/*! 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 38 If the density function of \(X\)... [FREE SOLUTION] | 91Ó°ÊÓ

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If the density function of \(X\) equals $$ f(x)=\left\\{\begin{array}{ll} c e^{-2 x}, & 02\\}\) ?

Short Answer

Expert verified
The constant \(c\) in the density function is 2, and the probability \(P\\{X > 2\\} = \frac{1}{e^4} \approx 0.01832\).

Step by step solution

01

Calculate the integral of the density function over its domain

To find the value of \(c\), we need to integrate the density function \(f(x)\) over its domain \([0, \infty]\) and set the result equal to 1. We know that the integral of a probability density function over its domain must be 1. So we have: \[ 1 = \int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^0 0 dx + \int_0^{\infty} c e^{-2x} dx \] As the first integral over the domain \((-\infty, 0)\) is just 0, we have that: \[ 1 = \int_0^{\infty} c e^{-2x} dx \]
02

Evaluate the integral and find the value of c

Now let's calculate the integral: \[ \int_0^{\infty} c e^{-2x} dx = -\frac{c}{2} e^{-2x} \Big|_0^{\infty} \] Evaluating at the limits, we get: \[ -\frac{c}{2}(\lim_{x \to \infty} e^{-2x} - e^0) = -\frac{c}{2}(0 - 1) \] Now we have \[ 1 = \frac{c}{2} \] So the value of \(c\) is 2. Now, we can rewrite the density function as: \[ f(x)=\left\\{\begin{array}{ll} 2 e^{-2 x}, & 0<x<\infty \\\ 0, & x<0 \end{array}\right. \]
03

Calculate the probability \(P\\{X>2\\}\)

Now that we have the density function , we can calculate the probability \(P\\{X>2\\}\): \[ P\\{X>2\\} = \int_{2}^{\infty} f(x) dx = \int_{2}^{\infty} 2 e^{-2x} dx \] Compute the integral: \[ \int_{2}^{\infty} 2 e^{-2x} dx = -e^{-2x} \Big|_2^{\infty} \] Evaluating at the limits, we get: \[ -e^{-2x} \Big|_2^{\infty} = -(\lim_{x \to \infty} e^{-2x} - e^{-4}) = -(- \frac{1}{e^4}) = \frac{1}{e^4} \] So, the probability \(P\\{X > 2\\} = \frac{1}{e^4} \approx 0.01832\).

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Key Concepts

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

Probability Density Function
A Probability Density Function (PDF) is a concept used in probability theory to describe the likelihood of a continuous random variable taking a specific value. Unlike discrete random variables, where probabilities can be assigned to individual outcomes, continuous variables require a different approach.

For continuous random variables, probabilities are defined over intervals rather than specific values. This is where the PDF comes in. It helps us understand the distribution of probabilities across a set range of values.

Here are some key points about PDFs:
  • The area under the curve of a PDF over the entire range of possible values (from \(-\infty\) to \(\infty\), for instance) is always equal to 1. This represents the total probability of all possible outcomes.
  • The probability that a variable falls within a specific range is given by the integral of the PDF over that interval.
  • The PDF can take any non-negative value, but the value itself isn't a probability. Instead, it can be thought of as a density. \(f(x)\) at a particular \(x\) tells you how likely values near \(x\) are, in a relative sense.
As seen in the original exercise, the PDF for the random variable \(X\) is \(f(x) = 2e^{-2x} \) for \(x>0\), emphasizing that probabilities related to \(X\) must be calculated over an interval using integration.
Exponential Distribution
The exponential distribution is a continuous probability distribution that is often used to model time until an event happens, like failure of a product or time between arrivals in a queue. If you think about its shape, it shows how rapidly probabilities decrease as you move further along the x-axis.

Its Probability Density Function is defined as follows: \[ f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for} \; x \geq 0 \] where \(\lambda\) is the rate parameter, which is the reciprocal of the mean. In our case, the given density function \(f(x) = 2e^{-2x} \) corresponds to \(\lambda = 2\).

**Key Characteristics of Exponential Distribution:**
  • Memoryless Property: The probability of an event occurring in a future segment of time is the same regardless of how much time has already passed.
  • Mean and Variance: For an exponential distribution with rate \(\lambda\), the mean is \(\frac{1}{\lambda}\) and the variance is \(\frac{1}{\lambda^2}\).
  • Real-world Applications: Common for modeling times between events in various contexts, like customer service systems, reliability analysis, and telecommunications.
Using this distribution, we often calculate probabilities by integrating the PDF over a certain range as seen in the solution when determining \(P\{X>2\}\).
Integration in Calculus
Integration is a fundamental concept in calculus that allows us to calculate areas under curves, among many other applications. When dealing with probability density functions, integration is crucial as it helps us determine the probability of certain outcomes over specified intervals.

In probability theory, the integral of a PDF over a given range will provide the probability that a random variable falls within that range. In mathematical terms, if \(f(x)\) is a PDF, then the probability of \(X\) lying between \([a, b]\) is represented by: \[P\{a \leq X \leq b\} = \int_a^b f(x) \, dx\]

The original step-by-step solution involved integrating the density function \(f(x) = 2e^{-2x}\) over \( [0, \infty)\) to determine the constant, \(c\), and then over \([2, \infty)\) to find \(P\{X>2\}\).

Let's look into the steps generally used in integration related to PDFs:
  • **Definite Integrals:** Calculate the integral of the function over a defined interval. This is what gives us the probability between two bounds.
  • **Finding antiderivatives:** Before evaluating a definite integral, find the antiderivative of the function. This involves determining what function would yield the given function when differentiated.
  • **Limits of Integration:** Apply the limits of the integration after finding the antiderivative by substituting the upper and lower bounds into the antiderivative and then taking the difference.
Mastering integration is indispensable for solving problems involving continuous random variables.

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