Question

Police estimate that \(80 \%\) of drivers now wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use. a) How many cars do they expect to stop before finding a driver whose seatbelt is not buckled? b) What's the probability that the first unbelted driver is in the 6 th car stopped? c) What's the probability that the first 10 drivers are all wearing their seatbelts? d) If they stop 30 cars during the first hour, find the mean and standard deviation of the number of drivers expected to be wearing seatbelts. e) If they stop 120 cars during this safety check, what's the probability they find at least 20 drivers not wearing their seatbelts?

Short answer

a) 5 cars; b) 0.0655; c) 0.1074; d) Mean: 24, SD: 2.19; e) Prob. \( \geq \) 20: 0.9115.

Step-by-step solution

Identify Known Values and Concepts

We know that 80% (\( p = 0.8 \)) of drivers wear seatbelts. We will apply geometric distribution for parts (a) and (b), binomial distribution for part (c), and normal approximation to binomial distribution for parts (d) and (e).

Solve Part a - Expected Number of Cars Stopped

For part (a), we are dealing with a geometric distribution where the probability of success (finding an unbelted driver) is \( 1 - 0.8 = 0.2 \). The expected number of trials before first success is given by \( \frac{1}{p} \). Therefore, \( \frac{1}{0.2} = 5 \). So, they expect to stop 5 cars before finding an unbelted driver.

Solve Part b - Probability for First Unbelted Driver in 6th Car

This is another geometric distribution problem. We calculate the probability that the first failure (unbuckled) happens on the 6th trial: \( (0.8)^5 \times 0.2 = 0.065536 \). Hence, the probability is 0.065536.

Solve Part c - Probability First 10 Drivers Wear Seatbelts

Here, we use the binomial distribution. The probability all 10 drivers wear seatbelts is \( 0.8^{10} \). Calculating, \( 0.8^{10} \approx 0.1073741824 \). Thus, the probability is approximately 0.1074.

Calculate Mean for Part d

In a binomial distribution, the mean \( \mu \) is given by \( np \), where \( n = 30 \) and \( p = 0.8 \). So, \( 30 \times 0.8 = 24 \). The mean number of drivers wearing seatbelt is 24.

Calculate Standard Deviation for Part d

The standard deviation \( \sigma \) for a binomial distribution is given by \( \sqrt{np(1-p)} \). Here, \( n = 30 \), \( p = 0.8 \), so \( \sigma = \sqrt{30 \times 0.8 \times 0.2} \approx \sqrt{4.8} \approx 2.19 \).

Solve Part e - Probability of At Least 20 Not Wearing Seatbelts

We need to find the probability of x being at least 20, where \( x \) follows a binomial distribution with \( n = 120 \) and \( p = 0.2 \). We calculate the mean \( np = 24 \) and standard deviation \( \sqrt{np(1-p)} \approx 4.38 \). Using a normal approximation, convert to z-score where \( z = \frac{x - 24}{4.38} \). Find \( P(x \geq 20) \): \( P(z \geq \frac{19.5 - 24}{4.38}) \). Using standard normal distribution tables, the probability is approximately 0.9115.

Key concepts

Binomial Distribution

The binomial distribution forms a fundamental concept in probability and statistics. It deals with scenarios where there are fixed numbers of trials, each with two possible outcomes: success or failure. In the context of the exercise, a driver wearing a seatbelt is considered a "success," and not wearing one is a "failure."

To use the binomial distribution effectively, certain conditions should be met:

• The number of trials, denoted as \( n \), is fixed.
• Each trial is independent of the others.
• The probability of success, denoted as \( p \), is constant for each trial.

In the given problem, when assessing the probability that all 10 drivers wear their seatbelts, the number of trials \( n \) is 10, and \( p = 0.8 \). You calculate it as \( 0.8^{10} \), reflecting the probability that each driver wears a seatbelt simultaneously.

Probability Calculations

Probability calculations are at the core of making informed decisions based on statistical data. In simpler terms, probability is the measure of how likely an event is to occur. It is expressed as a number between 0 and 1, where 0 means the event cannot happen and 1 indicates certainty.

Let's break down a key aspect through the problem: probability for the first unbelted driver being in the 6th car. Using the geometric distribution, we find this probability by calculating the success (wearing a seatbelt) raised to the power of 5 (number of prior trials), followed by a failure (the 6th car where the driver is unbelted). This gives us \( (0.8)^5 \times 0.2 = 0.065536 \). Thus, approximately 6.55% chance exists for the 6th driver to be the first unbelted one.

For different types of probability calculations, consider:

• Independent and dependent events impact the probabilities.
• Combining probabilities for multiple events might use multiplication or addition rules.
• Understanding properties, such as complements and intersections, enhances efficient calculations.

Normal Approximation

Normal approximation is a statistical technique used to simplify the calculations associated with binomial distributions when the number of trials (\( n \)) becomes large.

A binomial distribution can be approximated to a normal distribution when \( np \) and \( n(1-p) \) are both larger than 5, providing a way to estimate probabilities more easily with the aid of the standard normal distribution (known as the Z-distribution).

In the exercise's context, when finding the probability of at least 20 drivers not wearing seatbelts out of 120 stopped cars, normal approximation simplifies this process. Given \( n = 120 \) and probability \( p = 0.2 \) for a driver being unbelted, we calculate the mean \( np = 24 \) and standard deviation \( \sqrt{np(1-p)} \approx 4.38 \). This converts into a z-score, allowing us to find this probability using standard normal tables.

This method, called the Central Limit Theorem in statistical theory, is invaluable in real-life applications for handling large datasets or computational tasks.