In one city,police estimate that 78% of drivers wear their seat belts.They set up a safety roadblock and they stop drivers to check for seat belt use.If 160 drivers are stopped,what is the probability they find at least 30 not wearing their seat belts?

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Quizplus · Accepted Answer

The probability of not wearing a seat belt is 1 - 0.78 = 0.22. The problem can be approached using the binomial distribution or normal approximation. For at least 30 not wearing seat belts, we calculate the probability of 29 or fewer wearing them and subtract from 1. Using normal approximation (since np and n(1-p) are both greater than 5), with mean = np = 160*0.22 = 35.2 and standard deviation = sqrt(np(1-p)) ≈ 5.35, we find the Z-score for 29.5 (continuity correction) and use standard normal distribution tables or a calculator. The calculation will show that the probability of finding at least 30 not wearing their seat belts is approximately 0.839.

In One City,police Estimate That 78% of Drivers Wear Their