Question

Assume that police estimate that 12​% of drivers do not wear their seatbelts. They set up a safety​ roadblock, stopping cars to check for seatbelt use. They stop 50 cars during the first hour. a. Find the​ mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts. Use the fact that the mean of a geometric distribution is μ= 1 p and

Answer 1

Answer: mean = 8.33, variance =61.11 , standrad deviation = 7.82

Explanation:

For geometric distribution,

Mean =
`\frac1p`

variance =
`(1-p)/(p^2)`

Standard deviation
`=√(Variance)` , where p = probability of success in each trial.

Given: p= 0.12

Mean =
`\frac1{0.12}=8.33`

Variance =
`(1-0.12)/(0.12^2)=(0.88)/(0.0144)=61.11`

Standard deviation =
`√(61.11)\approx7.82`

Hence, mean = 8.33, variance =61.11 , standrad deviation = 7.82