Question

Assume that police estimate that 30% of drivers do not wear their seatbelts. They set up a safety roadblock, stopping cars

Answer 1

Using the binomial distribution, it is found that:

a)

• The mean is of 12.

• The variance is of 8.4.

• The standard deviation is of 2.9.

b) They expect to stop 3.3 cars before finding a driver whose seatbelt is not buckled.

For each driver, there are only two possible outcomes, either they wear their seatbelts, or they do not. The probability of a driver wearing their seatbelt is independent of any other driver, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem:

• 30% of drivers do not wear their seatbelts, hence $p = 0.3$.

• They stop 40 cars during the first hour, hence $n = 40$.

Item a:

$E(X) = np = 40(0.3) = 12$

$V(X) = np(1-p) = 40(0.3)(0.7) = 8.4$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{40(0.3)(0.7)} = 2.9$

Hence:

• The mean is of 12.

• The variance is of 8.4.

• The standard deviation is of 2.9.

Item b:

The number of trials until q successes of a binomial variable is modeled by a geometric distribution, with expected value given by:

$E(X) = \frac{q}{p}$

In this problem, one success, hence $q = 1$ and:

$E(X) = \frac{q}{p} = \frac{1}{0.3} = 3.3$

They expect to stop 3.3 cars before finding a driver whose seatbelt is not buckled.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377