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Tailgating Do you tailgate the car in front of you? About \(35 \%\) of all drivers will tailgate before passing, thinking they can make the car in front of them go faster (Source: Bernice Kanner, Are You Normal?, St. Martin's Press). Suppose that you are driving a considerable distance on a two-lane highway and are passed by 12 vehicles. (a) Let \(r\) be the number of vehicles that tailgate before passing. Make a histogram showing the probability distribution of \(r\) for \(r=0\) through \(r=12\). (b) Compute the expected number of vehicles out of 12 that will tailgate. (c) Compute the standard deviation of this distribution.

Short Answer

Expert verified
The expected number of tailgaters is 4.2 with a standard deviation of approximately 1.65. Create a histogram using the binomial probabilities for each possible number of tailgaters from 0 to 12.

Step by step solution

01

Define the Problem

We want to find the probability distribution of the number of vehicles that tailgate, where each vehicle has a probability of 35% of tailgating. This problem involves a binomial distribution since there's a fixed number of trials, each with two possible outcomes (tailgate or not). Let the random variable \( r \) denote the number of vehicles that tailgate.
02

Binomial Distribution Formula

Since each vehicle has two possible outcomes, the distribution of \( r \) can be defined using the binomial probability formula: \[ P(r) = \binom{n}{r} p^r (1-p)^{n-r} \] where \( n = 12 \) is the total number of vehicles, \( p = 0.35 \) is the probability of tailgating, and \( \binom{n}{r} \) is the binomial coefficient.
03

Calculate Probabilities for Each \( r \)

Calculate the probability \( P(r) \) for each \( r \) from 0 to 12 using the formula from Step 2. For example, for \( r = 0 \), \[ P(0) = \binom{12}{0} (0.35)^0 (0.65)^{12} \]. Do this calculation for each \( r \) to complete the distribution.
04

Create the Histogram

Use the probabilities calculated in Step 3 to create a histogram. The x-axis represents the number of vehicles that tailgate (\( r \)), and the y-axis represents the probability of each \( r \). This visual representation shows the probability distribution.
05

Calculate Expected Value

The expected number of vehicles that tailgate is given by \[ E(r) = n \cdot p = 12 \cdot 0.35 = 4.2 \]. This means that, on average, out of 12 vehicles, 4.2 vehicles are expected to tailgate.
06

Calculate Standard Deviation

The standard deviation of this binomial distribution is calculated using the formula \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{12 \cdot 0.35 \cdot 0.65} \]. Compute this value to find the standard deviation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution
A probability distribution is a powerful concept in statistics that shows all the possible outcomes of a random experiment and the probability of each outcome. In the context of tailgating on the highway, we use a **binomial distribution** because there are two possible outcomes for each of the 12 vehicles: either they tailgate or they do not.

The binomial distribution helps us determine the probability of a certain number of vehicles tailgating. The binomial probability formula is given as: \[ P(r) = \binom{n}{r} p^r (1-p)^{n-r} \] where:
  • \( n \) is the total number of vehicles (in our case, 12).
  • \( r \) is the number of vehicles tailgating.
  • \( p \) is the probability of tailgating for each vehicle (0.35).
  • \( 1-p \) is the probability of not tailgating (0.65).
By calculating the probability for each \( r \) (0 through 12), we form a complete probability distribution, showcasing the likelihood of each possible number of vehicles tailgating.
Expected Value
The concept of expected value can be thought of as the average outcome we expect in a probabilistic scenario. It's like predicting the most "usual" case outcome.

In our highway scenario, the expected number of vehicles that tailgate is calculated by multiplying the total number of vehicles \( n \) by the probability of a single vehicle tailgating \( p \). This is done through the formula: \[ E(r) = n \cdot p \] For our case: \[ E(r) = 12 \cdot 0.35 = 4.2 \] This tells us that out of 12 cars passing by, we can "expect" about 4.2 vehicles to tailgate, on average. This doesn't mean you can have 0.2 of a vehicle, but rather that in many repeated scenarios, the average approaches this number.
Standard Deviation
Standard deviation is a key measure that tells us how much variation or "spread" exists from the average or expected value in our data. In simpler terms, it shows how spread out the tailgating behavior is among the 12 vehicles.

For a binomial distribution, the standard deviation \( \sigma \) is found using the formula: \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \] Substituting in our values:\[ \sigma = \sqrt{12 \cdot 0.35 \cdot 0.65} \] By calculating this, we find the standard deviation, which quantifies the variation in the number of vehicles that tailgate. A smaller standard deviation means the results are close to the expected value, while a larger one suggests more variety in outcomes.
Histogram in Statistics
A histogram is a graphical representation of data that uses bars to show the frequency or probability of different possible outcomes of a dataset. In statistics, it is particularly useful to understand the probability distribution of a random variable, such as tailgating in this example.

To create the histogram:
  • The x-axis will represent the random variable \( r \), the number of vehicles tailgating, ranging from 0 to 12.
  • The y-axis will show the probability of each of these outcomes, calculated using the binomial formula.
Each bar on the histogram represents the probability of that particular number of vehicles tailgating, giving us a clear visual of how likely each scenario is. Seeing this distribution in a histogram format makes it easier to understand which outcomes are more probable and how the data is spread.

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Most popular questions from this chapter

For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

The Wall Street Journal reported that approximately \(25 \%\) of the people who are told a product is improved will believe that it is, in fact, improved. The remaining \(75 \%\) believe that this is just hype (the same old thing with no real improvement). Suppose a marketing study consists of a random sample of eight people who are given a sales talk about a new, improved product. (a) Make a histogram showing the probability that \(r=0\) to 8 people believe the product is, in fact, improved. (b) Compute the mean and standard deviation of this probability distribution. (c) Quota Problem How many people are needed in the marketing study to be \(99 \%\) sure that at least one person believes the product to be improved? (Hint: Note that \(P(r \geq 1)=0.99\) is equivalent to \(1-P(0)=0.99\), or \(P(0)=0.01 .)\)

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Poisson Approximation to Binomial: Comparisons (a) For \(n=100, p=0.02\), and \(r=2\), compute \(P(r)\) using the formula for the binomial distribution and your calculator: $$ P(r)=C_{n, t} p^{r}(1-p)^{n-r} $$ (b) For \(n=100, p=0.02\), and \(r=2\), estimate \(P(r)\) using the Poisson approximation to the binomial. (c) Compare the results of parts (a) and (b). Does it appear that the Poisson distribution with \(\lambda=n p\) provides a good approximation for \(P(r=2) ?\) (d) Repeat parts (a) to \((c)\) for \(r=3\)

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