Question

assume that police estimate that 18% of drivers do not wear their seatbelts
they set up a safety roadblock, stopping cars to check for seatbelt use. they
stop 30 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 \\( \mu = \frac{1}{p} \\) and the variance is \\( \sigma^2 = \frac{q}{p^2} \\).
b. how many cars do they expect to stop before finding a driver whose
seatbelt is not buckled?

the mean is 5.6. (round to the nearest tenth as needed.)
the variance is 25.3. (round to the nearest tenth as needed.)
the standard deviation is 5.0.
(do not round until the final answer. then round to the nearest tenth
as needed.)
they expect to stop \\( \square \\) cars before finding a driver whose seatbelt is not
buckled.
(round to the nearest tenth as needed.)

Explanation:

Step1: Identify the distribution and parameter

This is a geometric distribution problem. The probability \( p \) that a driver does not wear a seatbelt is \( 0.18 \), and \( q = 1 - p = 1 - 0.18 = 0.82 \). The mean of a geometric distribution is given by \( \mu=\frac{1}{p} \).

Step2: Calculate the mean

Substitute \( p = 0.18 \) into the formula for the mean of a geometric distribution: \( \mu=\frac{1}{0.18}\approx5.6 \)? Wait, no, wait. Wait, the geometric distribution for the number of trials until the first success (here, success is finding a driver without a seatbelt). Wait, but the formula given in the problem for the mean of a geometric distribution is \( \mu = \frac{1}{p} \), but wait, actually, the geometric distribution can be defined in two ways: one counts the number of trials until the first success, and the other counts the number of failures before the first success. But in the problem, it says "how many cars do they expect to stop before finding a driver whose seatbelt is not buckled". Wait, "before finding" – so if we let \( X \) be the number of cars stopped before finding a non - seatbelted driver, then \( X \) follows a geometric distribution where \( p = 0.18 \), and the mean of \( X \) (number of failures before first success) is \( \frac{q}{p}=\frac{1 - p}{p} \)? Wait, no, let's check the formula given in the problem. The problem says "the mean of a geometric distribution is \( \mu=\frac{1}{p} \)" and variance \( \sigma^{2}=\frac{q}{p^{2}} \). Wait, maybe the problem's definition of geometric distribution is the number of trials until the first success (including the success). But the question is "how many cars do they expect to stop before finding a driver whose seatbelt is not buckled". So if \( Y \) is the number of cars stopped until finding a non - seatbelted driver (including that car), then the number of cars stopped before is \( Y - 1 \). But according to the formula given in the problem, the mean of \( Y \) (geometric distribution with \( p = 0.18 \)) is \( \mu=\frac{1}{p}=\frac{1}{0.18}\approx5.6 \). But "before finding" would be \( Y - 1 \)? Wait, no, maybe the problem's formula is for the number of trials until the first success, and the question is asking for the number of trials until the first success. Wait, let's re - read the question: "How many cars do they expect to stop before finding a driver whose seatbelt is not buckled?" Wait, maybe there's a misinterpretation. Wait, if we consider that "before finding" means the number of cars with seatbelts (failures) before the first non - seatbelt (success), then the mean number of failures before first success in a geometric distribution is \( \frac{q}{p}=\frac{1 - p}{p} \). But the problem's formula says mean is \( \frac{1}{p} \). Wait, maybe the problem's geometric distribution is defined as the number of trials until the first success (including the success), so if we want the number of cars stopped before, it's \( \frac{1}{p}-1 \)? But no, the formula given in the problem for the mean is \( \frac{1}{p} \). Wait, the problem statement says "Use the fact that the mean of a geometric distribution is \( \mu=\frac{1}{p} \) and the variance is \( \sigma^{2}=\frac{q}{p^{2}} \)". So we have \( p = 0.18 \), so \( \mu=\frac{1}{0.18}\approx5.6 \)? But that seems to be the mean number of trials until the first success (including the success). But the question is "how many cars do they expect to stop before finding a driver whose seatbelt is not buckled". Wait, maybe there's a mistake in my understanding. Wait, let's check the numbers. If \( p = 0.18 \), then \( \frac{1}{p}=\frac{1}{0.18}\approx5.555\cdots\approx5.6 \). But if we want the number of cars stopped before, it would be \( \frac{1}{p}-1 \approx5.6 - 1=4.6 \)? But that contradicts. Wait, no, maybe the problem's question is actually asking for the number of cars stopped until finding a non - seatbelted driver (including that car), and the wording "before" is a bit off. Wait, the formula given in the problem for the mean of the geometric distribution is \( \mu=\frac{1}{p} \). So with \( p = 0.18 \), \( \mu=\frac{1}{0.18}\approx5.6 \). But let's calculate it properly: \( \frac{1}{0.18}=\frac{100}{18}=\frac{50}{9}\approx5.6 \). Wait, but if we use the formula given in the problem, the mean is \( \frac{1}{p} \). So the expected number of cars to stop (including the one with no seatbelt) is \( \frac{1}{0.18}\approx5.6 \). But the question says "before finding", which is a bit confusing. Wait, maybe the problem has a typo, and it's supposed to be "until finding". Given that the formula in the problem is \( \mu=\frac{1}{p} \), we use that. So \( \frac{1}{0.18}\approx5.6 \)? Wait, no, \( \frac{1}{0.18}=5.555\cdots\approx5.6 \). Wait, but let's do the calculation: \( 1\div0.18 = 5.555\cdots\approx5.6 \).

Answer:

\( 5.6 \) (Wait, but according to the formula in the problem, the mean of the geometric distribution is \( \frac{1}{p} \), so with \( p = 0.18 \), \( \frac{1}{0.18}\approx5.6 \). So the expected number of cars to stop (until finding a non - seatbelted driver) is approximately \( 5.6 \). If we consider "before" as in "excluding" the non - seatbelted driver, then it would be \( \frac{1}{p}-1=\frac{1 - p}{p}=\frac{0.82}{0.18}\approx4.6 \), but the formula in the problem says the mean is \( \frac{1}{p} \), so we follow the problem's formula. So the answer is \( 5.6 \).)