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.
Accepted Solution
A:
Answer:0.8802Step-by-step 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 = 140Each car is independent of the otherHence X no of cars with drivers wearing seat belts is binomial with p = 0.85Required probability = the probability they find at least 27 drivers not wearing their seatbeltsSince normal approximation is required we can approximate to X is Normal with mean = np = [tex]140(0.84)\\=117.6[/tex]std dev = [tex]\sqrt{npq} =4.338[/tex]Required probability =atelast 27 drivers not wearing their seatbelts= P(X>(140-27))= P(X>113)[tex]=P(X>112.5)\\=1- 0.1198\\=0.8802[/tex]