91Ó°ÊÓ

The number \(N\) of residential homes that a fire company can serve depends on the distance \(r\) (in city blocks) that a fire engine can cover in a specified (fixed) period of time. If we assume that \(N\) is proportional to the area of a circle \(R\) blocks from the firehouse, then \(N=C \pi R^{2},\) where \(C\) is a constant, \(\pi=3.1416 \ldots,\) and \(R,\) a random variable, is the number of blocks that a fire engine can move in the specified time interval. For a particular fire company, \(C=8\), the probability distribution for \(R\) is as shown in the accompanying table, and \(p(r)=0\) for \(r \leq 20\) and \(r \geq 27\). $$\begin{array}{c|cccccc}r & 21 & 22 & 23 & 24 & 25 & 26 \\\\\hline \rho(r) & .05 & .20 & .30 & .25 & .15 & .05 \\\\\hline\end{array}$$ Find the expected value of \(N\), the number of homes that the fire department can serve.

Step by step solution

01

Understanding the Problem

We need to find the expected number of homes, \(N\), that the fire department can serve. \(N\) is given by \(N = C \pi R^2\), where \(C = 8\), \(\pi\) is approximately 3.1416, and \(R\) is a random variable representing blocks. The task involves calculating the expected value of \(N\) using the given probability distribution of \(R\).

02

Expressing the Expected Value of N

The expected value of \(N\), \(E(N)\), can be expressed in terms of \(R\) as \(E(N) = E(C \pi R^2) = C \pi E(R^2)\). Therefore, we first need to calculate \(E(R^2)\), the expected value of \(R^2\).

03

Calculating E(R^2)

To find \(E(R^2)\), we use the definition of expected value for a discrete random variable: \[ E(R^2) = \sum_{r=21}^{26} r^2 \cdot \rho(r) \]. Evaluate \(r^2\) for each value of \(r\) and multiply by its respective probability \(\rho(r)\).

04

Computing Each Term for E(R^2)

Calculate each term in the sum for \(E(R^2)\): - For \(r = 21\), \(21^2 \cdot 0.05 = 22.05\).- For \(r = 22\), \(22^2 \cdot 0.20 = 96.8\).- For \(r = 23\), \(23^2 \cdot 0.30 = 158.7\).- For \(r = 24\), \(24^2 \cdot 0.25 = 144.0\).- For \(r = 25\), \(25^2 \cdot 0.15 = 93.75\).- For \(r = 26\), \(26^2 \cdot 0.05 = 33.8\).

05

Summing to Find E(R^2)

Add all the computed terms together to find \(E(R^2)\): \[ E(R^2) = 22.05 + 96.8 + 158.7 + 144.0 + 93.75 + 33.8 = 549.1 \].

06

Calculating E(N)

Now that we have \(E(R^2) = 549.1\), use the formula \(E(N) = C \pi E(R^2) = 8 \times 3.1416 \times 549.1\) to find \(E(N)\).

07

Final Computation for E(N)

Compute the final multiplication to get \(E(N)\): \[ E(N) = 8 \times 3.1416 \times 549.1 \approx 13776.9 \].

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Key Concepts

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

Discrete Random Variable

In probability theory, a discrete random variable can only take on a finite or countably infinite set of values. These values are typically integers, and each has a corresponding probability. In our problem, the random variable is the distance \( R \) in blocks that a fire engine can travel within a given time. The distance \( R \) is expressed as discrete values from 21 to 26 blocks, making it a discrete random variable.

• Each possible outcome (21, 22, 23, 24, 25, 26) is tied to a probability \( \rho(r) \).
• The sum of all these probabilities equals 1.

This allows us to create a probability distribution which is essential for further calculations.

Probability Distribution

A probability distribution provides the probabilities of all possible outcomes of a discrete random variable. It is usually presented in a tabular form for ease of understanding and calculation. In our example, the table gives the probabilities for each region \( r \) that a fire engine might cover. These probabilities tell us:

• For \( r = 21 \), \( \rho(r) = 0.05 \)
• For \( r = 22 \), \( \rho(r) = 0.20 \)
• For \( r = 23 \), \( \rho(r) = 0.30 \)
• For \( r = 24 \), \( \rho(r) = 0.25 \)
• For \( r = 25 \), \( \rho(r) = 0.15 \)
• For \( r = 26 \), \( \rho(r) = 0.05 \)

These probabilities are crucial for calculating the expected value since they weigh the contribution of each possible outcome.

Calculation of Expected Values

The expected value is a critical concept in statistics, particularly in the context of random variables. It represents the average outcome we anticipate over many repetitions of a random process. For our problem, we focus on \( E(R^2) \), the expected value of the square of the random variable \( R \). The calculation involves:

• Finding \( r^2 \) for each \( r \) value.
• Multiplying each \( r^2 \) by its probability \( \rho(r) \).
• Summing the results to get \[ E(R^2) = 22.05 + 96.8 + 158.7 + 144.0 + 93.75 + 33.8 = 549.1 \].

After obtaining \( E(R^2) \), we determine \( E(N) \) using the formula \( E(N) = C \pi E(R^2) \). This simple yet powerful method is deeply rooted in mathematical statistics.

Mathematical Statistics

Mathematical statistics offers tools and methods for analyzing data, including random variables and probability distributions. In our exercise, we apply statistical techniques to estimate the average number of homes a fire engine can serve. Key steps include:

• Utilizing the probability distribution to compute expected values.
• Understanding how expected values relate to real-world problems, like fire service coverage.
• Recognizing how mathematical statistics enables precise predictions and decision-making in various sectors.

This structured approach showcases the power of statistics in solving practical problems, allowing organizations to optimize resources effectively.