# Given Information - The probability density function (PDF) of a random variable \( X \) is: \[ f(x) = \begin{cases} kx, & \leq x \leq 2 \\ , & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used - The total area under the PDF over its domain must be 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) falls within an interval \([a, b]\) is: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its domain and set the result to 1: \[ \int_^2 kx \, dx = 1 \] \[ k \int_^2 x \, dx = 1 \] \[ k \left[ \frac{x^2}{2} \right]_^2 = 1 \] \[ k \left( \frac{2^2}{2} - \frac{^2}{2} \right) = 1 \] \[ k \left( \frac{4}{2} \right) = 1 \] \[ k \cdot 2 = 1 \] \[ k = \frac{1}{2} \] To find \( P(1 \leq x \leq 2) \), integrate the PDF over \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substitute \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2}x \, dx \] \[ = \frac{1}{2} \int_1^2 x \, dx \] \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is: \[ f(x) = \begin{cases} k(2 - x), & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(0 \leq x \leq 1) \). # Definition or Concept Used - The total area under the PDF over its domain must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) falls within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its defined domain and set the result to 1: \[ \int_0^2 k(2 - x) \, dx = 1 \] \[ k \int_0^2 (2 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 2x - \frac{x^2}{2} \right]_0^2 \] \[ = k \left( 2(2) - \frac{2^2}{2} - (0) \right) \] \[ = k \left( 4 - 2 \right) \] \[ = k \cdot 2 = 1 \] Solving for \( k \): \[ k = \frac{1}{2} \] Next, to find \( P(0 \leq x \leq 1) \), integrate the PDF over \([0, 1]\): \[ P(0 \leq x \leq 1) = \int_0^1 k(2 - x) \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_0^1 \frac{1}{2}(2 - x) \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ 2x - \frac{x^2}{2} \right]_0^1 \] \[ = \frac{1}{2} \left( 2(1) - \frac{1^2}{2} - (0) \right) \] \[ = \frac{1}{2} \left( 2 - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(0 \leq x \leq 1) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} k(3 - x), & 0 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - Determine the value of \( k \). - Calculate the probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used - The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] - The probability that \( X \) lies within the interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution Begin by determining \( k \) by integrating \( f(x) \) over its valid range and setting the result equal to 1: \[ \int_0^3 k(3 - x) \, dx = 1 \] This leads to: \[ k \int_0^3 (3 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 3x - \frac{x^2}{2} \right]_0^3 \] Evaluating the boundaries: \[ = k \left( 3(3) - \frac{3^2}{2} - (0) \right) \] \[ = k \left( 9 - \frac{9}{2} \right) \] \[ = k \left( \frac{18}{2} - \frac{9}{2} \right) = k \cdot \frac{9}{2} \] Setting this equal to 1 gives: \[ k \cdot \frac{9}{2} = 1 \] Solving for \( k \): \[ k = \frac{2}{9} \] Next, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 k(3 - x) \, dx \] Substituting \( k = \frac{2}{9} \): \[ = \int_1^2 \frac{2}{9}(3 - x) \, dx \] Calculating the integral: \[ = \frac{2}{9} \left[ 3x - \frac{x^2}{2} \right]_1^2 \] Evaluating at the boundaries: \[ = \frac{2}{9} \left( 3(2) - \frac{2^2}{2} - (3(1) - \frac{1^2}{2}) \right) \] \[ = \frac{2}{9} \left( 6 - 2 - (3 - \frac{1}{2}) \right) \] \[ = \frac{2}{9} \left( 4 - 2.5 \right) \] \[ = \frac{2}{9} \cdot 1.5 \] \[ = \frac{3}{9} = \frac{1}{3} \] # Summary - Value of \( k \): \( \frac{2}{9} \) - \( P(1 \leq x \leq 2) = \frac{1}{3} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} k(3 - x), & 0 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - Determine the value of \( k \). - Calculate the probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within the interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate \( f(x) \) over its valid range and set the result equal to 1: \[ \int_0^3 k(3 - x) \, dx = 1 \] This simplifies to: \[ k \int_0^3 (3 - x) \, dx = 1 \] Calculating the integral: \[ = k \left[ 3x - \frac{x^2}{2} \right]_0^3 \] Evaluating the boundaries: \[ = k \left( 3(3) - \frac{3^2}{2} \right) \] \[ = k \left( 9 - \frac{9}{2} \right) = k \left( \frac{18}{2} - \frac{9}{2} \right) = k \cdot \frac{9}{2} \] Setting this equal to 1 gives: \[ k \cdot \frac{9}{2} = 1 \] Solving for \( k \): \[ k = \frac{2}{9} \] Now, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 k(3 - x) \, dx \] Substituting \( k = \frac{2}{9} \): \[ = \int_1^2 \frac{2}{9}(3 - x) \, dx \] Calculating the integral: \[ = \frac{2}{9} \left[ 3x - \frac{x^2}{2} \right]_1^2 \] Evaluating at the boundaries: \[ = \frac{2}{9} \left( \left( 3(2) - \frac{2^2}{2} \right) - \left( 3(1) - \frac{1^2}{2} \right) \right) \] \[ = \frac{2}{9} \left( 6 - 2 - (3 - 0.5) \right) \] \[ = \frac{2}{9} \left( 4 - 2.5 \right) \] \[ = \frac{2}{9} \cdot 1.5 = \frac{3}{9} = \frac{1}{3} \] # Summary - Value of \( k \): \( \frac{2}{9} \) - \( P(1 \leq x \leq 2) = \frac{1}{3} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To determine \( k \), integrate \( f(x) \) over its valid domain and set the result equal to 1: \[ \int_0^2 kx \, dx = 1 \] Calculating the integral: \[ k \int_0^2 x \, dx = 1 \] This leads to: \[ k \left[ \frac{x^2}{2} \right]_0^2 = 1 \] Evaluating the boundaries: \[ k \left( \frac{2^2}{2} - 0 \right) = 1 \] \[ k \cdot 2 = 1 \] Solving for \( k \): \[ k = \frac{1}{2} \] Next, to find \( P(1 \leq x \leq 2) \), integrate the PDF over the interval \([1, 2]\): \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the boundaries: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is expressed as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The requirement that the total area under the PDF is equal to 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) falls within the interval \([a, b]\) can be computed as: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate the PDF over its specified range and set the area equal to 1: \[ \int_0^2 kx \, dx = 1 \] This expands to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral yields: \[ k \left[ \frac{x^2}{2} \right]_0^2 = 1 \] Substituting the limits gives: \[ k \left( \frac{2^2}{2} - 0 \right) = 1 \] \[ k \cdot 2 = 1 \] Solving for \( k \) results in: \[ k = \frac{1}{2} \] Next, to determine \( P(1 \leq x \leq 2) \), integrate the PDF over the interval from 1 to 2: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting in the value of \( k \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating this integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the limits: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) \] \[ = \frac{1}{2} \cdot \frac{3}{2} \] \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is described as: \[ f(x) = \begin{cases} kx, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The area under the PDF must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within an interval \([a, b]\) is given by: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find the value of \( k \), integrate the PDF over its valid range and equate the result to 1: \[ \int_0^2 kx \, dx = 1 \] This simplifies to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral: \[ = k \left[ \frac{x^2}{2} \right]_0^2 \] Evaluating the integral gives: \[ = k \left( \frac{2^2}{2} - 0 \right) = k \cdot 2 \] Setting this equal to 1 results in: \[ k \cdot 2 = 1 \] Thus, solving for \( k \) yields: \[ k = \frac{1}{2} \] Next, to calculate \( P(1 \leq x \leq 2) \), we integrate the PDF over the specified interval: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating the integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the integral at the boundaries: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} \] Thus, we have: \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)

# Given Information - The probability density function (PDF) of a random variable \( X \) is defined as: \[ f(x) = \begin{cases} kx, & \text{if } 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] # What To Find - The value of the constant \( k \). - The probability \( P(1 \leq x \leq 2) \). # Definition or Concept Used 1. The total area under the PDF must be equal to 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] 2. The probability that \( X \) lies within the interval \([a, b]\) is expressed as: \[ P(a \leq x \leq b) = \int_a^b f(x) \, dx \] # Solution To find \( k \), integrate the PDF across its valid range and set the resulting area equal to 1: \[ \int_0^2 kx \, dx = 1 \] This expands to: \[ k \int_0^2 x \, dx = 1 \] Calculating the integral: \[ = k \left[ \frac{x^2}{2} \right]_0^2 \] Substituting the limits gives: \[ = k \left( \frac{2^2}{2} - 0 \right) = k \cdot 2 \] Setting this equal to 1 results in: \[ k \cdot 2 = 1 \] Thus, solving for \( k \) provides: \[ k = \frac{1}{2} \] Next, to compute \( P(1 \leq x \leq 2) \), we integrate the PDF over the range from 1 to 2: \[ P(1 \leq x \leq 2) = \int_1^2 kx \, dx \] Substituting \( k = \frac{1}{2} \): \[ = \int_1^2 \frac{1}{2} x \, dx \] Calculating this integral: \[ = \frac{1}{2} \left[ \frac{x^2}{2} \right]_1^2 \] Evaluating the limits: \[ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) \] This simplifies to: \[ = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} \] Thus, we find: \[ = \frac{3}{4} \] # Summary - Value of \( k \): \( \frac{1}{2} \) - \( P(1 \leq x \leq 2) = \frac{3}{4} \)