Question

Assume that police estimate that 11% 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 2 and the variance is 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 b. How many cars do they expect to stop before finding a driver whose seatbelt is not buckled?

Answer 1

To find the mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts, we can use the geometric distribution.

Step-by-step explanation:

To find the mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts, we can use the geometric distribution. In a geometric distribution, the probability of success on each trial is the same, and each trial is independent.

The mean of a geometric distribution, denoted by m, is calculated as m = 1/p, where p is the probability of success. In this case, p = 0.11, so the mean is m = 1/0.11 = 9.09.

The variance of a geometric distribution, denoted by var, is calculated as var = (1-p)/p^2. In this case, var = (1-0.11)/0.11^2 = 8.26.

The standard deviation of a geometric distribution is the square root of the variance, so the standard deviation is sqrt(8.26) = 2.87.