Question

Police estimate that 77% of drivers wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use. If they stop 110 cars, what is the probability they find at least 27 drivers not wearing their seatbelts? Use a Normal approximation. The probability is (Type an integer or a decimal rounded to four decimal places as needed.)

Answer 1

Answer)

N = 110

P = 0.33 { as 77% wears so 100-77 = 33% do not}

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 36.3

N*(1-p) = 84.7

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 36.3

S.d = √{n*p*(1-p)} = 4.41372858250

We need to find

P(x>=27)

By continuity correction

P(x>26.5)

Z = (26.5 - 36.3)/4.41372858250 = -2.22

From z table

P(z>-2.22) = 0.9868