/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 The number of years a radio func... [FREE SOLUTION] | 91Ó°ÊÓ

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The number of years a radio functions is exponentially distributed with parameter \(\lambda=\frac{1}{8} .\) If Jones buys a used radio, what is the probability that it will be working after an additional 8 years?

Short Answer

Expert verified
The probability that the used radio will be working after an additional 8 years is approximately \(0.3679\) or \(36.79\%\).

Step by step solution

01

Identify the exponential distribution function

The exponential distribution function has the probability density function (pdf) given by: \[f(x) = \lambda e^{-\lambda x},\] where \(x \ge 0\), and the Cumulative Distribution Function (CDF) given by: \[F(x) = 1 - e^{-\lambda x},\] where \(x \ge 0\). In this problem, we have \(\lambda = \frac{1}{8}\).
02

Calculate the probability of the radio working after 8 years

We are interested in finding the probability that the radio will be working after an additional 8 years, i.e., we want to find the probability that it will function for at least 8 years: \[P(X > 8).\] Using the CDF, we can find this probability as: \[P(X > 8) = 1 - P(X \le 8).\] Now, we can use the CDF formula: \[P(X > 8) = 1 - (1 - e^{-\lambda * 8}).\]
03

Substitute the parameter value into the equation

Substitute \(\lambda = \frac{1}{8}\) into the equation: \[P(X > 8) = 1 - (1 - e^{-\frac{1}{8} * 8}).\]
04

Calculate the probability

Now, we compute the value of the probability: \[P(X > 8) = 1 - (1 - e^{-1})\] \[P(X > 8) = 1 - (1 - e^{-1})\] \[P(X > 8) = e^{-1}\]
05

Evaluate the exponential term

Evaluate the exponential term: \[P(X > 8) \approx 0.3679\] The probability that the used radio will be working after an additional 8 years is approximately 0.3679 or 36.79%.

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Key Concepts

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

Probability Density Function
When exploring the world of statistics, the Probability Density Function (PDF) is an essential concept, especially in the context of continuous random variables. Imagine a graph, where the area under the curve represents the likelihood of a random variable falling within a particular range. For the exponential distribution, the curve is described mathematically as:
\[f(x) = \lambda e^{-\lambda x},\]
where \( \lambda \) is the rate parameter. The exponential distribution's PDF is unique in the way it quantifies occurrences, and in our exercise, it is used to determine the likelihood of an event within a continuous interval, such as the lifespan of a radio.
The primary purpose of the PDF is to provide a visual and mathematical understanding of how probable different outcomes are. It is essential to remember that the PDF itself cannot give us the probability directly; instead, we use the area under the curve between two values of \(x\) to find the probability in that interval.
Cumulative Distribution Function
Building on the Probability Density Function, there's a way to calculate the probability that our variable of interest, say 'the time until a radio stops working', will be less than or equal to a specific value. This is where the Cumulative Distribution Function (CDF) comes into play.
It is defined for the exponential distribution as:
\[F(x) = 1 - e^{-\lambda x},\]
where \(x\) is the value of interest and \(\lambda\) is as before, the rate parameter. The useful feature of the CDF is its ability to give us directly the probability \(P(X \le x)\), which is the probability that the random variable \(X\) is less than or equal to \(x\).
In our radio example, we were able to calculate the probability that the radio would still function after 8 more years using the complement of the CDF. CDFs are particularly valuable when dealing with 'at least' or 'greater than' type probabilities, which are common in real-life scenarios.
Exponential Decay
Exponential decay is a phenomenon that shows a decrease in quantity over time, at a rate proportional to the current amount. In many real-world scenarios, such as radioactive decay, population decay, and yes, even the reliability of electronic components like a radio, the concept is fittingly applied.
Its mathematical representation in our context involved the formula \(e^{-\lambda x}\), where \(x\) is the time and \(\lambda\) is the decay constant. The exponent's negative sign indicates that the function is decreasing over time.
The wider implication of exponential decay is that the amount decreases by the same percentage over equal time intervals, creating a characteristic 'halving-time' or 'mean life' depending on the context. This concept is essential to comprehend when interpreting the exponential distribution's PDF and CDF, as it directly relates to how quickly the probability of a radio's functional lifespan decreases as time goes on.
Reliability Engineering
The field of reliability engineering is concerned with the probability of equipment or a system performing its required functions without failure for a specific period under stated conditions. The concept is intertwined with our study of the exponential distribution, where the focus is on the time until an event, such as the failure of a radio, occurs.
In reliability engineering, the exponential distribution is particularly favored for its memoryless property, meaning the failure rate is constant over time, which can simplify analysis and modeling. By applying the principles of the exponential distribution, engineers can estimate product lifetimes, plan maintenance, and improve system designs.
The probability that a used radio will work for an additional 8 years, as calculated in our exercise, is a typical reliability engineering problem. Such calculations help inform decisions on warranties, spare parts inventory, and even the design of new products.

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