Question

Assume that police estimate that 30% of drivers do not wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use. They stop 25 cars during the first hour. Find the mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts.

Answer 1

The mean number of drivers expected not to wear seatbelts is 7.5, the variance is 5.25, and the standard deviation is approximately 2.2913, based on the binomial distribution with a probability of 0.30 and 25 trials.

Step-by-step explanation:

The student is working on a problem related to the binomial distribution, which is a topic in statistics, a branch of mathematics. To find the mean, variance, and standard deviation of the number of drivers not wearing seatbelts, we use the fact that in a binomial distribution, the mean (μ) is n × p, and the variance (σ2) is n × p × (1-p), where n is the number of trials and p is the probability of success (in this case, not wearing a seatbelt).

Given that the probability of not wearing a seatbelt p is 0.30, and n, the number of cars stopped, is 25:

• Mean (μ) = 25 × 0.30 = 7.5
• Variance (σ2) = 25 × 0.30 × (1 - 0.30) = 25 × 0.30 × 0.70 = 5.25
• Standard Deviation (σ) = √5.25 = 2.2913

Therefore, the mean of drivers expected not to wear seatbelts is 7.5, variance is 5.25, and the standard deviation is approximately 2.2913.