Problem 21
Question
In each of Exercises \(17-22,\) the probability density function \(f\) of a random variable \(X\) with range \(I=[a, b]\) is given. Calculate \(P(\alpha \leq X \leq \beta)\) for the given subinterval \(J=[\alpha, \beta]\) of \(I .\) $$ f(x)=e^{1-x} /(e-1) \quad I=[0,1] \quad J=[0,1 / 2] $$
Step-by-Step Solution
Verified Answer
Probability is \( \frac{e - e^{0.5}}{e-1} \).
1Step 1: Understand the Probability Density Function (PDF)
The given probability density function (PDF) is \( f(x) = \frac{e^{1-x}}{e-1} \) and it is defined on the interval \( I = [0, 1] \). This function describes how the probability is distributed over the interval \( I \).
2Step 2: Identify the Subinterval of Interest
The subinterval \( J \) for which we need to calculate the probability is \( [0, \frac{1}{2}] \). This means we are interested in finding \( P(0 \leq X \leq \frac{1}{2}) \).
3Step 3: Set Up the Probability Calculation
To find \( P(0 \leq X \leq \frac{1}{2}) \), we need to compute the integral of the PDF \( f(x) \) over the subinterval \([0, \frac{1}{2}]\). The probability is given by the integral: \[ P(0 \leq X \leq \frac{1}{2}) = \int_{0}^{1/2} \frac{e^{1-x}}{e-1} \, dx \].
4Step 4: Evaluate the Integral
First, find the antiderivative of \( \frac{e^{1-x}}{e-1} \). We have:\[ \int e^{1-x} \, dx = -e^{1-x} + C \]Now, evaluate the definite integral:\[ \int_{0}^{1/2} \frac{e^{1-x}}{e-1} \, dx = \left[ -\frac{e^{1-x}}{e-1} \right]_{0}^{1/2} = \left( -\frac{e^{1-1/2}}{e-1} + \frac{e^{1-0}}{e-1} \right) \].
5Step 5: Compute the Probability
Substitute the values into the expression:\[ \left( -\frac{e^{1-1/2}}{e-1} + \frac{e^{1-0}}{e-1} \right) = \left( -\frac{e^{0.5}}{e-1} + \frac{e}{e-1} \right) \].Calculate the result:\[ \frac{e - e^{0.5}}{e-1} \] is the probability that \( X \) falls in the interval \([0, \frac{1}{2}]\).
6Step 6: Simplify and Finalize the Calculation
Simplify the expression as necessary. The calculated probability represents the area under the curve of the PDF from 0 to 0.5. With the evaluations completed, this integral provides the exact probability for the specified subinterval.
Key Concepts
Integral CalculusProbability DistributionDefinite Integral
Integral Calculus
Integral calculus is a fundamental aspect of calculus that focuses on accumulation and finding the area under curves. When working with probability density functions (PDFs), integral calculus allows us to determine probabilities over specific intervals.
In terms of PDFs, a probability is represented by the integral of the function over a given interval. The integral, in this context, sums up all the infinitesimal probabilities (areas) over that interval. This effectively gives us the total probability of a random variable, say \(X\), falling within a certain range.
The definite integral for this probability calculation is set up as:
In terms of PDFs, a probability is represented by the integral of the function over a given interval. The integral, in this context, sums up all the infinitesimal probabilities (areas) over that interval. This effectively gives us the total probability of a random variable, say \(X\), falling within a certain range.
The definite integral for this probability calculation is set up as:
- The integral of the PDF over the interval of interest provides the probability.
- Use the limits of integration to define the start and finish of the interval for the variable.
Probability Distribution
Probability distributions are statistical functions that describe all the possible values and likelihoods that a random variable can take within a given range. A probability density function (PDF) is a type of probability distribution.
The PDF gives us a formula to compute the likelihood of a continuous random variable assuming a specific value. However, since the probability of a continuous random variable assuming an exact value is almost always zero, we look at intervals. The PDF allows us to model and understand how values are distributed across intervals.
The PDF gives us a formula to compute the likelihood of a continuous random variable assuming a specific value. However, since the probability of a continuous random variable assuming an exact value is almost always zero, we look at intervals. The PDF allows us to model and understand how values are distributed across intervals.
- Provides the probability that a value falls within a certain range.
- The total area under the PDF curve across all possible values equals 1, representing total certainty.
Definite Integral
A definite integral is a key concept in calculus used to compute the area under the curve of a function across a specific interval. In probability, definite integrals allow for the precise calculation of probabilities over intervals.
The definite integral involves:
The definite integral involves:
- Identifying limits of integration to demarcate the interval of interest.
- Establishing the integral of the function, representing our PDF.
- Evaluating the integral using antiderivatives, which are expressions that describe the accumulated area when integrated.
- Performing a subtraction of the evaluated upper and lower bounds of the integral to find the total area (probability).
Other exercises in this chapter
Problem 20
Calculate the length of the given parametric curve. $$ x=\sin \left(\pi t^{2}\right) \quad y=\cos \left(\pi t^{2}\right) \quad 0 \leq t \leq 1 $$
View solution Problem 21
In each of Exercises \(17-28,\) solve the given initial value problem.\ $$ \frac{d y}{d x}+\frac{2}{x} y=7 \sqrt{x} \quad y(4)=17 $$
View solution Problem 21
Find the solution of the given initial value problem. $$ y^{\prime}(x)=\cos (x) \quad y(0)=2 $$
View solution Problem 21
Calculate the length of the given parametric curve. $$ x=3 t^{2} \quad y=2 t^{3} \quad 0 \leq t \leq 1 $$
View solution