Question

Police estimate that 84​% of drivers wear their seatbelts. They set up a safety​ roadblock, stopping cars to check for seatbelt use. If they stop 140 ​cars,

what is the probability they find at least 27 drivers not wearing their​ seatbelts?
Use a Normal approximation.

Answer 1

0.8802

Explanation:

given that the Police estimate that 84​% of drivers wear their seatbelts.

when they stop 140 cars, no of trials = no of cars checked = 140

Each car is independent of the other

Hence X no of cars with drivers wearing seat belts is binomial with p = 0.85

Required probability =

the probability they find at least 27 drivers not wearing their​ seatbelts

Since normal approximation is required we can approximate to

X is Normal with mean = np =
`140(0.84)\\=117.6`

std dev =
`√(npq) =4.338`

Required probability =atelast 27 drivers not wearing their​ seatbelts

= P(X>(140-27))

= P(X>113)

`=P(X>112.5)\\=1- 0.1198\\=0.8802`