in Example 4, you can use a TI-84 Plus to find the probability once the mean, standard deviation, and continuity correction are calculated. (Use 10,000 for the upper bound.) normalcdf(99.5,1000,104, 7.07) 0.7377722394
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