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Assume that police estimate that 30% of drivers do not wear their seatbelts. They set up a safety roadblock, stopping cars
to check for seatbelt use. They stop 40 cars during the first hour.
a. Find the mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts. Use the
fact that the mean of a geometric distribution is j = and the variance is 02 - 9
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b. How many cars do they expect to stop before finding a driver whose seatbelt is not buckled?
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물
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The mean is
(Round to the nearest tenth as needed.)
The variance is
(Round to the nearest tenth as needed.)
The standard deviation is
(Do not round until the final answer. Then round to the nearest tenth as needed.)
They expect to stop cars before finding a driver whose seatbelt is not buckled.
(Round to the nearest tenth as needed.)

Assume that police estimate that 30 of drivers do not wear their seatbelts They set up a safety roadblock stopping cars to check for seatbelt use They stop 40 c class=

Respuesta :

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:

[tex]E(X) = np[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

In this problem:

  • 30% of drivers do not wear their seatbelts, hence [tex]p = 0.3[/tex].
  • They stop 40 cars during the first hour, hence [tex]n = 40[/tex].

Item a:

[tex]E(X) = np = 40(0.3) = 12[/tex]

[tex]V(X) = np(1-p) = 40(0.3)(0.7) = 8.4[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{40(0.3)(0.7)} = 2.9[/tex]

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:

[tex]E(X) = \frac{q}{p}[/tex]

In this problem, one success, hence [tex]q = 1[/tex] and:

[tex]E(X) = \frac{q}{p} = \frac{1}{0.3} = 3.3[/tex]

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 https://brainly.com/question/24863377

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