/*! 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 20 The probability of a transistor ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The probability of a transistor failing between \(t=a\) months and \(t=b\) months is given by \(c \int_{a}^{b} e^{-c t} d t,\) for some constant \(c\). (a) If the probability of failure within the first six months is \(10 \%,\) what is \(c ?\). (b) Given the value of \(c\) in part \((a),\) what is the probability the transistor fails within the second six months?

Short Answer

Expert verified
(a) \(c \approx 0.022314\). (b) Probability \(\approx 9.5\%\).

Step by step solution

01

Understand the problem

We are given a probability distribution function for the failure of a transistor within certain time limits, which is determined by the constant \(c\). We need to find \(c\) assuming a known probability for an interval, and use it to find the probability over another interval.
02

Set up the integral for probability

The probability for a transistor to fail within the first 6 months \( (t=0 \text{ to } t=6) \) is given by \(\int_{0}^{6} c e^{-c t} dt = 0.10\).
03

Solve the integral

Calculate the integral: \[ \int c e^{-ct} dt = -e^{-ct} + C \] over the limits from \(0\) to \(6\). Thus, \[ \left[c(-e^{-ct})\right]_0^6 = c(-e^{-6c} + 1) = 0.10 \]
04

Solve for the constant c

Set the equation from the integral result: \[ c(1 - e^{-6c}) = 0.10 \]Solving this equation involves numerical or iterative methods to estimate \(c\). It turns out that \(c \approx 0.022314\).
05

Apply the constant to the second problem

To find the probability that the transistor fails within the second six months \((t=6 \text{ to } t=12)\) with found \(c\), compute \[ \int_{6}^{12} 0.022314 e^{-0.022314 t} dt \].
06

Solve the integral for the second interval

Evaluate the new integral: \[ \int_{6}^{12} 0.022314 e^{-0.022314 t} dt = 0.022314(-e^{-0.022314 t}) \Big|_6^{12} \]} Compute this numerically to find \(\approx 0.095\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Decay
Exponential decay is a fundamental concept in probability and statistics used to describe processes that reduce exponentially over time. In this context, a parameter or event decreases at a rate proportional to its current value. This behavior is modeled by the function \(e^{-ct}\), where \(c\) is a positive constant. The larger the value of \(c\), the faster the decay process happens. In real-world applications, you often find this concept applied to cell phone battery discharge, radioactive decay, or as in our exercise, the probability of a transistor failing over time.

With exponential decay, the crucial takeaway is understanding how changes in the decay constant \(c\) affect the rate at which the probability or measurement declines. An increase in \(c\) means the process unfolds faster, while a decrease results in a slower progression. This understanding helps predict and manage various scenarios where decay processes occur.
Probability Distribution Function
A probability distribution function (PDF) represents the likelihood of a random variable taking specific values within a defined range. It enables us to quantify and visualize how probable different outcomes of an event might be. In continuous random variables, such as the one used in our exercise, the PDF is a smooth curve that illustrates probability density.

The function \(c e^{-ct}\) used in the problem is an example of an exponential probability distribution. This distribution is especially convenient in modeling the time until an event occurs, like the time until a transistor fails. It highlights how systems behave under the influence of random processes that are memoryless, meaning that the probability of an event occurring at a future time is independent of any past occurrence.
  • The total area under the PDF curve over its range is 1, representing the certainty of one of the outcomes happening.
  • The PDF is integrated over the desired interval to find the probability of the event occurring within that timeframe, much like our exercise when calculating probabilities for specific months.
Integral Calculus
Integral Calculus is a branch of mathematics dealing with integrals and their applications. Integrals are used to calculate the area under a curve, among other things. This tool becomes essential when determining the probability over a continuous interval, as seen in the exercise.

Let's break down how integrals are applied in probability calculations:
  • Integrate the PDF over the specified range to determine the probability of an event occurring within that time span.
  • The definite integral \( \int_{a}^{b} c e^{-ct} \, dt \) from \(t=a\) to \(t=b\) finds the precise probability of failure between those two points in time. It considers the weight of each moment across the given interval.

Understanding how to set up and solve these integrals is crucial for answering more complex probability questions accurately. Integrals allow you to compile and interpret data over a continuous time frame, reflecting more realistic real-world scenarios compared to discrete calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The speeds of cars on a road are approximately normally distributed with a mean \(\mu=58 \mathrm{km} / \mathrm{hr}\) and standard deviation \(\sigma=4 \mathrm{km} / \mathrm{hr}\) (a) What is the probability that a randomly selected car is going between 60 and \(65 \mathrm{km} / \mathrm{hr} ?\) (b) What fraction of all cars are going slower than 52 \(\mathrm{km} / \mathrm{hr} ?\)

Let \(p(t)=-0.0375 t^{2}+0.225 t\) be the density function for the shelf life of a brand of banana which lasts up to 4 weeks. Time, \(t\), is measured in weeks and \(0 \leq t \leq 4\) Find the median shelf life of a banana using \(p(t) .\) Plot the median on a graph of \(p(t) .\) Does it look like half the area is to the right of the median and half the area is to the left?

Suppose \(F(x)\) is the cumulative distribution function for heights (in meters) of trees in a forest. (a) Explain in terms of trees the meaning of the statement \(F(7)=0.6\) (b) Which is greater, \(F(6)\) or \(F(7)\) ? Justify your answer in terms of trees.

A group of people have received treatment for cancer. Let \(t\) be the survival time, the number of years a person lives after the treatment. The density function giving the distribution of \(t\) is \(p(t)=C e^{-C t}\) for some positive constant \(C .\) What is the practical meaning of the cumulative distribution function \(P(t)=\int_{0}^{t} p(x) d x ?\)

Let \(p(x)\) be the density function for annual family income, where \(x\) is in thousands of dollars. What is the meaning of the statement \(p(70)=0.05 ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.