🧠:<think>Okay, so I'm trying to solve this problem about detecting adulterated honey using spectral signatures. It's divided into two parts, and I need to tackle them one by one. Let's start with the first part.Problem 1: Deriving the Mahalanobis Distance and Finding a ThresholdAlright, the first part says that genuine honey samples form a cluster approximated by a multivariate Gaussian distribution with mean vector μ and covariance matrix Σ. I need to derive the Mahalanobis distance formula to determine if a sample vector x is an outlier. Then, find a threshold distance such that only 1% of genuine samples fall outside this threshold.Hmm, I remember that the Mahalanobis distance is a measure of distance between a point and a distribution. It's like the generalization of the Euclidean distance to account for the covariance structure of the data. The formula, if I recall correctly, is:D² = (x - μ)ᵀ Σ⁻¹ (x - μ)Yes, that's the squared Mahalanobis distance. So, the distance itself would be the square root of that, but often we just use the squared version for computational simplicity.Now, to find a threshold such that only 1% of genuine samples are outside. Since the samples are from a multivariate Gaussian distribution, the Mahalanobis distance squared follows a chi-squared distribution with degrees of freedom equal to the dimensionality of the data, right?So, if we denote the squared Mahalanobis distance as D², then D² ~ χ²(d), where d is the number of dimensions.We need to find a threshold c such that P(D² > c) = 0.01. That means c is the 99th percentile of the chi-squared distribution with d degrees of freedom.Therefore, the threshold distance would be the square root of c, but since we often work with squared distances, maybe we can just use c as the threshold for D².Wait, but the question says "threshold distance", so maybe they want the actual distance, not squared. So, if c is the 99th percentile of χ²(d), then the threshold distance would be sqrt(c). But in practice, people often use the squared distance because it's computationally easier, especially in high dimensions.But the question specifically says "threshold distance", so probably sqrt(c). However, I need to confirm.Alternatively, maybe the threshold is on the squared distance. Hmm.But regardless, the key point is that we need to find the value c such that the probability that D² exceeds c is 1%, which is the 99th percentile of the chi-squared distribution.So, to summarize, the Mahalanobis distance formula is D = sqrt[(x - μ)ᵀ Σ⁻¹ (x - μ)], and the threshold c is the 99th percentile of χ²(d). So, any sample with D² > c is considered an outlier.Wait, but in practice, sometimes people use the squared distance for the threshold because it's more straightforward. So, maybe the threshold is c, and we compare D² to c.I think either way is acceptable, but since the question asks for the threshold distance, it's safer to provide the square root, so the actual distance. So, the threshold distance is sqrt(c), where c is the 99th percentile of χ²(d).But I should check if the chi-squared distribution applies here. Yes, for a multivariate Gaussian, the squared Mahalanobis distance follows a chi-squared distribution with d degrees of freedom, where d is the number of variables.So, I think that's the answer for part 1.Problem 2: Developing a Discriminant Function and Decision BoundaryNow, the second part is about developing a discriminant function to classify a new sample vector y as genuine or adulterated. The adulterated samples are modeled by another Gaussian with mean ν and covariance Ψ.So, we have two Gaussian distributions:Genuine: N(μ, Σ)Adulterated: N(ν, Ψ)We need to develop a discriminant function. I think this is a classic case of quadratic discriminant analysis (QDA) since the covariances are different.The discriminant function for QDA is based on the log-likelihood ratio. So, for a new sample y, we compute the log-likelihood under each model and classify it to the one with higher log-likelihood.The discriminant function can be written as:g(y) = (y - μ)ᵀ Σ⁻¹ (y - μ) - (y - ν)ᵀ Ψ⁻¹ (y - ν) + log(|Σ| / |Ψ|)Wait, actually, the discriminant function is derived from the ratio of the two densities. Let me recall.The log-likelihood ratio is:log [f1(y)/f2(y)] = log [ (1/(2π)^(d/2) |Σ|^(1/2) exp(-0.5 (y-μ)ᵀ Σ⁻¹ (y-μ)) ) / (1/(2π)^(d/2) |Ψ|^(1/2) exp(-0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν)) ) ]Simplifying, the constants cancel out:= -0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) - 0.5 log |Σ| - [ -0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Ψ| ]= -0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) + 0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Σ| + 0.5 log |Ψ|So, setting this ratio greater than 0 for classification:-0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) + 0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Σ| + 0.5 log |Ψ| > 0Multiplying both sides by 2:- (y-μ)ᵀ Σ⁻¹ (y-μ) + (y-ν)ᵀ Ψ⁻¹ (y-ν) - log |Σ| + log |Ψ| > 0Rearranging:(y-ν)ᵀ Ψ⁻¹ (y-ν) - (y-μ)ᵀ Σ⁻¹ (y-μ) > log |Σ| - log |Ψ|So, the discriminant function can be written as:g(y) = (y-ν)ᵀ Ψ⁻¹ (y-ν) - (y-μ)ᵀ Σ⁻¹ (y-μ) - log(|Σ| / |Ψ|) > 0If g(y) > 0, classify as adulterated; else, classify as genuine.Alternatively, sometimes it's written as:(y - μ)ᵀ Σ⁻¹ (y - μ) - (y - ν)ᵀ Ψ⁻¹ (y - ν) + log(|Ψ| / |Σ|) < 0But the exact form depends on how you set up the inequality.In any case, the decision boundary is where g(y) = 0, which is a quadratic equation in y. So, the decision boundary is a quadratic surface in the high-dimensional space.Now, the question asks to determine the decision boundary and discuss its sensitivity to the covariance matrices Σ and Ψ.So, the decision boundary is quadratic because it's derived from the difference of two quadratic forms. It's sensitive to the covariance matrices because changes in Σ and Ψ affect the quadratic terms and the determinant terms in the discriminant function.Specifically, if Σ and Ψ are similar, the decision boundary might be less distinct, making classification harder. Conversely, if they are very different, the boundary might be more pronounced.Moreover, the determinant terms log(|Σ| / |Ψ|) affect the bias in the discriminant function. If one covariance matrix is larger (in determinant) than the other, it can influence the classification towards one class.Also, if the covariance matrices are not accurately estimated, the decision boundary can be misestimated, leading to higher classification errors. So, the sensitivity is high; accurate estimation of Σ and Ψ is crucial for good classification performance.In summary, the discriminant function is quadratic, and the decision boundary is a quadratic surface. The boundary is sensitive to the covariance matrices because they influence both the shape of the quadratic terms and the constant term involving the determinants.Wait, let me make sure I didn't make any mistakes in the discriminant function derivation.Starting from the log-likelihood ratio:log [f1(y)/f2(y)] = -0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) + 0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Σ| + 0.5 log |Ψ|Set this greater than 0 for classification to genuine or adulterated.Wait, actually, the classification depends on which log-likelihood is higher. If log f1 > log f2, classify as genuine; else, classify as adulterated.So, the discriminant function is:g(y) = log f1(y) - log f2(y) = [ -0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) - 0.5 log |Σ| ] - [ -0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Ψ| ]= -0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) + 0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Σ| + 0.5 log |Ψ|So, if g(y) > 0, classify as genuine; else, as adulterated.So, the decision boundary is where g(y) = 0:-0.5 (y-μ)ᵀ Σ⁻¹ (y-μ) + 0.5 (y-ν)ᵀ Ψ⁻¹ (y-ν) - 0.5 log |Σ| + 0.5 log |Ψ| = 0Multiplying both sides by 2:- (y-μ)ᵀ Σ⁻¹ (y-μ) + (y-ν)ᵀ Ψ⁻¹ (y-ν) - log |Σ| + log |Ψ| = 0Rearranged:(y-ν)ᵀ Ψ⁻¹ (y-ν) - (y-μ)ᵀ Σ⁻¹ (y-μ) = log |Σ| - log |Ψ|So, that's the equation of the decision boundary.Yes, that seems correct.So, the discriminant function is quadratic in y, and the decision boundary is quadratic.As for sensitivity, if Σ and Ψ are close, the quadratic terms might not differ much, making the boundary less clear. Also, if the determinants are similar, the constant term might not influence much. But if Σ and Ψ are very different, the boundary can be more distinct.Moreover, if the covariance matrices are misspecified, the boundary could be misplaced, leading to higher misclassification rates.So, that's my understanding.Final Answer1. The Mahalanobis distance is given by ( D = sqrt{(mathbf{x} - mu)^top Sigma^{-1} (mathbf{x} - mu)} ). The threshold distance is the square root of the 99th percentile of the chi-squared distribution with ( d ) degrees of freedom, where ( d ) is the dimensionality of the data. Thus, the threshold is ( sqrt{chi^2_{0.99}(d)} ).2. The discriminant function is ( g(mathbf{y}) = (mathbf{y} - nu)^top Psi^{-1} (mathbf{y} - nu) - (mathbf{y} - mu)^top Sigma^{-1} (mathbf{y} - mu) - logleft(frac{|Sigma|}{|Psi|}right) ). The decision boundary is where ( g(mathbf{y}) = 0 ), which is a quadratic surface. The boundary is sensitive to the covariance matrices as changes in ( Sigma ) and ( Psi ) affect the quadratic terms and the determinant ratio.boxed{D = sqrt{(mathbf{x} - mu)^top Sigma^{-1} (mathbf{x} - mu)}}boxed{g(mathbf{y}) = (mathbf{y} - nu)^top Psi^{-1} (mathbf{y} - nu) - (mathbf{y} - mu)^top Sigma^{-1} (mathbf{y} - mu) - logleft(frac{|Sigma|}{|Psi|}right)}</think>

🧠:<think>Alright, so I've got this problem about a law student researching how legal correspondents' interviews affect public opinion. The student uses a function P(t) = 1 / (1 + e^{-kt}) to model the probability that a viewer's opinion changes after watching t minutes of an interview. There are two sub-problems here, and I need to tackle them one by one.Starting with the first sub-problem: The student has data showing that after 20 minutes, the probability is 0.8. I need to find the constant k. Hmm, okay, so let me write down what I know.Given:P(t) = 1 / (1 + e^{-kt})At t = 20, P(20) = 0.8So, plugging in t = 20 into the equation:0.8 = 1 / (1 + e^{-20k})I need to solve for k. Let me rearrange this equation step by step.First, take the reciprocal of both sides to get rid of the fraction:1 / 0.8 = 1 + e^{-20k}Which simplifies to:1.25 = 1 + e^{-20k}Subtract 1 from both sides:1.25 - 1 = e^{-20k}0.25 = e^{-20k}Now, to solve for k, I can take the natural logarithm of both sides:ln(0.25) = ln(e^{-20k})Which simplifies to:ln(0.25) = -20kI know that ln(0.25) is the same as ln(1/4), which is equal to -ln(4). So,-ln(4) = -20kMultiply both sides by -1:ln(4) = 20kTherefore, k = ln(4) / 20Calculating ln(4): I remember that ln(2) is approximately 0.6931, so ln(4) is ln(2^2) = 2*ln(2) ≈ 2*0.6931 ≈ 1.3862So, k ≈ 1.3862 / 20 ≈ 0.06931Let me double-check my steps to make sure I didn't make a mistake.1. Plugged t=20 and P=0.8 into the equation.2. Took reciprocal correctly: 1/0.8 = 1.253. Subtracted 1: 0.25 = e^{-20k}4. Took natural log: ln(0.25) = -20k5. Recognized ln(0.25) = -ln(4)6. So, ln(4) = 20k7. Calculated ln(4) ≈ 1.3862, so k ≈ 0.06931Seems solid. Maybe I can verify by plugging k back into the original equation.Compute P(20) with k ≈ 0.06931:P(20) = 1 / (1 + e^{-0.06931*20})Calculate exponent: 0.06931*20 ≈ 1.3862So, e^{-1.3862} ≈ e^{-ln(4)} ≈ 1/4 ≈ 0.25Thus, P(20) = 1 / (1 + 0.25) = 1 / 1.25 = 0.8, which matches the given data. Perfect, so k is approximately 0.06931.Moving on to the second sub-problem: The student models the average time T viewers spend watching an interview with a probability density function f(t) = λ e^{-λ t}, where λ = 0.1 and t ≥ 0. I need to find the expected value E[T] and interpret it.Okay, so f(t) is an exponential distribution with parameter λ = 0.1. I remember that for an exponential distribution, the expected value E[T] is 1 / λ. Let me confirm that.Yes, the exponential distribution has the PDF f(t) = λ e^{-λ t} for t ≥ 0, and its expected value is indeed 1 / λ. So, plugging in λ = 0.1:E[T] = 1 / 0.1 = 10So, the expected value is 10 minutes. Interpreting this, it means that on average, viewers spend 10 minutes watching the correspondent's interviews. It's a measure of central tendency, indicating the typical time a viewer engages with the content.Just to make sure I didn't overlook anything, let me recall the formula for expected value for a continuous distribution. It's the integral from 0 to infinity of t*f(t) dt. For f(t) = λ e^{-λ t}, this integral evaluates to 1 / λ. So, yes, E[T] = 10 minutes.Alternatively, if I didn't remember the formula, I could compute it:E[T] = ∫₀^∞ t * λ e^{-λ t} dtLet me compute this integral to verify.Let’s set λ = 0.1 for clarity.E[T] = ∫₀^∞ t * 0.1 e^{-0.1 t} dtThis integral can be solved by integration by parts. Let me set u = t, dv = 0.1 e^{-0.1 t} dtThen, du = dt, and v = ∫0.1 e^{-0.1 t} dt = -e^{-0.1 t}So, integration by parts formula: ∫u dv = uv - ∫v duThus,E[T] = [ -t e^{-0.1 t} ] from 0 to ∞ + ∫₀^∞ e^{-0.1 t} dtEvaluate the first term:As t approaches infinity, e^{-0.1 t} approaches 0, so -t e^{-0.1 t} approaches 0 (since exponential decay dominates polynomial growth). At t=0, it's 0. So, the first term is 0 - 0 = 0.Now, the second integral:∫₀^∞ e^{-0.1 t} dt = [ (-1/0.1) e^{-0.1 t} ] from 0 to ∞ = (-10) [0 - 1] = 10Therefore, E[T] = 0 + 10 = 10 minutes.Yep, that confirms it. So, the expected value is 10 minutes, meaning on average, viewers watch the interviews for 10 minutes. This is significant because it tells us about the typical engagement time, which can be useful for understanding how much exposure viewers have to the correspondent's arguments, thereby influencing the probability of opinion change as modeled by P(t).Just to tie it back to the first part, if viewers typically watch for 10 minutes, we can plug t=10 into P(t) with k≈0.06931 to find the probability of opinion change at the average viewing time.P(10) = 1 / (1 + e^{-0.06931*10}) = 1 / (1 + e^{-0.6931}) ≈ 1 / (1 + 0.5) = 1 / 1.5 ≈ 0.6667So, about a 66.67% chance of opinion change at the average viewing time. That might be useful information for the law student to consider in their research.But wait, the second sub-problem only asks for the expected value and its interpretation, so I don't need to go into that extra calculation unless it's part of the problem. But it's good to see how the two parts connect.In summary, for the first part, k is approximately 0.06931, and for the second part, the expected viewing time is 10 minutes, indicating the typical engagement duration.Final Answer1. The value of ( k ) is boxed{0.06931}.2. The expected value ( E[T] ) is boxed{10} minutes.</think>

🧠:<think>Okay, so I have this problem about a vlogger who is trying to estimate the number of unique fish species in a coral reef. They’re using underwater drones to capture footage in different regions, and they want to use some statistical methods to figure out the total species. The problem has two parts, and I need to tackle them one by one.Starting with part 1: They say that the vlogger uses a sampling method where the drone captures footage over n different regions. For each region i, S_i is the number of unique species observed. The observed data follows a Poisson distribution with parameter λ_i for each region. I need to express the likelihood function L(λ₁, λ₂, ..., λₙ) for the observed data (S₁, S₂, ..., Sₙ).Alright, so I remember that the likelihood function is the probability of observing the data given the parameters. For a Poisson distribution, the probability mass function is P(S_i = k) = (λ_i^k e^{-λ_i}) / k! So, for each region i, the probability of observing S_i species is (λ_i^{S_i} e^{-λ_i}) / S_i!.Since the regions are independent, the likelihood function is the product of the probabilities for each region. So, the overall likelihood L is the product from i=1 to n of [(λ_i^{S_i} e^{-λ_i}) / S_i!].Therefore, L(λ₁, λ₂, ..., λₙ) = product_{i=1}^n [ (λ_i^{S_i} e^{-λ_i}) / S_i! ]I think that's it for part 1. It seems straightforward because each region is independent, so we just multiply the individual likelihoods.Moving on to part 2: Now, they assume that λ_i = λ is the same for all regions because the reef is homogeneous. So, all regions have the same Poisson parameter λ. Given the total observed species count S = sum_{i=1}^n S_i, I need to derive the maximum likelihood estimator (MLE) for λ.Hmm, okay. So, since all λ_i are equal to λ, the likelihood function simplifies. Instead of having different λ_i for each region, we have the same λ for all. So, the likelihood function becomes the product from i=1 to n of [ (λ^{S_i} e^{-λ}) / S_i! ]Which can be written as [λ^{sum S_i} e^{-nλ}] / (product S_i! )Because when you multiply all the λ^{S_i}, it's λ^{sum S_i}, and e^{-λ} multiplied n times is e^{-nλ}. The denominator is the product of S_i! for each i.So, the likelihood function L(λ) = [λ^S e^{-nλ}] / (product S_i! ), where S = sum S_i.To find the MLE, we usually take the log-likelihood, differentiate it with respect to λ, set the derivative equal to zero, and solve for λ.So, let's compute the log-likelihood:log L(λ) = sum [S_i log λ - λ - log S_i! ]Which simplifies to S log λ - nλ - sum log S_i!But since we're maximizing with respect to λ, the terms that don't involve λ can be ignored. So, effectively, the log-likelihood is proportional to S log λ - nλ.Taking the derivative of the log-likelihood with respect to λ:d/dλ [log L(λ)] = (S / λ) - nSet this equal to zero to find the critical point:(S / λ) - n = 0Solving for λ:S / λ = n => λ = S / nSo, the MLE for λ is the total number of species observed divided by the number of regions.Wait, let me double-check that. So, the derivative is (S / λ) - n, setting to zero gives λ = S / n. That seems right.Is there a second derivative to check concavity? The second derivative would be -S / λ², which is negative, so the function is concave, meaning that this critical point is indeed a maximum.Therefore, the MLE for λ is S / n.So, summarizing:1. The likelihood function is the product over each region of (λ_i^{S_i} e^{-λ_i}) / S_i!.2. When all λ_i are equal to λ, the MLE for λ is the total number of species S divided by the number of regions n.I think that's the solution.Final Answer1. The likelihood function is boxed{prod_{i=1}^{n} frac{lambda_i^{S_i} e^{-lambda_i}}{S_i!}}.2. The maximum likelihood estimator for (lambda) is boxed{frac{S}{n}}.</think>

🧠:<think>Alright, so I have this problem about a former cricketer who's now a commentator, and his witty commentary increases viewership exponentially. The function given is ( V(t) = V_0 cdot e^{kt} ), where ( V_0 ) is the initial viewership, ( k ) is a constant based on his humor index, and ( t ) is time in hours. There are two parts to this problem. Let me tackle them one by one.Problem 1: Expected Viewership at Time T = 3 HoursOkay, so the first part asks for the expected viewership ( E[V(T)] ) at time ( T = 3 ) hours. The humor index ( k ) varies according to a normal distribution with mean ( mu = 0.5 ) and standard deviation ( sigma = 0.1 ). The initial viewership ( V_0 ) is 1000.Hmm, so I need to find the expectation of ( V(3) ). Since ( V(t) = V_0 cdot e^{kt} ), then ( V(3) = 1000 cdot e^{3k} ). Therefore, ( E[V(3)] = 1000 cdot E[e^{3k}] ).Now, ( k ) is normally distributed with mean 0.5 and standard deviation 0.1. So, ( k sim N(0.5, 0.1^2) ). I remember that for a normally distributed random variable ( X sim N(mu, sigma^2) ), the expectation ( E[e^{aX}] ) is ( e^{amu + frac{1}{2}a^2sigma^2} ). This is because the moment generating function of a normal distribution is ( M_X(a) = e^{mu a + frac{1}{2}sigma^2 a^2} ).So, applying this to our case where ( a = 3 ), we have:( E[e^{3k}] = e^{3mu + frac{1}{2}(3)^2sigma^2} ).Plugging in the values:( mu = 0.5 ), ( sigma = 0.1 ), so ( sigma^2 = 0.01 ).Calculating the exponent:( 3 * 0.5 = 1.5 )( (3)^2 * 0.01 = 9 * 0.01 = 0.09 )So, the exponent is ( 1.5 + 0.09/2 ). Wait, no, hold on. The formula is ( amu + frac{1}{2}a^2sigma^2 ). So, it's ( 3 * 0.5 + 0.5 * 9 * 0.01 ).Wait, let me clarify:( E[e^{3k}] = e^{3mu + frac{1}{2}(3)^2sigma^2} )So, that's ( e^{3*0.5 + 0.5*9*0.01} ).Calculating each term:( 3*0.5 = 1.5 )( 0.5*9 = 4.5 )( 4.5*0.01 = 0.045 )So, total exponent is ( 1.5 + 0.045 = 1.545 ).Therefore, ( E[e^{3k}] = e^{1.545} ).Calculating ( e^{1.545} ). Let me compute that.I know that ( e^{1.6} ) is approximately 4.953, and ( e^{1.5} ) is approximately 4.4817. Since 1.545 is between 1.5 and 1.6, let's see:Compute ( e^{1.545} ):Using a calculator, but since I don't have one, maybe approximate it.Alternatively, use Taylor series or linear approximation.But perhaps I can compute it step by step.Alternatively, use natural logarithm properties.Wait, maybe I can use the fact that ( e^{1.545} = e^{1.5 + 0.045} = e^{1.5} cdot e^{0.045} ).We know ( e^{1.5} approx 4.4817 ).Compute ( e^{0.045} ). Since 0.045 is small, we can approximate it as ( 1 + 0.045 + (0.045)^2/2 + (0.045)^3/6 ).Calculating:( 0.045^2 = 0.002025 ), so divided by 2 is 0.0010125.( 0.045^3 = 0.000091125 ), divided by 6 is approximately 0.0000151875.Adding up: 1 + 0.045 + 0.0010125 + 0.0000151875 ≈ 1.0460276875.So, ( e^{0.045} ≈ 1.0460276875 ).Therefore, ( e^{1.545} ≈ 4.4817 * 1.0460276875 ).Compute 4.4817 * 1.0460276875.First, 4 * 1.046 = 4.1840.4817 * 1.046 ≈ 0.4817 + 0.4817*0.046 ≈ 0.4817 + 0.02215 ≈ 0.50385So total is approximately 4.184 + 0.50385 ≈ 4.68785.But let me do a more precise multiplication:4.4817 * 1.0460276875Multiply 4.4817 by 1.0460276875.First, 4.4817 * 1 = 4.48174.4817 * 0.04 = 0.1792684.4817 * 0.006 = 0.02689024.4817 * 0.0000276875 ≈ 0.0001238Adding all together:4.4817 + 0.179268 = 4.6609684.660968 + 0.0268902 ≈ 4.68785824.6878582 + 0.0001238 ≈ 4.687982So, approximately 4.688.Therefore, ( e^{1.545} ≈ 4.688 ).Thus, ( E[e^{3k}] ≈ 4.688 ).Therefore, ( E[V(3)] = 1000 * 4.688 = 4688 ).Wait, but let me double-check my calculations because 4.688 seems a bit low given the exponent.Alternatively, maybe I should use a calculator for better precision, but since I don't have one, perhaps I can recall that ( ln(4.688) ) is approximately 1.545, which is correct.Wait, actually, ( e^{1.545} ) is approximately 4.688, so that seems correct.Hence, the expected viewership is approximately 4688.But let me think again: the formula is ( E[e^{3k}] = e^{3mu + (9/2)sigma^2} ).Wait, hold on. Is it ( frac{1}{2}a^2sigma^2 ), so with ( a = 3 ), it's ( frac{1}{2} * 9 * 0.01 = 0.045 ). So, exponent is 1.5 + 0.045 = 1.545, as I had before.Therefore, yes, ( e^{1.545} ≈ 4.688 ), so ( E[V(3)] ≈ 4688 ).Alternatively, maybe I can use a calculator for more precise value.But since I don't have one, I think 4688 is a reasonable approximation.Problem 2: Maximum Duration ( t_{text{max}} ) with 95% ConfidenceThe second part is a bit trickier. It says that during a critical match, a new regulation requires that the total viewership should not exceed 5000 at any point to prevent server overload. We need to determine the maximum duration ( t_{text{max}} ) for which he can maintain his commentary without exceeding this limit, with 95% confidence, given his humor index is normally distributed as described.So, we need to find ( t_{text{max}} ) such that ( V(t) leq 5000 ) with 95% probability.Given ( V(t) = 1000 cdot e^{kt} ), so ( 1000 cdot e^{kt} leq 5000 ).Divide both sides by 1000: ( e^{kt} leq 5 ).Take natural logarithm: ( kt leq ln(5) ).So, ( k leq frac{ln(5)}{t} ).We need ( P(k leq frac{ln(5)}{t}) geq 0.95 ).Since ( k ) is normally distributed with mean 0.5 and standard deviation 0.1, we can write:( Pleft( frac{k - 0.5}{0.1} leq frac{frac{ln(5)}{t} - 0.5}{0.1} right) geq 0.95 ).Let me denote ( Z = frac{k - 0.5}{0.1} ), which is a standard normal variable.So, ( P(Z leq frac{ln(5)/t - 0.5}{0.1}) geq 0.95 ).We need the value of ( t ) such that the probability that ( Z ) is less than or equal to ( frac{ln(5)/t - 0.5}{0.1} ) is at least 0.95.From standard normal distribution tables, the 95th percentile is approximately 1.645. So, we set:( frac{ln(5)/t - 0.5}{0.1} = 1.645 ).Solving for ( t ):Multiply both sides by 0.1:( ln(5)/t - 0.5 = 0.1645 )Add 0.5 to both sides:( ln(5)/t = 0.6645 )Therefore, ( t = ln(5)/0.6645 ).Compute ( ln(5) ). I know that ( ln(5) ≈ 1.6094 ).So, ( t ≈ 1.6094 / 0.6645 ≈ ).Calculating that:Divide 1.6094 by 0.6645.Let me compute 1.6094 / 0.6645.First, 0.6645 * 2 = 1.329, which is less than 1.6094.0.6645 * 2.4 = 1.5948, which is close to 1.6094.Compute 0.6645 * 2.42:0.6645 * 2 = 1.3290.6645 * 0.4 = 0.26580.6645 * 0.02 = 0.01329Adding up: 1.329 + 0.2658 = 1.5948 + 0.01329 = 1.60809.So, 0.6645 * 2.42 ≈ 1.60809, which is very close to 1.6094.The difference is 1.6094 - 1.60809 = 0.00131.So, to get the remaining 0.00131, we can compute how much more t is needed.0.6645 * x = 0.00131x ≈ 0.00131 / 0.6645 ≈ 0.00197.So, total t ≈ 2.42 + 0.00197 ≈ 2.42197.Therefore, ( t ≈ 2.422 ) hours.So, approximately 2.422 hours.But let me check my calculations again.Wait, so we have:( t = ln(5) / 0.6645 ≈ 1.6094 / 0.6645 ≈ 2.422 ) hours.Yes, that seems correct.But let me verify the steps again to ensure I didn't make a mistake.We started with ( V(t) = 1000 e^{kt} leq 5000 ).So, ( e^{kt} leq 5 ), take ln: ( kt leq ln(5) ).Thus, ( k leq ln(5)/t ).Since ( k sim N(0.5, 0.1^2) ), we want ( P(k leq ln(5)/t) = 0.95 ).So, the z-score corresponding to 0.95 is 1.645.Thus, ( (ln(5)/t - 0.5)/0.1 = 1.645 ).Solving for ( t ):( ln(5)/t - 0.5 = 0.1645 )( ln(5)/t = 0.6645 )( t = ln(5)/0.6645 ≈ 1.6094 / 0.6645 ≈ 2.422 ) hours.Yes, that seems consistent.So, the maximum duration ( t_{text{max}} ) is approximately 2.422 hours.But let me think about whether this is correct.Wait, so with 95% confidence, the viewership will not exceed 5000. So, we're finding the time t such that 95% of the time, ( V(t) leq 5000 ).Alternatively, 5% of the time, it could exceed 5000, but we want to ensure it doesn't exceed with 95% confidence.So, yes, we set the probability that ( k leq ln(5)/t ) to be 0.95, which corresponds to the 95th percentile of the normal distribution.Therefore, the calculation seems correct.But let me check if I used the right z-score. For 95% confidence, the z-score is 1.645 for one-tailed test, which is correct because we're looking at the upper tail.Yes, so that's correct.Therefore, the maximum duration is approximately 2.422 hours, which is about 2 hours and 25 minutes.But the question asks for the maximum duration ( t_{text{max}} ), so we can present it as approximately 2.42 hours.Alternatively, if we want more decimal places, we can compute it more precisely.But given that 0.6645 * 2.422 ≈ 1.6094, as we saw earlier, so 2.422 is accurate enough.Summary of Thoughts1. For the first part, I recognized that the expectation of an exponential function of a normal variable can be found using the moment generating function. I applied the formula correctly and computed the exponent step by step, approximating the result.2. For the second part, I translated the viewership constraint into a probability statement about the humor index ( k ). I then converted this into a standard normal variable and solved for ( t ) using the z-score corresponding to the 95th percentile. The calculations were a bit involved, but breaking them down step by step helped ensure accuracy.I think both parts are handled correctly, but let me just verify if I didn't make any arithmetic errors.Verification for Problem 1:- ( E[e^{3k}] = e^{3*0.5 + 0.5*(3)^2*(0.1)^2} = e^{1.5 + 0.5*9*0.01} = e^{1.5 + 0.045} = e^{1.545} ≈ 4.688 ).Yes, that's correct.Verification for Problem 2:- ( t = ln(5)/0.6645 ≈ 1.6094 / 0.6645 ≈ 2.422 ) hours.Yes, that's correct.Therefore, I'm confident in my solutions.Final Answer1. The expected viewership at ( T = 3 ) hours is boxed{4688}.2. The maximum duration ( t_{text{max}} ) is approximately boxed{2.42} hours.</think>