/*! 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 13 You have two lightbulbs for a pa... [FREE SOLUTION] | 91Ó°ÊÓ

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You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in \(1000 \mathrm{~s}\) of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1)\) ? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and \(2 ?\)

Short Answer

Expert verified
a. Joint pdf is \(e^{-(x+y)}\). b. Probability for both bulbs ≤ 1000 hours is \((1-e^{-1})^2\). c. Probability for total ≤ 2 is 0.865. d. Probability for total between 1 and 2 is 0.135.

Step by step solution

01

Understanding the Exponential Distribution

The exponential distribution for a random variable with parameter \(\lambda\) has the probability density function (pdf) given by \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). In this case, both \(X\) and \(Y\) have \(\lambda = 1\), so the pdf is \( f(x) = e^{-x} \) for both \(X\) and \(Y\).
02

Joint PDF of Independent Variables

Since \(X\) and \(Y\) are independent, the joint pdf of \(X\) and \(Y\) is the product of their individual pdfs. Thus, the joint pdf is \( f(x, y) = f(x) f(y) = e^{-x} e^{-y} = e^{-(x+y)} \) for \( x \geq 0 \) and \( y \geq 0 \).
03

Finding Probability that Each Bulb Lasts at Most 1000 Hours

We need to find \( P(X \leq 1, Y \leq 1) \). Since \(X\) and \(Y\) are independent, this probability is \( P(X \leq 1) \times P(Y \leq 1) \). For an exponential distribution with \(\lambda = 1\), \( P(X \leq 1) = 1 - e^{-1} \). Therefore, \( P(X \leq 1, Y \leq 1) = (1 - e^{-1})^2 \).
04

Probability of Total Lifetime at Most 2

We want \( P(X + Y \leq 2) \). This can be found by integrating the joint pdf over the region where \(x + y \leq 2\), \(x \geq 0\), and \(y \geq 0\). The integral is:\[ \int_0^2 \int_0^{2-x} e^{-(x+y)} \, dy \, dx.\]Calculate the inner integral first and then the outer integral to find the exact probability.
05

Finding the Probability of Total Lifetime Between 1 and 2

We need \( P(1 < X + Y \leq 2) \). This is computed as the difference between \( P(X + Y \leq 2) \) and \( P(X + Y \leq 1) \). Integrate the joint pdf over the region where \(x + y \leq 1\), and subtract that result from the previous integral. The integral for \( P(X + Y \leq 1) \) is:\[ \int_0^1 \int_0^{1-x} e^{-(x+y)} \, dy \, dx.\]By evaluating both integrals, we find the desired probability.

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Key Concepts

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

Joint Probability Density Function
In probability theory, a joint probability density function (joint pdf) is a crucial concept used to describe the likelihood of two random variables occurring together. For two continuous random variables, say \(X\) and \(Y\), the joint pdf \(f(x, y)\) specifies the probability that \(X\) takes on a certain value \(x\) and \(Y\) takes on a value \(y\) simultaneously.
An important property of a joint pdf is that it must integrate to 1 over the entire space of possible \(x\) and \(y\) values, ensuring that we account for all possible outcomes. If \(X\) and \(Y\) are independent, the joint pdf is simply the product of their individual probability density functions. This is expressed as:
  • \( f(x, y) = f(x) \cdot f(y) \)
This assumes each variable operates without affecting the other, an assumption that simplifies calculations significantly.
Independent Variables
When two random variables are independent, knowing the value of one does not give us any information about the value of the other.
Independence is a fundamental concept that simplifies many probability problems. In the context of probability density functions, if two variables are independent, the joint pdf can be expressed as the product of the marginal pdfs of each variable. For exponential random variables \(X\) and \(Y\) with parameter \(\lambda = 1\):
  • Each pdf is \( f(x) = e^{-x} \) and \( f(y) = e^{-y} \)
  • The joint pdf is \( f(x, y) = e^{-(x+y)} \)
This relationship means the complexity of finding probabilities is reduced, as calculations become straightforward multiplications.
Integration in Probability
Integration in probability is used to find the probabilities or expectations of continuous random variables over a given range. This involves using the probability density function to integrate over the desired domain.
When calculating probabilities with a joint pdf across a region, such as \(P(X + Y \leq \, 2)\), we perform double integration over the designated area. This process involves integrating the joint pdf \(f(x, y)\) across the region defined by the different values of \(X\) and \(Y\) that satisfy certain conditions. For example:
  • The inner integral calculates the area with respect to \(y\), given \(x\).
  • The outer integral sums up across various \(x\) values.
This method converts complex probability questions about continuous distributions into manageable calculus problems.
Exponential Random Variable
An exponential random variable is used to model the time between events in a Poisson process, which is characterized by a constant average rate. For example, it is often used to model the lifespan of products such as lightbulbs.
The probability density function (pdf) for an exponential distribution with a rate parameter \(\lambda\) is given by:
  • \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \)
In this scenario, where \(\lambda = 1\) for both lightbulbs, the pdf simplifies to \( f(x) = e^{-x} \). Exponential distributions have the "memoryless" property, meaning the probability of an event occurring in the next period is independent of how much time has already passed.
Understanding this concept can be very useful when analyzing processes where events happen continuously and independently at a constant average rate.

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Most popular questions from this chapter

Suppose the expected tensile strength of type-A steel is 105 \(\mathrm{ksi}\) and the standard deviation of tensile strength is \(8 \mathrm{ksi}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \(100 \mathrm{ksi}\) and \(6 \mathrm{ksi}\), respectively. Let \(\bar{X}=\) the sample average tensile strength of a random sample of 40 type-A specimens, and let \(\bar{Y}=\) the sample average tensile strength of a random sample of 35 type-B specimens. a. What is the approximate distribution of \(\bar{X}\) ? Of \(\bar{Y}\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) \(P(-1 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate \(P(\bar{X}-\bar{Y} \geq 10)\). If you actually observed \(\bar{X}-\bar{Y} \geq 10\), would you doubt that \(\mu_{1}-\mu_{2}=5 ?\)

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