/*! 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 5 The lifetime of a radio is expon... [FREE SOLUTION] | 91Ó°ÊÓ

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The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?

Short Answer

Expert verified
The probability that Jones' 10-year-old radio will work after an additional 10 years is approximately 36.79%.

Step by step solution

01

Understand the Exponential Distribution

The exponential distribution is a continuous probability distribution with a probability density function given by: \(f(x;\lambda) = \lambda e^{-\lambda x}\) where x is the time (in this case the age of the radio) and λ is the hazard rate parameter. The mean of the exponential distribution is equal to \(\frac{1}{\lambda}\), so given that the mean lifetime is 10 years, we can calculate λ: \(10 = \frac{1}{\lambda} \)
02

Calculate the hazard rate parameter λ

Solve for λ using the mean given in the problem: \(\lambda = \frac{1}{10}\)
03

Understand the survival function

The survival function, S(x), is the probability that the radio survives (is still working) beyond age x. The survival function is the complement of the cumulative distribution function (the probability that the radio has already failed before age x). For an exponential distribution, the survival function is given by: \(S(x) = 1 - F(x)\) Where S(x) is the survival function and F(x) is the cumulative distribution function. For an exponential distribution, the cumulative distribution function F(x) is given by: \(F(x;\lambda) = 1 - e^{-\lambda x}\)
04

Calculate survival probability

We are interested in the survival probability after an additional 10 years (x=10). Therefore, using the exponential distribution survival function, we can calculate the probability: \(S(10) = 1 - F(10)\) \(S(10) = 1 - (1 - e^{-\frac{1}{10} \times 10})\) \(S(10) = 1 - (1 - e^{-1})\) \(S(10) = e^{-1}\)
05

Find the probability

Calculate the probability: \(S(10) \approx 0.3679\) The probability that Jones' 10-year-old radio will work after an additional 10 years is approximately 36.79%.

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Key Concepts

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

Survival Function
Imagine owning a gadget, such as a radio, and wanting to know the likelihood that it will continue to work after a certain period. This is where the survival function comes into play in statistics, especially for products or systems with lifetimes that can be modeled by an exponential distribution.

The survival function, denoted as \(S(x)\), represents the probability that the gadget will 'survive' beyond a certain time \(x\). In simpler terms, if you are looking at a graph, the survival function gives you the odds of your gadget still being operational as you move along the timeline. The key property to remember is that it's the complement to the cumulative distribution function (CDF), which means it can be expressed as \(S(x) = 1 - F(x)\), where \(F(x)\) is the CDF.

The survival function is particularly useful in reliability engineering and life testing. For instance, if you buy a used radio with a known exponential survival function and you want to find out the probability it will work for another decade, the survival function can give you that precise estimate.
Hazard Rate Parameter
Delving into the exponential distribution, there's a term known as the 'hazard rate parameter', which is represented by \(\lambda\text{ (lambda)}\). This \(\lambda\) is not just an arbitrary variable; it holds significant meaning. It's the rate at which failures occur in the process or system over time.

In the context of our radio scenario, \(\lambda\) helps us determine the rate at which radios are expected to stop working per year. If the mean lifetime, \(\mu\), of radios is known, \(\lambda\) can be easily calculated as the inverse of \(\mu\), or \(\lambda = 1/\mu\).

You can think of the hazard rate parameter a bit like the 'rhythm' at which the exponential distribution 'beats'. The higher the \(\lambda\), the more rapid the 'pulse', leading to a higher likelihood of failure over a given interval. This concept is pivotal in predicting the time of events like mechanical failures or in planning the expected lifespan of products.
Cumulative Distribution Function
In probability and statistics, we use the cumulative distribution function (CDF), denoted as \(F(x)\), to measure the probability that a variable will take a value less than or equal to a specific point. For an exponential distribution, the CDF is crucial because it tells us exactly the probability that the event of interest (such as the failure of a radio) has occurred by a certain time point.

For an exponential distribution, the CDF is defined as \(F(x;\lambda) = 1 - e^{-\lambda x}\). What this essentially says is as time \(x\) progresses, the likelihood of our radio having already failed increases. The cumulative part of the name implies that the function is accumulative over the duration. So, as \(x\) increases, so does \(F(x)\).

To put it in a real-world context, if you wanted to know the probability that a new radio would fail within the first ten years, the CDF would give you that information. It represents the trajectory of the radios from perfect function to their eventual failure.
Probability Density Function
Finally, let's break down the probability density function (PDF) for an exponential distribution. The PDF, which we denote as \(f(x;\lambda)\), is a formula that provides the probabilities of the variable taking on a specific value.

In the case of the exponential distribution, the PDF has the form \(f(x;\lambda) = \lambda e^{-\lambda x}\) and is directly related to the hazard rate parameter \(\lambda\). It gives us a snapshot of the likelihood that a failure will occur exactly at time \(x\). The exponential nature of the function means that the radio is more likely to fail sooner rather than later, reflecting the constant failure rate assumption of the exponential distribution.

To visualize it, imagine a graph where the y-axis represents the probability density and the x-axis represents time. The graph will show a declining curve, starting high when the radio is new and tapering off as time increases. This curve helps us understand not just when most failures are likely to happen, but also the nature of the risk associated with continued use of the radio over time. It's a vital tool for quality control and product lifecycle management.

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Most popular questions from this chapter

Events occur accórding to a Poisson process with rate \(\lambda\). Each time an event occurs, we must decide whether or not to stop, with our objective being to stop at the last event to occur prior to some specified time \(T\), where \(T>1 / \lambda\). That is, if an event occurs at time \(t, 0 \leqslant t \leqslant T\), and we decide to stop, then we win if there are no additional events by time \(T\), and we lose otherwise. If we do not stop when an event occurs and no. additional events occur by time \(T\), then we lose. Also, if no events occur by time \(T\), then we lose. Consider the strategy that stops at the first event to occur after some fixed time \(s, 0 \leqslant s \leqslant T\). (a) Using this strategy, what is the probability of winning? (b) What value of \(s\) maximizes the probability of winning? (c) Show that on?'s probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu\). Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \quad \text { and } \quad X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(\dot{E}\left[X_{(1)}\right]\) (b) \(-\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[\dot{X}_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

An event independently occurs on each day with probability \(p\). Let \(N(n)\) denote the total number of events that occur on the first \(n\) days, and let \(T_{r}\) denote the day on which the \(r\) th event occurs. (a) What is the distribution of \(N(n) ?\) (b) What is the distribution of \(T_{1} ?\) (c) What is the distribution of \(T_{r} ?\) (d) Given that \(N(n)=r\), show that the unordered set of \(r\) days on which events occurred has the same distribution as a random selection (without replacement) of \(r\) of the values \(1,2, \ldots, n\).

If \(X_{1}\) and \(X_{2}\) are independent nonnegative continuous random variables, show that $$ P\left\\{X_{1}

Consider an infinite server queueing system in which customers arrive in accordance with a Poisson process and where the service distribution is exponential with rate \(\mu\). Let \(X(t)\) denote the number of customers in the system at time \(t\). Find (a) \(E[X(t+s) \mid X(s)=n]\) (b) \(\operatorname{Var}[X(t+s) \mid X(s)=n]\) Hint: Divide the customers in the system at time \(t+s\) into two groups, one consisting of "old" customers and the other of "new" customers.

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