Question

Police estimate that 81 ​% of drivers wear their seatbelts. They set up a safety​ roadblock, stopping cars to check for seatbelt use. If they stop 100 ​cars, what is the probability they find at least 16 drivers not wearing their​ seatbelts? Use a Normal approximation. ​(Type an integer or a decimal rounded to four decimal places as​ needed.)

Answer 1

Given that,

p = 1 - 0.81 = 0.19

q = 1 - p = 1 - 0.19 = 0.81

n = 100

Using binomial distribution,

= n * p = 100 * 0.19 = 19

= n * p * q = 100 * 0.19 * 0.81 = 3.923

Using continuity correction ,

P(x 16 - 0.5) = P(x 15.5)

= 1 - P(x 15.5)

= 1 - P((x - ) / (15.5 - 19) / 3.923)

= 1 - P(z -0.89)

= 1 - 0.1867   

= 0.8133

Probability = 0.8133