Let's solve each step by step. --- ## 1. Find the value of \( c \) Since \( P(X = k) \) is a probability mass function (pmf), the total probability must sum to 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] First, let's simplify \( \frac{1}{k(k+1)} \): \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] So, \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = c \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] This is a telescoping series: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 \] So, \[ c \cdot 1 = 1 \implies c = 1 \] --- ## 2. Find \( E(X) \) The expectation is: \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] \[ = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This is the harmonic series starting from \( k = 2 \): \[ \sum_{k=1}^{\infty} \frac{1}{k+1} = \sum_{m=2}^{\infty} \frac{1}{m} \] This sum **diverges** (goes to infinity). Thus, \[ E(X) = \infty \] --- ## **Summary** 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Solution Let's analyze the given discrete random variable \( X \) step by step. --- ## 1. Find the value of \( c \) Given the probability mass function: \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] To determine \( c \), we must ensure the total probability sums to 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] This becomes: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] We can simplify \( \frac{1}{k(k+1)} \): \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] Thus, the sum can be restructured as: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] This is a telescoping series: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 \] Therefore, we have: \[ c \cdot 1 = 1 \implies c = 1 \] --- ## 2. Find \( E(X) \) The expected value \( E(X) \) is defined as: \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] This simplifies to: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This series starts at \( k = 2 \): \[ \sum_{k=1}^{\infty} \frac{1}{k+1} = \sum_{m=2}^{\infty} \frac{1}{m} \] This series diverges, indicating that: \[ E(X) = \infty \] --- ## **Summary** 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information Let \( X \) be a discrete random variable with the probability mass function: \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] ## What We Need to Find 1. The value of \( c \). 2. The expected value \( E(X) \). ## Definitions/Concepts Used - **Probability Mass Function (PMF):** A function that gives the probability of each possible value of a discrete random variable. - **Expected Value:** The long-term average or mean of a random variable, calculated as \( E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) \). --- ## Step-by-Step Solution To find the value of \( c \), we need to ensure that the total probability sums to 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] Substituting the given PMF: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] We simplify \( \frac{1}{k(k+1)} \) using partial fractions: \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This transforms our sum into a telescoping series: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] The telescoping series simplifies to: \[ 1 \] Thus, we have: \[ c \cdot 1 = 1 \implies c = 1 \] Next, we calculate the expected value \( E(X) \): \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] This simplifies to: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This series diverges, as it is equivalent to the harmonic series starting from \( k = 2 \): \[ \sum_{m=2}^{\infty} \frac{1}{m} \] Thus, we conclude that: \[ E(X) = \infty \] --- ## Summary of Final Answers 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information Let \( X \) be a discrete random variable defined by its probability mass function (PMF): \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] ## What We Need to Find 1. The value of \( c \). 2. The expected value \( E(X) \). ## Definitions/Concepts Used - **Probability Mass Function (PMF):** A function that defines the probability of a discrete random variable taking on specific values. - **Expected Value:** The mean of a random variable, calculated as \( E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) \). --- ## Step-by-Step Solution To find the value of \( c \), we start with the requirement that the total probability must equal 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] Substituting the PMF into the equation gives: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] Next, we simplify \( \frac{1}{k(k+1)} \) using partial fraction decomposition: \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This leads to: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] This series is telescoping and simplifies to: \[ 1 \] Thus, we have: \[ c \cdot 1 = 1 \implies c = 1 \] Next, we calculate the expected value \( E(X) \): \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] This simplifies to: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] The series diverges, as it is analogous to the harmonic series, expressed as: \[ \sum_{m=2}^{\infty} \frac{1}{m} \] Thus, we conclude: \[ E(X) = \infty \] --- ## Summary of Final Answers 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information Let \( X \) be a discrete random variable with the probability mass function (PMF): \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] ## What We Need to Find 1. The constant \( c \). 2. The expected value \( E(X) \). ## Definitions/Concepts Used - **Probability Mass Function (PMF):** A function that assigns probabilities to each possible value of a discrete random variable. - **Expected Value:** The average value of a random variable, calculated as \( E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) \). --- ## Step-by-Step Solution To determine the value of \( c \), we need to ensure that the total probability sums to 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] Substituting the PMF into this equation gives: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] Next, we can simplify \( \frac{1}{k(k+1)} \) through partial fractions: \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This results in a telescoping series: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] The telescoping nature of this series simplifies to: \[ 1 \] Thus, we find: \[ c \cdot 1 = 1 \implies c = 1 \] Next, we will compute the expected value \( E(X) \): \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] This simplifies to: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This series diverges, as it corresponds to the harmonic series starting from \( k = 2 \): \[ \sum_{m=2}^{\infty} \frac{1}{m} \] Hence, we conclude that: \[ E(X) = \infty \] --- ## Summary of Final Answers 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information Let \( X \) be a discrete random variable with the probability mass function (PMF): \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] ## What We Need to Find 1. The constant \( c \). 2. The expected value \( E(X) \). ## Definitions/Concepts Used - **Probability Mass Function (PMF):** A function that describes the probability of each discrete outcome for a random variable. - **Expected Value:** The mean of a random variable, calculated as \( E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) \). --- ## Step-by-Step Solution To find the value of \( c \), we begin with the requirement that the total probability must equal 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] Substituting the PMF into this equation gives: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] Next, we simplify the term \( \frac{1}{k(k+1)} \) using partial fractions: \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This transformation leads to a telescoping series: \[ \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) \] The telescoping nature of this series simplifies as follows: - The first few terms cancel each other out, leading to: \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots = 1 \] Thus, we conclude: \[ c \cdot 1 = 1 \implies c = 1 \] Next, we compute the expected value \( E(X) \): \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] This expression simplifies to: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This series represents the harmonic series starting from \( k = 2 \): \[ \sum_{m=2}^{\infty} \frac{1}{m} \] The harmonic series diverges, thus leading to: \[ E(X) = \infty \] --- ## Summary of Final Answers 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information Let \( X \) be a discrete random variable defined by the probability mass function (PMF): \[ P(X = k) = \frac{c}{k(k+1)} \quad \text{for } k = 1, 2, \ldots \] ## What We Need to Find 1. The constant \( c \). 2. The expected value \( E(X) \). ## Definitions/Concepts Used - **Probability Mass Function (PMF):** A function that gives the probability of a discrete random variable taking on specific values. - **Expected Value:** The average value of a random variable, calculated as \( E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) \). --- ## Step-by-Step Solution To determine the value of \( c \), we need to ensure that the total probability sums to 1: \[ \sum_{k=1}^{\infty} P(X = k) = 1 \] Substituting the PMF into this equation results in: \[ \sum_{k=1}^{\infty} \frac{c}{k(k+1)} = 1 \] Next, we simplify \( \frac{1}{k(k+1)} \) using the method of partial fractions: \[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \] This leads to the summation: \[ \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right) \] This series is telescoping: - The terms cancel sequentially, yielding: \[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots \] The sum converges to 1: \[ \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 \] Thus, we have: \[ c \cdot 1 = 1 \implies c = 1 \] Next, we compute the expected value \( E(X) \): \[ E(X) = \sum_{k=1}^{\infty} k \cdot P(X = k) = \sum_{k=1}^{\infty} k \cdot \frac{1}{k(k+1)} \] Simplifying further: \[ E(X) = \sum_{k=1}^{\infty} \frac{1}{k+1} \] This series diverges, as it corresponds to the harmonic series starting from \( k = 2 \): \[ \sum_{m=2}^{\infty} \frac{1}{m} \] Thus, we conclude: \[ E(X) = \infty \] --- ## Summary of Final Answers 1. **Value of \( c \):** \( \boxed{1} \) 2. **Expected value \( E(X) \):** \( \boxed{\infty} \) (diverges)

# Given Information - Heat removed from the cold chamber, \( Q_c = 800 \, \text{kJ/min} \) - Power consumed by the refrigerator, \( W = 5 \, \text{kW} \) ## What We Need to Find - The Coefficient of Performance (COP) of the refrigerator. ## Definitions/Concepts Used - **Coefficient of Performance (COP):** For a refrigerator, the COP is defined as: \[ \text{COP} = \frac{Q_c}{W} \] where \( Q_c \) is the heat removed from the cold space and \( W \) is the work input (power consumed). --- ## Step-by-Step Solution 1. **Convert Heat Removed to kW:** Since the power consumed is in kW, we should convert \( Q_c \) from kJ/min to kW: \[ Q_c = 800 \, \text{kJ/min} = \frac{800 \, \text{kJ}}{60 \, \text{s}} = \frac{800}{60} \, \text{kW} \approx 13.33 \, \text{kW} \] 2. **Calculate COP:** Using the COP formula: \[ \text{COP} = \frac{Q_c}{W} = \frac{13.33 \, \text{kW}}{5 \, \text{kW}} = 2.666 \] --- ## Summary of Final Answer - **COP of the Refrigerator:** \( \boxed{2.67} \) (rounded to two decimal places)

# Given Information A national economic research organization studies annual income in three metropolitan sectors: - **Urban Core (U)**: 50% of the employed population - **Suburban Belt (S)**: 30% of the employed population - **Industrial Zone (I)**: 20% of the employed population Annual incomes are normally distributed in each sector: - \( Y_U \sim N(\mu_U, \sigma_U^2) \) - \( Y_S \sim N(\mu_S, \sigma_S^2) \) - \( Y_I \sim N(\mu_I, \sigma_I^2) \) ## Means The means for each sector are calculated as follows: \[ \mu_U = 70 + 15 - 5 = 80 \quad (\text{in thousands of dollars}) \] \[ \mu_S = 60 + 20 - 10 = 70 \quad (\text{in thousands of dollars}) \] \[ \mu_I = 50 + 25 - 5 = 70 \quad (\text{in thousands of dollars}) \] ## Variances The variances are given as: - \( \sigma_U^2 = 10^2 \) - \( \sigma_S^2 = 12^2 \) - \( \sigma_I^2 = 8^2 \) ## What We Need to Find - The expected annual income \( E(Y) \) of a randomly selected employed individual from the entire population. ## Definitions/Concepts Used - **Expected Value of a Mixture Distribution:** The expected value for a mixture distribution can be computed as the weighted average of the expected values of the individual distributions. --- ## Step-by-Step Solution To find the expected income \( E(Y) \): 1. **Calculate the expected income for each sector:** - For Urban Core: \[ E(Y_U) = \mu_U = 80 \, \text{(thousands of dollars)} \] - For Suburban Belt: \[ E(Y_S) = \mu_S = 70 \, \text{(thousands of dollars)} \] - For Industrial Zone: \[ E(Y_I) = \mu_I = 70 \, \text{(thousands of dollars)} \] 2. **Combine the expected values using the population proportions:** - The expected annual income \( E(Y) \) is given by: \[ E(Y) = 0.5 \cdot E(Y_U) + 0.3 \cdot E(Y_S) + 0.2 \cdot E(Y_I) \] Substituting the expected values: \[ E(Y) = 0.5 \cdot 80 + 0.3 \cdot 70 + 0.2 \cdot 70 \] 3. **Calculate the total:** \[ E(Y) = 40 + 21 + 14 = 75 \, \text{(thousands of dollars)} \] --- ## Summary of Final Answer - **Expected Annual Income of a Randomly Selected Employed Individual:** \( \boxed{75} \) (thousands of dollars)