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Binomial Random Variable According to the CDC, in 2015 20% of high school students rode with a driver (in the last 30 days) who had been drinking alcohol. A random sample 16 high school students was chosen. Assume the distribution is normal. Use the Binomial Distribution Table (PDF, 739 KB) (opens in new window) to find the probabilities. Please note, this question is specifically assessing your ability to use the table to find the probability. You may get a slightly different answer due to rounding if you use a calculator or other technology. 1. At least 15 have ridden with a drunk driver. P(r 2 + v 15) = 0 2. Less than 4 have ridden with a drunk driver. P(r < ¢ v 4) = 0.5981 3. No more than 2 have ridden with a drunk driver. P(rse V 2) =| 0.3518 4. Exactly 11 have ridden with a drunk driver. P(r = v 11) 0.029 5. At least 1 have ridden with a drunk driver. P(r 2 + v 1) =| 0.9719 6. Between 2 and 4 (exclusive) have ridden with a drunk driver. P(2 < v r V 4) = 0.24629

Binomial Random Variable According to the CDC, in 2015 20% of high school students rode with a driver (in the last 30 days) who had been drinking alcohol. A random sample 16 high school students was chosen. Assume the distribution is normal. Use the Binomial Distribution Table (PDF, 739 KB) (opens in new window) to find the probabilities. Please note, this question is specifically assessing your ability to use the table to find the probability. You may get a slightly different answer due to rounding if you use a calculator or other technology. 1. At least 15 have ridden with a drunk driver. P(r 2 + v 15) = 0 2. Less than 4 have ridden with a drunk driver. P(r < ¢ v 4) = 0.5981 3. No more than 2 have ridden with a drunk driver. P(rse V 2) =| 0.3518 4. Exactly 11 have ridden with a drunk driver. P(r = v 11) 0.029 5. At least 1 have ridden with a drunk driver. P(r 2 + v 1) =| 0.9719 6. Between 2 and 4 (exclusive) have ridden with a drunk driver. P(2 < v r V 4) = 0.24629

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Binomial Random Variable According to the CDC, in 2015 20% of high school students rode with a driver (in the last 30 days) who had been drinking alcohol. A random sample 16 high school students was chosen. Assume the distribution is normal. Use the Binomial Distribution Table (PDF, 739 KB) (opens in new window) to find the probabilities. Please note, this question is specifically assessing your ability to use the table to find the probability. You may get a slightly different answer due to rounding if you use a calculator or other technology. 1. At least 15 have ridden with a drunk driver. P(r 2 + v 15) =0 2. Less than 4 have ridden with a drunk driver. P(r < ¢ v 4) = 0.5981 3. No more than 2 have ridden with a drunk driver. P(rs+ V 2) = 0.3518 4. Exactly 11 have ridden with a drunk driver. P(r = v 11) 0.029 5. At least 1 have ridden with a drunk driver. P(r 2 + v 1) = 0.9719 6. Between 2 and 4 (exclusive) have ridden with a drunk driver. P(2 < v r v 4) = 0.24629
Transcribed Image Text:Binomial Random Variable According to the CDC, in 2015 20% of high school students rode with a driver (in the last 30 days) who had been drinking alcohol. A random sample 16 high school students was chosen. Assume the distribution is normal. Use the Binomial Distribution Table (PDF, 739 KB) (opens in new window) to find the probabilities. Please note, this question is specifically assessing your ability to use the table to find the probability. You may get a slightly different answer due to rounding if you use a calculator or other technology. 1. At least 15 have ridden with a drunk driver. P(r 2 + v 15) =0 2. Less than 4 have ridden with a drunk driver. P(r < ¢ v 4) = 0.5981 3. No more than 2 have ridden with a drunk driver. P(rs+ V 2) = 0.3518 4. Exactly 11 have ridden with a drunk driver. P(r = v 11) 0.029 5. At least 1 have ridden with a drunk driver. P(r 2 + v 1) = 0.9719 6. Between 2 and 4 (exclusive) have ridden with a drunk driver. P(2 < v r v 4) = 0.24629
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