Question

Police estimate that 73% of drivers wear their seatbelts They set up a safety roadblock, stopping cars to check for seatbelt use. If they stop 120 cars, what is the probability they find at least 28 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

Solution :

Given that,

Using binomial distribution,

$\mu$ = n * p = 120 * 0.27 = 32.4

$\sigma$ = $\sqrt$ n * p * q = $\sqrt$ 120 * 0.27 * 0.73 = 4.8633

Using continuity correction ,

P($\bar x$$\geq$ 27.5) = 1 - P($\bar x$$\leq$ 27.5)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}$\leq$ (27.5 - 32.4) / 4.8633]

= 1 - P(z $\leq$ -1.01)

= 0.8438