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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
(a) The histogram shows a binomial distribution with probabilities calculated using the formula. (b) The expected number of tailgaters is 4.2. (c) The standard deviation is approximately 1.67.

Step by step solution

01

Determine the Distribution Type

The problem involves a fixed number of trials (12 vehicles) with only two possible outcomes for each vehicle (tailgate or not tailgate). Therefore, this is a Binomial distribution with parameters: number of trials \( n = 12 \) and probability of success (tailgating) \( p = 0.35 \).
02

Probability Mass Function (PMF) Calculation

The probability of exactly \( r \) vehicles tailgating is given by the binomial probability formula: \[P(r) = {n \choose r} p^r (1-p)^{n-r}\]where \({n \choose r}\) is the binomial coefficient. Calculate \( P(r) \) for each \( r \) from 0 to 12.
03

Construct the Histogram

Using the calculated probabilities from Step 2, plot \( r \) values on the x-axis and their corresponding probabilities \( P(r) \) on the y-axis to create a histogram.
04

Calculate the Expected Value

The expected number of vehicles that will tailgate is given by the formula for the expected value of a Binomial distribution:\[E(r) = n \cdot p = 12 \cdot 0.35 = 4.2.\]
05

Calculate the Standard Deviation

The standard deviation of a Binomial distribution is calculated using the formula:\[\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{12 \cdot 0.35 \cdot 0.65} \approx 1.67.\]

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Key Concepts

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

Probability Distribution
The concept of probability distribution can be quite fascinating, especially when it relates to real-life scenarios such as the behavior of tailgating on the highway. In this context, we're dealing with a Binomial Distribution, which is ideal when you have a fixed number of trials and each trial has exactly two possible outcomes. For instance, in our example, each car either tailgates or it doesn't, which perfectly fits this model.We define a binomial distribution with two parameters:
  • n, the number of trials—in this case, 12 vehicles;
  • p, the probability of success (a car tailgating), here given as 0.35.
To find the probability of exactly r vehicles tailgating, you use the binomial probability formula: \[P(r) = {n \choose r} p^r (1-p)^{n-r}\] This formula helps in constructing a complete probability mass "histogram" showing the likelihood of each possible result from 0 to 12 cars tailgating.This distribution visually illustrates how likely it is for a specific number of cars to "choose" tailgating as their travel option.
Expected Value
Expected Value is a fundamental concept in probability and statistics, capturing the idea of the mean or average outcome you might expect in the long run from a random process.For a binomial distribution, the expected value can be determined rather straightforwardly. The formula used is \[E(r) = n \cdot p\] where:
  • n is the total number of trials, here 12 cars, and
  • p is the probability of an individual car tailgating, which is 0.35 in our scenario.
Applying these numbers gives us:\[E(r) = 12 \times 0.35 = 4.2\]This means, on average, you would expect about 4.2 of the 12 vehicles to exhibit tailgating behavior.This calculation gives a predictive sense of what you might expect over repeated trials or instances. In our daily life, especially in traffic, an understanding of expected values can be quite revealing, helping you plan or interpret behaviors over time.
Standard Deviation
Standard Deviation adds depth to our understanding by quantifying the amount of variance or "spread" in a set of values.In a Binomial Distribution, the formula for the standard deviation is:\[\sigma = \sqrt{n \cdot p \cdot (1-p)}\]Substituting the parameters of our distribution:
  • n is 12,
  • p is 0.35, and
  • the term (1-p) (or the probability of a car not tailgating) is 0.65.
This calculation yields:\[\sigma \approx \sqrt{12 \times 0.35 \times 0.65} \approx 1.67\]The standard deviation gives insight into how much variation from the average (expected value) we might expect to observe in the number of tailgating cars. A standard deviation of 1.67 implies that in most instances, the number of tailgating cars will likely be within 1.67 cars above or below the expected value of 4.2.Understanding standard deviation empowers you to anticipate and interpret variability in real-world phenomena, such as the varied nature of drivers on a busy highway.

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Most popular questions from this chapter

Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1. (b) Explain why \(P(r \geq 1)=1-P(0)\) (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\) (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

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Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in Western Civilization each term. Let \(n=1\) \(2,3, \ldots\) represent the number of times a student takes western civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\) (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass western civilization? (e) What is the expected number of attempts at western civilization Susan must make to have her (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

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