## Given Information - US Politics News: 26% - World News: 23% ## Have to Find - The difference between the percentage of US Politics News and World News. ## Definition or Concept Used To find the difference between two percentages, subtract the smaller percentage from the larger percentage. ## Step by Step Solution US Politics News = 26% World News = 23% Difference = 26% - 23% Difference = 3% ## Final Answers Only as Summary The difference between US Politics News (26%) and World News (23%) is **3%**.

# Given Information - **US Politics News**: 26% - **World News**: 23% # Have to Find - The difference between the percentages of US Politics News and World News. # Definition or Concept Used To determine the difference between two percentages, we apply the formula: \[ \text{Difference} = \text{Larger Percentage} - \text{Smaller Percentage} \] This concept is fundamental in comparative analysis, allowing us to quantify the gap between two data points. # Step by Step Solution 1. Identify the percentages of interest: - US Politics News: 26% - World News: 23% 2. Apply the formula for difference: - Since 26% (US Politics News) is greater than 23% (World News), we will subtract the latter from the former. 3. Perform the subtraction: \[ \text{Difference} = 26\% - 23\% = 3\% \] This calculation reveals how much more significant the US Politics News category is compared to World News. # Final Answers Only as Summary The difference between US Politics News (26%) and World News (23%) is **3%**.

# Given Information - **US Politics News**: 26% - **World News**: 23% # Have to Find - The difference between the percentages of US Politics News and World News. # Definition or Concept Used The difference between two percentages can be calculated by: \[ \text{Difference} = \text{Percentage A} - \text{Percentage B} \] This formula helps in understanding the comparative size of two data categories in a dataset. # Detailed Step by Step Solution - **Identify the values**: - US Politics News = 26% - World News = 23% - **Determine the larger and smaller percentages**: - Larger Percentage: US Politics News (26%) - Smaller Percentage: World News (23%) - **Perform the subtraction**: \[ \text{Difference} = 26\% - 23\% \] - **Calculate the result**: \[ \text{Difference} = 3\% \] This calculation indicates that US Politics News has a higher representation in the dataset compared to World News by 3%. # Final Answers Only as Summary The difference between US Politics News (26%) and World News (23%) is **3%**.

# Given Information - **US Politics News**: 26% - **World News**: 23% # Have to Find - The difference between the percentages of US Politics News and World News. # Definition or Concept Used To compute the difference between two percentages, we can use the following formula: \[ \text{Difference} = \text{Percentage A} - \text{Percentage B} \] This method allows us to assess the relative prominence of one category compared to another. # Detailed Step by Step Solution - **Define the Categories**: - Percentage of US Politics News: 26% - Percentage of World News: 23% - **Identify which is larger**: - US Politics News is greater than World News. - **Apply the formula**: - Using the identified values in the formula, we compute the difference: \[ \text{Difference} = 26\% - 23\% \] - **Execute the calculation**: - Performing the subtraction yields: \[ \text{Difference} = 3\% \] This indicates that US Politics News has a 3% higher representation in the news dataset compared to World News. # Final Answers Only as Summary The difference between US Politics News (26%) and World News (23%) is **3%**.

# Given Information - The mean time between router failures: \( \mu = 12 \) hours - The random variable \( T \): time (in hours) until the next router failure occurs # Have to Find - The probability that the next router failure occurs more than \( 18 \) hours after the previous failure. # Definition or Concept Used The time between failures follows an exponential distribution, characterized by the probability density function: \[ f(t) = \lambda e^{-\lambda t} \] where \( \lambda \) is the rate parameter, defined as: \[ \lambda = \frac{1}{\mu} \] For our case, \( \mu = 12 \) hours, thus: \[ \lambda = \frac{1}{12} \text{ hours}^{-1} \] The cumulative distribution function (CDF) for the exponential distribution is given by: \[ P(T \leq t) = 1 - e^{-\lambda t} \] To find the probability that the time until the next failure is more than \( t \) hours, we calculate: \[ P(T > t) = 1 - P(T \leq t) = e^{-\lambda t} \] # Step by Step Solution - **Identify parameters**: - Mean \( \mu = 12 \) hours - Rate \( \lambda = \frac{1}{12} \) hours\(^{-1}\) - Time \( t = 18 \) hours - **Calculate the probability**: \[ P(T > 18) = e^{-\lambda \cdot 18} \] Substituting \( \lambda \): \[ P(T > 18) = e^{-\frac{1}{12} \cdot 18} \] - **Simplify the exponent**: \[ P(T > 18) = e^{-1.5} \] - **Calculate \( e^{-1.5} \)**: Using a calculator or tables: \[ P(T > 18) \approx 0.2231 \] # Final Answers Only as Summary The probability that the next router failure occurs more than \( 18 \) hours after the previous failure is approximately **0.2231** or **22.31%**.

# Given Information - Number of servers: \( n = 40 \) - Probability of failure per server in 12-hour period: \( p = 0.05 \) - Random variable \( X \): number of servers that fail during the 12-hour test period # Have to Find - The probability that no more than four servers experience failure during the test period, i.e., \( P(X \leq 4) \). # Definition or Concept Used Since we have a large number of trials (40 servers) and a small probability of success (failure), the binomial distribution can be approximated by a Poisson distribution. The Poisson parameter \( \lambda \) is given by: \[ \lambda = n \cdot p \] The probability mass function for the Poisson distribution is: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] To find \( P(X \leq 4) \), we will compute: \[ P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \] # Step by Step Solution - **Calculate \( \lambda \)**: \[ \lambda = 40 \cdot 0.05 = 2 \] - **Calculate individual probabilities** using the Poisson formula: \[ P(X = k) = \frac{2^k e^{-2}}{k!} \] 1. For \( k = 0 \): \[ P(X = 0) = \frac{2^0 e^{-2}}{0!} = e^{-2} \approx 0.1353 \] 2. For \( k = 1 \): \[ P(X = 1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2} \approx 0.2707 \] 3. For \( k = 2 \): \[ P(X = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} \approx 0.2707 \] 4. For \( k = 3 \): \[ P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8e^{-2}}{6} \approx 0.1804 \] 5. For \( k = 4 \): \[ P(X = 4) = \frac{2^4 e^{-2}}{4!} = \frac{16e^{-2}}{24} \approx 0.0902 \] - **Sum the probabilities**: \[ P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \] \[ P(X \leq 4) \approx 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 \approx 0.9473 \] # Final Answers Only as Summary The probability that the server cluster will satisfy the operational stability requirement (i.e., no more than four servers experience failure) is approximately **0.9473** or **94.73%**.

# Given Information - Traffic volume (\( x \)): 28,000 vehicles (or 28 in thousands) - Regression equation: \[ \hat{y} = 6.5 + 0.42x \] where \( \hat{y} \) represents the predicted average travel time in minutes. # Have to Find - The predicted average travel time for a traffic volume of 28,000 vehicles. # Definition or Concept Used The regression equation allows us to predict the dependent variable (\( \hat{y} \), average travel time) based on the independent variable (\( x \), traffic volume). By substituting the value of \( x \) into the equation, we can calculate \( \hat{y} \). # Detailed Step by Step Solution - **Convert the traffic volume** to thousands: \[ x = 28,000 \text{ vehicles} = 28 \text{ (in thousands)} \] - **Substitute \( x \) into the regression equation**: \[ \hat{y} = 6.5 + 0.42 \cdot 28 \] - **Calculate the product**: \[ 0.42 \cdot 28 = 11.76 \] - **Add to the constant term**: \[ \hat{y} = 6.5 + 11.76 = 18.26 \] Thus, the predicted average travel time for a day with a traffic volume of 28,000 vehicles is approximately 18.26 minutes. # Final Answers Only as Summary The predicted average travel time for a traffic volume of 28,000 vehicles is **18.26 minutes**.

# Given Information - Score of the child: \( X = 200 \) - Mean of the test scores: \( \mu = 100 \) - Standard deviation of the test scores: \( \sigma = 15 \) # Have to Find - The deviation IQ of the child. # Definition or Concept Used The deviation IQ is calculated using the formula: \[ \text{Deviation IQ} = 100 + 15 \times \left( \frac{X - \mu}{\sigma} \right) \] This formula standardizes the score based on the mean and standard deviation of the test scores. # Step by Step Solution - **Substitute the values into the formula**: \[ \text{Deviation IQ} = 100 + 15 \times \left( \frac{200 - 100}{15} \right) \] - **Calculate the difference**: \[ 200 - 100 = 100 \] - **Divide by the standard deviation**: \[ \frac{100}{15} \approx 6.67 \] - **Multiply by 15**: \[ 15 \times 6.67 \approx 100.05 \] - **Add to 100**: \[ \text{Deviation IQ} \approx 100 + 100.05 \approx 200.05 \] # Final Answers Only as Summary The deviation IQ of the child is approximately **200.05**.

# Given Information - Probability of positive return on a trading day: \( p = 0.55 \) - Probability of negative return on a trading day: \( 1 - p = 0.45 \) - Total number of trading days: \( n = 10 \) - Random variable \( X \): number of days with positive returns # Have to Find - The probability that the portfolio has positive returns on at least 7 out of 10 trading days, i.e., \( P(X \geq 7) \). # Definition or Concept Used The number of days with positive returns can be modeled using the binomial distribution. The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) successes in \( n \) trials, - \( p \) is the probability of success, - \( (1-p) \) is the probability of failure. To find \( P(X \geq 7) \), we calculate: \[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \] # Step by Step Solution 1. **Calculate individual probabilities** for \( X = 7, 8, 9, 10 \): - For \( k = 7 \): \[ P(X = 7) = \binom{10}{7} (0.55)^7 (0.45)^{3} \] \[ = 120 \cdot (0.55)^7 \cdot (0.45)^3 \approx 120 \cdot 0.0197 \cdot 0.0911 \approx 0.2162 \] - For \( k = 8 \): \[ P(X = 8) = \binom{10}{8} (0.55)^8 (0.45)^{2} \] \[ = 45 \cdot (0.55)^8 \cdot (0.45)^2 \approx 45 \cdot 0.0108 \cdot 0.2025 \approx 0.0983 \] - For \( k = 9 \): \[ P(X = 9) = \binom{10}{9} (0.55)^9 (0.45)^{1} \] \[ = 10 \cdot (0.55)^9 \cdot (0.45)^{1} \approx 10 \cdot 0.0059 \cdot 0.45 \approx 0.0265 \] - For \( k = 10 \): \[ P(X = 10) = \binom{10}{10} (0.55)^{10} (0.45)^{0} \] \[ = 1 \cdot (0.55)^{10} \approx 0.0032 \] 2. **Sum the probabilities**: \[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \] \[ P(X \geq 7) \approx 0.2162 + 0.0983 + 0.0265 + 0.0032 \approx 0.3442 \] # Final Answers Only as Summary The probability that the portfolio produces positive returns on at least seven of the ten trading days is approximately **0.3442** or **34.42%**.