Police estimate that 80% of drivers wear their seatbelts. they set up a safety r
ID: 3230384 • Letter: P
Question
Police estimate that 80% of drivers wear their seatbelts. they set up a safety roadblock, stopping cars to check for seatbelt use.
a) how many cars do they expect to stop before finding a driver whose seatbelt is not buckled?
b) what is the probability that the first unbelted driver is the 6th car stopped?
c) what is the probability that the first 10 drivers are all wearing their seatbelts?
d) if they stop 30 cars during the first hours, find the mean and standard deviation of the number of drivers expected to be wearing seatbelts.
e) what is the probability that they will have to stop at least FOUR drivers until they find a driver who is not wearing the seatbelt?
Explanation / Answer
P(wearing seatbelts) = 0.8
P(not wearing seat belts) = 1 - 0.8 = 0.2
a) Expected number of cars to stop before finding a driver whose seatbelt is not buckled = 1 / 0.2 = 5
b) P(1st ubelted driver is 6th car) = (1 - 0.2)5 * 0.2 = 0.0655
c) P(1st 10 driver wearing seatbelts) = 0.810 = 0.1074
d) n = 30
p = 0.8
mean = n * p = 30 * 0.8 = 24
standard deviation = sqrt(n * p * (1 - p)) = sqrt(30 * 0.8 * 0.2) = 2.19
e) P(stopping at least 4 cars to find a driver not wearing seatbelts) = 1 - [P(stopping 1 car to find a driver not wearing seatbelts) + P(stopping 2 cars to find a driver not wearing seatbelts) + P(stopping 3 cars to find a driver not wearing seatbelts) ]
= 1 - [0.2 + 0.8 * 0.2 + 0.82 * 0.2]
= 1 - 0.488
= 0.512