## Given Information The probability density function (PDF) of a random variable is\[ f(x) = \begin{cases} kx & \text{if } \leq x \leq 2 \\ & \text{otherwise} \end{cases} \] We are asked to find: 1. The value of \( k \) 2. \( P(1 \leq X \leq 2) \) --- ## What To Find - The value of \( k \) so that \( f(x) \) is a valid PDF. - The probability that \( X \) lies between 1 and 2. --- ## Definition or Concept Used A probability density function must satisfy: 1. \( f(x) \geq \) for all \( x \) 2. The total area under the curve must be 1: \(\int_{-\infty}^{\infty} f(x) \, dx = 1\) To find the probability that \( X \) lies within an interval \([a, b]\), you integrate the PDF over that interval: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Step-by-Step Solution To find \( k \), use the rule that the total area under the PDF must be 1. Since \( f(x) = \) outside \( [, 2] \), integrate from to 2: \[ \int_{}^{2} kx \, dx = 1 \] Calculate the integral: \[ k \int_{}^{2} x \, dx = k \left[ \frac{x^2}{2} \right]_{}^{2} \] Substitute the limits: \[ k \left( \frac{2^2}{2} - \frac{^2}{2} \right) = k \left( \frac{4}{2} \right) = k \times 2 \] Set equal to 1: \[ k \times 2 = 1 \implies k = \frac{1}{2} \] Now, to find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) dx = \int_{1}^{2} kx \, dx \] Substitute \( k = \frac{1}{2} \): \[ \int_{1}^{2} \frac{1}{2}x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Calculate the values: \[ \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - .5 \right) = \frac{1}{2} \times 1.5 = .75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - \( P(1 \leq X \leq 2) = .75 \).

# Solution to the Problem ## Given Information The probability density function (PDF) of a random variable is defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to find: 1. The value of \( k \) 2. The probability \( P(1 \leq X \leq 2) \) --- ## What To Find - Determine the constant \( k \) such that \( f(x) \) is a valid PDF. - Calculate the probability that \( X \) falls within the interval from 1 to 2. --- ## Definition or Concept Used A probability density function must adhere to the following rules: 1. \( f(x) \geq 0 \) for all \( x \) 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability that \( X \) lies within an interval \([a, b]\), use: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find the value of \( k \), we need to ensure that the total area under the PDF equals 1. Given that \( f(x) = kx \) for \( 0 < x < 2 \), we compute the integral: \[ \int_0^2 kx \, dx = 1 \] Calculate the integral: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluate the limits: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = k \times 2 \] Set this equal to 1: \[ k \times 2 = 1 \implies k = \frac{1}{2} \] Next, to find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculate the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluate the limits: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We are provided with the probability density function (PDF) for a random variable, which is defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to determine: 1. The constant \( k \) 2. The probability \( P(1 \leq X \leq 2) \) --- ## What To Find - Find the value of \( k \) so that \( f(x) \) represents a valid probability density function. - Calculate the probability that the random variable \( X \) lies within the interval from 1 to 2. --- ## Definition or Concept Used A valid probability density function must satisfy the following conditions: 1. \( f(x) \geq 0 \) for all \( x \) 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) being in the interval \([a, b]\), use the formula: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To compute \( k \), we start by ensuring that the total area under the PDF equals 1. Since \( f(x) = kx \) is applicable for \( 0 < x < 2 \), we integrate: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral yields: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the limits gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = k \times 2 \] Setting the equation equal to 1: \[ k \times 2 = 1 \implies k = \frac{1}{2} \] Next, we find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting the value of \( k \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information The probability density function (PDF) for a random variable is expressed as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to determine: 1. The value of \( k \) 2. The probability \( P(1 \leq X \leq 2) \) --- ## What To Find - Calculate the value of \( k \) to ensure \( f(x) \) is a valid PDF. - Determine the probability that \( X \) falls within the range from 1 to 2. --- ## Definition or Concept Used A valid probability density function must meet these criteria: 1. The function must be non-negative: \( f(x) \geq 0 \) for all \( x \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) being in the interval \([a, b]\), the formula is: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find \( k \), we enforce that the total area under the PDF equals 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we compute the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral yields: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating this gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = k \times 2 \] Setting this equal to 1: \[ k \times 2 = 1 \implies k = \frac{1}{2} \] Next, we find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We have a probability density function (PDF) defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] Our goals are: 1. To find the constant \( k \). 2. To calculate the probability \( P(1 \leq X \leq 2) \). --- ## What To Find - The appropriate value of \( k \) that makes \( f(x) \) a valid PDF. - The probability that the random variable \( X \) is between 1 and 2. --- ## Definition or Concept Used A probability density function must satisfy two main conditions: 1. It must be non-negative across its range: \( f(x) \geq 0 \) for all \( x \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) being within the interval \([a, b]\), we use: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To determine \( k \), we start by ensuring the total area under the curve of \( f(x) \) is 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we calculate the integral: \[ \int_0^2 kx \, dx = 1 \] Evaluating the integral gives: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Calculating the limits results in: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting the equation equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we find the probability \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We have a probability density function (PDF) defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to find: 1. The value of \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Determine the constant \( k \) that makes \( f(x) \) a valid probability density function. - Calculate the probability that the random variable \( X \) lies in the range from 1 to 2. --- ## Definition or Concept Used A valid probability density function must satisfy the following conditions: 1. \( f(x) \geq 0 \) for all \( x \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) being in the interval \([a, b]\), the formula is: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find the value of \( k \), we need the total area under the PDF to equal 1. Since \( f(x) = kx \) is valid for \( 0 < x < 2 \), we compute the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral results in: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating this yields: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting the total area equal to 1 gives: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral gives: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating the limits results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We are dealing with a probability density function (PDF) defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] The tasks are to determine: 1. The constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Calculate the value of \( k \) that ensures \( f(x) \) is a legitimate PDF. - Compute the probability that the random variable \( X \) is within the range of 1 to 2. --- ## Definition or Concept Used For a function to be a valid probability density function, it must satisfy these criteria: 1. The function must be non-negative: \( f(x) \geq 0 \) for all \( x \). 2. The total area under the PDF curve must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability that \( X \) falls within the interval \([a, b]\), we utilize: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To determine \( k \), we first ensure that the total area under the curve of the PDF equals 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we compute the integral: \[ \int_0^2 kx \, dx = 1 \] Evaluating the integral gives: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Calculating the limits results in: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we calculate \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this expression gives: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We have a probability density function (PDF) represented as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We are tasked with finding: 1. The value of \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Determine the constant \( k \) that ensures \( f(x) \) behaves as a valid probability density function. - Calculate the probability that the random variable \( X \) is within the interval from 1 to 2. --- ## Definition or Concept Used A valid probability density function must satisfy two essential conditions: 1. The function must be non-negative for all \( x \): \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) falling within the interval \([a, b]\), we utilize the formula: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find \( k \), we need to ensure that the total area under the PDF equals 1. The integration is performed over the range \( 0 < x < 2 \): \[ \int_0^2 kx \, dx = 1 \] Evaluating the integral yields: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Calculating this gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting the total area equal to 1, we have: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we calculate \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We are presented with a probability density function (PDF) defined as follows: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] Our objectives are to find: 1. The constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Calculate the value of \( k \) that ensures \( f(x) \) is a valid probability density function. - Determine the probability that the random variable \( X \) is within the range from 1 to 2. --- ## Definition or Concept Used A valid probability density function must meet two criteria: 1. The function must be non-negative: \( f(x) \geq 0 \) for all \( x \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To calculate the probability of \( X \) falling between the limits \( a \) and \( b \), we use: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find \( k \), we first ensure that the total area under the PDF equals 1. We integrate over the interval \( 0 < x < 2 \): \[ \int_0^2 kx \, dx = 1 \] Calculating the integral gives: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the limits results in: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we calculate \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this gives: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We are provided with a probability density function (PDF) defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] Our objectives are to determine: 1. The value of the constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Calculate the value of \( k \) that ensures \( f(x) \) is a proper PDF. - Compute the probability of the random variable \( X \) falling within the interval from 1 to 2. --- ## Definition or Concept Used A valid probability density function must meet the following criteria: 1. The function must be non-negative for all \( x \) in its domain: \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability that \( X \) lies within an interval \([a, b]\), the formula used is: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find the value of \( k \), we first need to ensure that the total area under the PDF equals 1. Since \( f(x) = kx \) is valid for \( 0 < x < 2 \), we set up the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the limits gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we compute the probability \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information The probability density function (PDF) is defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to determine: 1. The constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Identify the correct value of \( k \) that makes \( f(x) \) a valid probability density function. - Calculate the probability that the random variable \( X \) lies within the interval from 1 to 2. --- ## Definition or Concept Used A valid probability density function must satisfy two primary conditions: 1. The function must be non-negative for all \( x \): \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability of \( X \) being in the interval \([a, b]\), we use: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To determine the value of \( k \), we need the total area under the PDF to equal 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we set up the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the integral gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we calculate \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this gives: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information We are given a probability density function (PDF) defined as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] The tasks are to find: 1. The value of the constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Identify the value of \( k \) that ensures \( f(x) \) is a valid PDF. - Compute the probability that the random variable \( X \) falls within the interval from 1 to 2. --- ## Definition or Concept Used To be a valid probability density function, \( f(x) \) must satisfy the following conditions: 1. The function must be non-negative for all \( x \): \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To determine the probability that \( X \) lies within the interval \([a, b]\), the formula used is: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find the value of \( k \), we start by ensuring that the total area under the PDF equals 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we set up the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral gives us: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating this integral: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Now, to find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating the limits results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information The probability density function (PDF) is specified as: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We need to find: 1. The value of the constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Determine the constant \( k \) that makes \( f(x) \) a valid PDF. - Compute the probability that the random variable \( X \) falls within the range from 1 to 2. --- ## Definition or Concept Used For a function to qualify as a valid probability density function, it must satisfy these two conditions: 1. The function must be non-negative for all \( x \): \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability that \( X \) is in the interval \([a, b]\), we use: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To calculate \( k \), we need to ensure that the total area under the PDF equals 1. Since \( f(x) = kx \) is valid for \( 0 < x < 2 \), we set up the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating this gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equation equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Now, to find \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).

# Solution to the Problem ## Given Information The probability density function (PDF) is defined as follows: \[ f(x) = \begin{cases} kx & \text{if } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \] We are tasked with determining: 1. The value of the constant \( k \). 2. The probability \( P(1 \leq X \leq 2) \). --- ## What To Find - Calculate the constant \( k \) that ensures \( f(x) \) is a valid probability density function. - Compute the probability that the random variable \( X \) falls within the interval from 1 to 2. --- ## Definition or Concept Used For a function to be a valid probability density function, it must satisfy two essential criteria: 1. The function must be non-negative across its range: \( f(x) \geq 0 \). 2. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] To find the probability that \( X \) lies within the interval \([a, b]\), we use the formula: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] --- ## Solution Steps To find \( k \), we start by ensuring that the total area under the PDF equals 1. Since \( f(x) = kx \) for \( 0 < x < 2 \), we set up the integral: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the limits gives: \[ k \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{4}{2} \right) = 2k \] Setting this equal to 1: \[ 2k = 1 \implies k = \frac{1}{2} \] Next, we compute \( P(1 \leq X \leq 2) \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ P(1 \leq X \leq 2) = \int_{1}^{2} \frac{1}{2} x \, dx = \frac{1}{2} \int_{1}^{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} \] Evaluating this results in: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( 2 - 0.5 \right) = \frac{1}{2} \times 1.5 = 0.75 \] --- ## Summary (Final Answers) - The value of \( k \) is \( \frac{1}{2} \). - The probability \( P(1 \leq X \leq 2) = 0.75 \).