Seatbelts II Police estimate that 80 % of drivers now wear...

Seatbelts II 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 sixth 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?

Seatbelts II 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 sixth 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?

Key Concepts

Bernoulli Trials

Bernoulli trials are experiments with exactly two outcomes, typically labeled as success and failure, with a constant probability for each outcome in every trial. This concept is the foundation for both the geometric and binomial distributions, as it allows the modeling of individual trials, like each driver's seatbelt status check.

Expected Value and Standard Deviation

The expected value describes the average outcome of a random variable over many trials, while the standard deviation measures the variability or spread of the outcomes around this mean. In the context of distributions like the binomial, these parameters provide insights into the central tendency and dispersion of the number of successes, such as the expected number of seatbelt users and the variability around that expectation.

Geometric Distribution

The geometric distribution models the number of independent trials required to get the first success, where each trial has the same probability of success. In problems where you need to determine how many attempts are expected before a particular event occurs, such as finding the first driver not wearing a seatbelt, the geometric distribution is key.

Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent trials with the same probability of success on each trial. It is used to calculate the probability of a certain number of events occurring, such as the number of drivers wearing or not wearing seatbelts in a predetermined number of stops.

Transcript

00:01 So we're going to assume that 80 % of people wear a seatbelt.

00:04 And in our first one, how many cars would they expect to stop before finding one who is not buckled? so we want to find how many they stop to find not buckled.

00:15 So the success now is not buckled.

00:18 And that is a .2 chance.

00:21 So our expected value for that is 1 over 1 5th, which will end up being 5.

00:29 Part b, what is the probability that the first unbuckled is the sixth? so the first unbuckled is the sixth.

00:39 So that means you'll have five who are buckled, and then finally your sixth person will be unbuckled.

00:47 So that is .6, excuse me, .8 to the 5th power times .2.

00:54 And that is an answer to four decimal places of .0655.

00:59 Part c, what is the probability that the first 10 are all wearing their seatbelts? so the probability of the first 10 are all wearing the seatbelts is going to be .8 to the 10th power.

01:17 And that comes out to be, to four decimal places, .1074.

01:22 On part d, if they stop 30 cars during the first hour, find the mean of the number of people who are wearing seatbelts.

01:35 So the mean would be 30 times .8, which is 24.

01:41 And the standard deviation would be the square root of 30 times .8 times .2.

01:47 So i'm going to take that 24 times .2, and we'll square root it...

Please give Ace some feedback