Requirement: Provide a 96% confidence interval for . I used numerical integration (e.g., Monte Carlo integration) to approximate the value of I. Here is how I could use Monte Carlo integration: Generate a large number of random samples (xi) from a suitable distribution (e.g., a normal distribution with a large standard deviation). Calculate the function values for each sample. Estimate the integral by taking the average of these function values. Repeat the process multiple times to get a sample of integral estimates. import numpy as np import scipy.integrate as integrate def integrand(x): return np.exp(-x**2) * np.abs(np.sin(x)) # Numerical integration using scipy.integrate.quad result, error = integrate.quad(integrand, -np.inf, np.inf) print(f"Numerical integration result: {result}") # Monte Carlo integration def monte_carlo_integral(num_samples): samples = np.random.normal(0, 5, num_samples) # adjust standard deviation as needed function_values = integrand(samples) return np.mean(function_values)*np.sqrt(2*np.pi)*5 # adjust multiplier based on standard deviation. num_iterations = 1000 num_samples = 10000 estimates = [monte_carlo_integral(num_samples) for i in range(num_iterations)] mean_estimate = np.mean(estimates) std_estimate = np.std(estimates) print(f"Monte Carlo mean estimate: {mean_estimate}") print(f"Monte Carlo standard deviation estimate: {std_estimate}") Currently, I had a sample of estimates for I. I could use this sample to calculate a confidence interval. Since I had a reasonably large number of iterations (1000), I could use the normal distribution to approximate the confidence interval. Confidence level: 96%. z-score for 96% confidence: Find the z-score corresponding to (1 + 0.96) / 2 = 0.98. Using a z-table or calculator, the z-score is approximately 2.05. Margin of error (E): Confidence interval: (mean_estimate - E, mean_estimate + E) z_score = 2.05 margin_of_error = z_score * (std_estimate / np.sqrt(num_iterations)) confidence_interval = (mean_estimate - margin_of_error, mean_estimate + margin_of_error) print(f"96% Confidence Interval: {confidence_interval}") At the moment, I don't know how to solve this problem mathematically, so I hope anyone can support me to analyze it. Continue reading...