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The Acme Super light bulb is known to have a useful life described by the density function $$ f(t)=.01 e^{-.01 t} $$ where time \(t\) is measured in hours. (a) Find the failure rate of this bulb (see Exercise 2.2 .6 ). (b) Find the reliability of this bulb after 20 hours. (c) Given that it lasts 20 hours, find the probability that the bulb lasts another 20 hours. (d) Find the probability that the bulb burns out in the forty-first hour, given that it lasts 40 hours.

Short Answer

Expert verified
(a) 0.01; (b) ~0.8187; (c) ~0.8187; (d) ~0.0099

Step by step solution

01

Understanding Failure Rate

The failure rate, often called the hazard function, is defined as \( h(t) = \frac{f(t)}{R(t)} \), where \( R(t) \) is the reliability function, \( R(t) = 1 - F(t) \), and \( F(t) \) is the cumulative distribution function (CDF) of the density function \( f(t) = 0.01 e^{-0.01 t} \). First, we find the CDF by integrating \( f(t) \).
02

Calculating CDF and Reliability Function

Calculate the cumulative distribution function (CDF) \( F(t) \):\[ F(t) = \int_0^t 0.01 e^{-0.01 x} \, dx = 1 - e^{-0.01 t} \]Thus, the reliability function is:\[ R(t) = 1 - F(t) = e^{-0.01 t} \]
03

Calculating Failure Rate

Using the formula for failure rate, we have:\[ h(t) = \frac{f(t)}{R(t)} = \frac{0.01 e^{-0.01 t}}{e^{-0.01 t}} = 0.01 \]Thus, the failure rate is a constant 0.01.
04

Finding Reliability After 20 Hours

To find the reliability after 20 hours, substitute \( t = 20 \) into the reliability function:\[ R(20) = e^{-0.01 \times 20} = e^{-0.2} \approx 0.8187 \]
05

Conditional Reliability

Given the bulb lasts 20 hours, find the probability it lasts another 20 hours using the conditional reliability:\[ P(T > 40 | T > 20) = \frac{R(40)}{R(20)} \]Calculate \( R(40) = e^{-0.01 \times 40} = e^{-0.4} \approx 0.6703 \).Thus, the probability is:\[ P(T > 40 | T > 20) = \frac{0.6703}{0.8187} \approx 0.8187 \]
06

Probability of Failing in the Forty-First Hour

The probability that the bulb burns out in the forty-first hour, given it lasts 40 hours, is found by evaluating:\[ P(40 < T \leq 41 | T > 40) = P(T > 40) - P(T > 41) \]Calculate \( R(41) = e^{-0.01 \times 41} = e^{-0.41} \approx 0.6641 \).The probability is:\[ P(40 < T \leq 41 | T > 40) = \frac{R(41) - R(40)}{R(40)} = \frac{0.6641 - 0.6703}{0.6703} \approx 0.0099 \]

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Key Concepts

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

Hazard Function
In reliability theory, understanding the hazard function, also known as the failure rate or risk function, is crucial. This function quantifies the instantaneous failure rate of a component, such as a light bulb, at a specific time, given that the component has survived up to that time. It's like asking, "If the bulb hasn't failed yet, how likely is it to fail immediately?" The hazard function, denoted as \( h(t) \), is defined mathematically as:
  • \( h(t) = \frac{f(t)}{R(t)} \)
Where:
  • \( f(t) \) is the probability density function (PDF).
  • \( R(t) \) is the reliability function.
Both the PDF and reliability function are crucial for determining \( h(t) \). A critical point about the hazard function is that it can be constant, increasing, or decreasing over time. For our specific light bulb problem, the hazard function turns out to be a constant 0.01. This indicates that the failure rate of the bulb remains the same regardless of how long it has already been functioning.
Exponential Distribution
The exponential distribution is a common and simple probability distribution often used to model the time until an event, such as a failure or arrival, occurs. A notable feature of the exponential distribution is its memoryless property. This means the probability of failure in the next moment is always the same, no matter how long the bulb has been functioning. The density function of an exponential distribution is given by:
  • \( f(t) = \lambda e^{-ambda t} \)
The parameter \( \lambda \) is the rate parameter, which indicates the rate at which events occur. For our light bulb problem, \( \lambda = 0.01 \), meaning the expected life span of the bulb can be expected to follow this exponential pattern. This distribution frequently appears in reliability studies because it naturally models the randomness in time to failure events. The simplicity and familiarity of the exponential distribution make it particularly useful in practical applications.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is essential in probability theory and statistics as it gives the probability that a random variable is less than or equal to a certain value. In the context of reliability and failure systems, it gives insight into the life expectancy of products. To calculate the CDF, you integrate the probability density function (PDF) from 0 to \( t \):
  • \( F(t) = \int_0^t f(x) \, dx \)
For the exponential distribution involved in our light bulb example, the CDF is:
  • \( F(t) = 1 - e^{-0.01 t} \)
This function helps determine the reliability because it represents the probability of failure by a specific time, \( t \). The reliability function is consequently:
  • \( R(t) = 1 - F(t) = e^{-0.01 t} \)
The CDF is vital in problems like these as it forms the basis from which reliability, and thereby the hazard function, are calculated. It gives a comprehensive picture of how likely the bulb is to have failed by any given moment.

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Most popular questions from this chapter

Suppose you toss a dart at a circular target of radius 10 inches. Given that the dart lands in the upper half of the target, find the probability that (a) it lands in the right half of the target. (b) its distance from the center is less than 5 inches. (c) its distance from the center is greater than 5 inches. (d) it lands within 5 inches of the point (0,5) .

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