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Suppose that only \(25 \%\) of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible. What is the probability that, of 20 randomly chosen drivers coming to an intersection under these conditions, a. At most 6 will come to a complete stop? b. Exactly 6 will come to a complete stop? c. At least 6 will come to a complete stop? d. How many of the next 20 drivers do you expect to come to a complete stop?

Short Answer

Expert verified
a. Calculate \(P(X \leq 6)\) using the binomial distribution formula. b. \(P(X = 6)\). c. Compute \(1 - P(X \leq 5)\). d. Expect 5 drivers to stop.

Step by step solution

01

Define the Random Variable

Let the random variable \(X\) denote the number of drivers who come to a complete stop out of 20. \(X\) follows a binomial distribution with parameters \(n = 20\) and \(p = 0.25\), where \(n\) is the number of trials (drivers) and \(p\) is the probability of a success (a driver coming to a complete stop).
02

Calculate Probability for At Most 6 Stops (a)

We want \(P(X \leq 6)\). For a binomial distribution, the probability mass function is \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\). Calculate and sum \(P(X = k)\) for \(k = 0\) to \(k = 6\). This can be done using a statistical software or binomial probability table.
03

Calculate Probability for Exactly 6 Stops (b)

To find \(P(X = 6)\), use the binomial probability formula: \(P(X = 6) = \binom{20}{6} (0.25)^6 (0.75)^{14}\). This calculation gives the exact probability for 6 drivers stopping.
04

Calculate Probability for At Least 6 Stops (c)

We want \(P(X \geq 6)\). Since \(P(X \geq 6) = 1 - P(X \leq 5)\), first calculate \(P(X \leq 5)\) by summing up probabilities \(P(X = k)\) for \(k = 0\) to \(k = 5\) and then subtract from 1.
05

Calculate Expected Number of Stops (d)

The expected value \(E[X]\) of a binomial distribution is \(n \cdot p\). Thus, \(E[X] = 20 \cdot 0.25 = 5\). This means we expect 5 drivers to come to a complete stop.

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Key Concepts

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

Understanding Probability Calculations
In the context of the binomial distribution, probability calculations form the basis of predicting outcomes in situations like the one explained in the exercise. Here, we are looking at how likely different numbers of drivers are to stop at an intersection. This involves assessing probabilities for discrete events using a mathematical function known as a probability mass function (PMF).

For binomial distributions, the PMF is expressed as:
  • \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
In this formula, \(n\) is the number of trials, \(k\) is the number of successful trials (e.g., drivers stopping), and \(p\) is the probability of each success.

This formula allows us to compute the probability of specific outcomes, like exactly 6 out of 20 drivers stopping. And to find probabilities like 'at most 6 drivers stopping', we would sum the probabilities for 0 to 6 drivers. This process can be simplified using statistical software or binomial probability tables, making it more practical for larger computations.
Expected Value in Binomial Distribution
The expected value, or mean, in a binomial distribution offers a straightforward way to anticipate the average outcome over many trials. For drivers at a stop sign, we use the expected value to predict how many of the drivers are likely to stop.

The expected value \(E[X]\) in a binomial context is calculated using the formula:
  • \(E[X] = n \cdot p\)
where \(n\) represents the total number of drivers (20 in our exercise) and \(p\) is the probability of a single driver stopping (0.25 in this scenario). Thus, \(E[X] = 20 \times 0.25 = 5\), suggesting that on average, 5 drivers will come to a complete stop.

This expectation helps not just in understanding typical outcomes, but also in planning and decision-making, considering it reflects the average result across many repetitions of the event.
Defining a Random Variable
A random variable is a key concept in probability and statistics, representing a numerical outcome of a random process. In binomial distributions, random variables help us model and make predictions about real-world phenomena like drivers at a stop sign.

In the given exercise, the random variable \(X\) is defined as the number of drivers stopping a complete stop out of 20. Here, \(X\) can take on any integer value from 0 to 20, corresponding to the number of drivers surprised by the conditions at the intersection.

This random variable follows a binomial distribution because each driver's decision is independent, and the probability of stopping (0.25) remains constant. Understanding this concept is crucial, as it lays the groundwork for using statistical models and software to predict and calculate probabilities and expected outcomes.
Role of Statistical Software
In practice, statistical software plays a critical role in solving complex problems quickly and accurately, like the binomial distribution exercises presented. Using software, you can bypass manual calculations and directly obtain results.

Such software often includes functions to calculate probabilities for various statistical distributions, like the binomial distribution, where inputs such as number of trials (\(n\)), probability of success (\(p\)), and specific outcomes (like "at most 6 drivers stop") are entered. This provides an output with precise probability calculations.

Moreover, statistical software can graphically represent data, which aids in visual understanding and interpretation of statistical results. This ease of use and reliability makes these tools indispensable for students and professionals alike in analyzing and making data-driven decisions.

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Most popular questions from this chapter

Let \(X\) be the number of points earned by a randomly selected student on a 10 point quiz, with possible values \(0,1,2, \ldots, 10\) and pmf \(p(x)\), and suppose the distribution has a skewness of \(c\). Now consider reversing the probabilities in the distribution, so that \(p(0)\) is interchanged with \(p(10)\), \(p(1)\) is interchanged with \(p(9)\), and so on. Show that the skewness of the resulting distribution is \(-c\). [Hint: Let \(Y=10-X\) and show that \(Y\) has the reversed distribution. Use this fact to determine \(\mu_{Y}\) and then the value of skewness for the \(Y\) distribution.]

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Three brothers and their wives decide to have children until each family has two female children. Let \(X=\) the total number of male children born to the brothers. What is \(E(X)\), and how does it compare to the expected number of male children borm to each brother?

Some parts of Califomia are particularly earthquake-prone. Suppose that in one such area, \(30 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. One possible outcome is SFSS, with probability \((.3)(.7)(.3)(.3)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(1 \mathrm{mile}\). Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres.]

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