🧠:<think>Okay, so I have two problems to solve here. The first one is about calculating the probability that a randomly selected participant from the drug group has a higher improvement score than someone from the placebo group. The second one is about solving a differential equation to model the long-term impact of the drug on cognitive decline. Let me tackle them one by one.Starting with the first problem. We have two groups: the drug group and the placebo group. Both groups have their cognitive test score improvements modeled by normal distributions. For the drug group, the mean improvement is 15 points with a standard deviation of 5. For the placebo group, the mean improvement is 10 points with a standard deviation of 6. I need to find the probability that a randomly selected participant from the drug group has a higher improvement score than a randomly selected participant from the placebo group.Hmm, so this sounds like a problem where I need to compare two independent normal distributions. I remember that when comparing two independent normal variables, the difference between them is also normally distributed. So, if I let X be the improvement score for the drug group and Y be the improvement score for the placebo group, then X ~ N(15, 5²) and Y ~ N(10, 6²). I need to find P(X > Y). To do this, I can consider the difference D = X - Y. Then, D will also be normally distributed with mean μ_D = μ_X - μ_Y = 15 - 10 = 5. The variance of D will be Var(D) = Var(X) + Var(Y) since X and Y are independent. So, Var(D) = 5² + 6² = 25 + 36 = 61. Therefore, the standard deviation σ_D is sqrt(61).So, D ~ N(5, 61). Now, I need to find P(D > 0), which is the probability that X - Y > 0, or equivalently, X > Y. To find this probability, I can standardize D. Let me compute the Z-score for D = 0. The Z-score is (0 - μ_D)/σ_D = (0 - 5)/sqrt(61) ≈ (-5)/7.81 ≈ -0.64.So, P(D > 0) = P(Z > -0.64). Looking at the standard normal distribution table, P(Z > -0.64) is equal to 1 - P(Z < -0.64). From the table, P(Z < -0.64) is approximately 0.2611. Therefore, P(Z > -0.64) = 1 - 0.2611 = 0.7389.So, the probability that a randomly selected participant from the drug group has a higher improvement score than a randomly selected participant from the placebo group is approximately 73.89%.Wait, let me double-check my calculations. The mean difference is 5, which is positive, so the probability should be more than 50%, which 73.89% is. The standard deviation is sqrt(25 + 36) = sqrt(61) ≈ 7.81. Then, the Z-score is (0 - 5)/7.81 ≈ -0.64. The area to the right of -0.64 is indeed about 0.7389. Yeah, that seems correct.Moving on to the second problem. The executive wants to model the long-term impact of the drug on cognitive decline using a differential equation. Without the drug, the rate of cognitive decline C(t) is modeled by dC/dt = -kC(t), where k is a positive constant. When the drug is administered, the rate becomes dC/dt = -kC(t) + r, where r is a positive constant representing the drug's effect. We need to solve this differential equation with the initial condition C(0) = C₀.Alright, so this is a linear first-order differential equation. The standard form is dC/dt + P(t)C = Q(t). Let me rewrite the equation:dC/dt + kC = r.Here, P(t) = k and Q(t) = r. Since P(t) and Q(t) are constants, this is a linear ODE with constant coefficients. The integrating factor method should work here.The integrating factor μ(t) is given by exp(∫P(t) dt) = exp(∫k dt) = e^{kt}.Multiplying both sides of the differential equation by the integrating factor:e^{kt} dC/dt + k e^{kt} C = r e^{kt}.The left side is the derivative of (e^{kt} C) with respect to t. So, we can write:d/dt (e^{kt} C) = r e^{kt}.Now, integrate both sides with respect to t:∫ d/dt (e^{kt} C) dt = ∫ r e^{kt} dt.This gives:e^{kt} C = (r / k) e^{kt} + constant.Let me solve for C(t):C(t) = (r / k) + constant * e^{-kt}.Now, apply the initial condition C(0) = C₀. When t = 0,C(0) = (r / k) + constant * e^{0} = (r / k) + constant = C₀.Therefore, the constant is C₀ - (r / k).So, the solution is:C(t) = (r / k) + (C₀ - r / k) e^{-kt}.Simplify this, we can write:C(t) = C₀ e^{-kt} + (r / k)(1 - e^{-kt}).Alternatively, it can be expressed as:C(t) = (C₀ - r / k) e^{-kt} + r / k.Either form is correct, but perhaps the second form is more insightful because it shows the steady-state term r/k and the transient term (C₀ - r/k) e^{-kt}.Let me verify this solution by plugging it back into the differential equation.Compute dC/dt:dC/dt = -k (C₀ - r / k) e^{-kt} + 0.So, dC/dt = -k C(t) + r.Which matches the original differential equation. So, yes, the solution is correct.Therefore, the solution to the differential equation is C(t) = (C₀ - r/k) e^{-kt} + r/k.So, summarizing my answers:1. The probability is approximately 73.89%.2. The solution to the differential equation is C(t) = (C₀ - r/k) e^{-kt} + r/k.Final Answer1. The probability is boxed{0.7389}.2. The solution is boxed{C(t) = left(C_0 - frac{r}{k}right) e^{-kt} + frac{r}{k}}.</think>

🧠:<think>Okay, so I have this problem about Sarah studying the growth of her local congregation. It's modeled by the differential equation dN/dt = kN, which I remember is the exponential growth model. Let me try to figure this out step by step.First, part 1 asks me to find the value of k. They give me that at time t=0, the number of members N(0) is 150. After 3 years, at t=3, the number of members is 450. So, I need to use this information to find k.I recall that the solution to the differential equation dN/dt = kN is N(t) = N0 * e^(kt), where N0 is the initial amount. So in this case, N0 is 150. Therefore, the equation becomes N(t) = 150 * e^(kt).Now, they tell me that after 3 years, N(3) is 450. So I can plug that into the equation:450 = 150 * e^(3k)Hmm, okay, so I can divide both sides by 150 to simplify:450 / 150 = e^(3k)That simplifies to 3 = e^(3k)Now, to solve for k, I need to take the natural logarithm of both sides. Remember, ln(e^x) = x. So:ln(3) = ln(e^(3k)) => ln(3) = 3kTherefore, k = ln(3) / 3Let me compute that. I know that ln(3) is approximately 1.0986, so dividing that by 3 gives k ≈ 0.3662 per year. But since they might want an exact value, I should keep it as ln(3)/3.Wait, let me double-check my steps. I started with N(t) = 150e^(kt). Plugged in t=3, N=450. Divided both sides by 150, got 3 = e^(3k). Took natural log, got ln(3)=3k, so k=ln(3)/3. Yeah, that seems right.Moving on to part 2, they want the expected number of members 10 years after the introduction. So t=10.Using the same formula N(t) = 150e^(kt). We already found k is ln(3)/3, so plug that in:N(10) = 150 * e^( (ln(3)/3)*10 )Simplify the exponent: (ln(3)/3)*10 = (10/3)ln(3) = ln(3^(10/3))So, N(10) = 150 * e^(ln(3^(10/3))) = 150 * 3^(10/3)Hmm, 3^(10/3) is the same as the cube root of 3^10. Let me compute 3^10 first. 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187, 3^8=6561, 3^9=19683, 3^10=59049.So, 3^(10/3) is the cube root of 59049. Let me compute that. The cube root of 59049. Since 3^3=27, 27^3=19683, which is 3^9. So 3^10 is 59049, so the cube root of 59049 is 3^(10/3) = 3^(3 + 1/3) = 3^3 * 3^(1/3) = 27 * cube root of 3.Wait, that might not be helpful. Alternatively, I can compute 3^(10/3) as e^( (10/3) ln 3 ). Let me compute that numerically.First, ln(3) is approximately 1.0986. So (10/3)*1.0986 ≈ (3.3333)*1.0986 ≈ 3.662.So e^3.662. Let me compute e^3 is about 20.0855, e^3.662 is a bit more. Let me see, e^3.662 ≈ e^(3 + 0.662) = e^3 * e^0.662.Compute e^0.662: e^0.6 is about 1.8221, e^0.662 is a bit higher. Let's approximate. The derivative of e^x is e^x, so around x=0.6, e^0.662 ≈ e^0.6 + (0.062)*e^0.6 ≈ 1.8221 + 0.062*1.8221 ≈ 1.8221 + 0.113 ≈ 1.935.So e^3.662 ≈ 20.0855 * 1.935 ≈ Let's compute that. 20 * 1.935 = 38.7, and 0.0855 * 1.935 ≈ 0.165. So total ≈ 38.7 + 0.165 ≈ 38.865.Therefore, N(10) ≈ 150 * 38.865 ≈ 150 * 38.865. Let me compute 150*38 = 5700, 150*0.865=129.75, so total ≈ 5700 + 129.75 ≈ 5829.75.So approximately 5830 members after 10 years.Wait, but let me check if I did that correctly. Alternatively, maybe I can compute 3^(10/3) more accurately.Since 3^(1/3) is approximately 1.4422. So 3^(10/3) = (3^(1/3))^10 = (1.4422)^10.Wait, that might not be the best approach. Alternatively, since 3^(10/3) is equal to e^(10/3 ln3) ≈ e^(3.662) as before.Alternatively, maybe I can use logarithms to compute 3^(10/3):Take log base 10: log10(3^(10/3)) = (10/3) log10(3) ≈ (10/3)(0.4771) ≈ 1.5903.So 10^1.5903 ≈ 10^1 * 10^0.5903 ≈ 10 * 3.88 ≈ 38.8. So 3^(10/3) ≈ 38.8, which matches my previous calculation.Therefore, N(10) = 150 * 38.8 ≈ 5820. So approximately 5820 members.Wait, but let me see if I can express it more precisely. Since 3^(10/3) is 3^(3 + 1/3) = 27 * 3^(1/3). And 3^(1/3) is approximately 1.4422, so 27 * 1.4422 ≈ 38.9394.So N(10) = 150 * 38.9394 ≈ 150 * 38.9394. Let's compute that:150 * 38 = 5700150 * 0.9394 ≈ 150 * 0.9 = 135, 150 * 0.0394 ≈ 5.91, so total ≈ 135 + 5.91 ≈ 140.91So total N(10) ≈ 5700 + 140.91 ≈ 5840.91, which is approximately 5841.Wait, that's a bit different from my previous estimate. Hmm, maybe I miscalculated earlier.Wait, 3^(10/3) is 3^(3 + 1/3) = 3^3 * 3^(1/3) = 27 * 1.4422 ≈ 38.9394.So 150 * 38.9394 = ?Compute 100 * 38.9394 = 3893.9450 * 38.9394 = 1946.97So total is 3893.94 + 1946.97 = 5840.91, which is approximately 5841.So, approximately 5841 members after 10 years.Alternatively, maybe I can express it exactly as 150 * 3^(10/3). But since they might want a numerical value, 5841 is a good approximation.Wait, let me check with another method. Let's use the formula N(t) = N0 * e^(kt). We have k = ln(3)/3, so N(10) = 150 * e^( (ln(3)/3)*10 ) = 150 * e^(10/3 ln3) = 150 * 3^(10/3).Alternatively, I can compute 3^(10/3) as e^( (10/3) ln3 ). Let me compute (10/3) ln3 ≈ (3.3333)(1.0986) ≈ 3.662. Then e^3.662 ≈ 38.9394 as before. So 150 * 38.9394 ≈ 5840.91, which rounds to 5841.So, I think that's the correct number.Wait, but let me make sure I didn't make a mistake in the exponent. The exponent is (ln3)/3 * 10, which is (10/3) ln3, which is correct.Alternatively, maybe I can compute it using semi-log plot. Since the growth is exponential, the number of members triples every 3 years because N(3) = 3*N(0). So, let's see:At t=0: 150t=3: 450 (tripled)t=6: 450*3=1350t=9: 1350*3=4050t=12: 4050*3=12150But wait, they want t=10, which is between t=9 and t=12.So, from t=9 to t=12, the growth is from 4050 to 12150, which is tripling over 3 years. So, the annual growth factor is 3^(1/3) ≈ 1.4422.So, from t=9 to t=10, it's 1 year, so the growth factor is 1.4422.So, N(10) = N(9) * 1.4422 ≈ 4050 * 1.4422 ≈ Let's compute that:4050 * 1 = 40504050 * 0.4 = 16204050 * 0.04 = 1624050 * 0.0022 ≈ 8.91Adding up: 4050 + 1620 = 5670, +162 = 5832, +8.91 ≈ 5840.91, which is the same as before. So, that's consistent.So, N(10) ≈ 5841.Therefore, the value of k is ln(3)/3, and the number of members after 10 years is approximately 5841.Wait, but let me check if I can write it more precisely. Since 3^(10/3) is exact, maybe I can leave it as 150 * 3^(10/3), but I think they want a numerical value.Alternatively, using more precise calculations:Compute (10/3) ln3:ln3 ≈ 1.098612289(10/3)*1.098612289 ≈ 3.662040963Now, compute e^3.662040963.We can use the Taylor series expansion around x=3.662040963, but that might be complicated. Alternatively, use a calculator approximation.But since I don't have a calculator here, I can use the fact that e^3.662040963 ≈ 38.9394 as before.So, 150 * 38.9394 ≈ 5840.91, which is approximately 5841.So, I think that's correct.Wait, let me check with another approach. Since the doubling time isn't directly applicable here, but the tripling time is 3 years. So, the growth factor is 3 every 3 years, which is a growth rate of ln(3)/3 per year, as we found.So, over 10 years, the number of tripling periods is 10/3 ≈ 3.3333. So, the growth factor is 3^(10/3), which is the same as before.So, N(10) = 150 * 3^(10/3) ≈ 150 * 38.9394 ≈ 5840.91 ≈ 5841.Therefore, the answers are:1. k = ln(3)/32. N(10) ≈ 5841 membersI think that's it. Let me just recap:1. Solved the differential equation to get N(t) = 150e^(kt).2. Plugged in t=3, N=450 to find k = ln(3)/3.3. For t=10, used N(10) = 150e^(10k) = 150 * 3^(10/3) ≈ 5841.Yes, that seems consistent.</think>

🧠:<think>Alright, so I have this problem about optimizing a machine learning algorithm's prediction error. It's divided into two parts. Let me tackle them one by one.Starting with part 1: The developer has a prediction error function E(theta) which is a convex function. It's given by the sum from i=1 to n of (a_i theta_i squared plus b_i theta_i plus c_i). The goal is to find the theta that minimizes E(theta) subject to the constraint that the sum of theta_i equals k. They want me to use the method of Lagrange multipliers.Okay, so Lagrange multipliers are used for optimization with constraints. The basic idea is to introduce a multiplier for the constraint and then take derivatives with respect to each variable and the multiplier.First, let me write down the function to minimize:E(theta) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i + c_i]But since c_i are constants, they don't affect the minimization, so I can ignore them for the purpose of finding theta.So, effectively, E(theta) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i]And the constraint is sum_{i=1}^n theta_i = k.To use Lagrange multipliers, I need to set up the Lagrangian function, which incorporates the constraint. The Lagrangian L is:L(theta, lambda) = E(theta) + lambda (sum_{i=1}^n theta_i - k)So substituting E(theta):L(theta, lambda) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i] + lambda (sum_{i=1}^n theta_i - k)Now, to find the minimum, I need to take the partial derivatives of L with respect to each theta_i and lambda, set them equal to zero, and solve.Let's compute the partial derivative of L with respect to theta_j:dL/dtheta_j = 2 a_j theta_j + b_j + lambda = 0And the partial derivative with respect to lambda is:dL/dlambda = sum_{i=1}^n theta_i - k = 0So, for each j, we have:2 a_j theta_j + b_j + lambda = 0And sum(theta_i) = kSo, from the first equation, we can solve for theta_j:theta_j = (-b_j - lambda) / (2 a_j)But since a_j is a constant, and for convexity, a_j must be positive, right? Because E(theta) is convex, so each quadratic term must be convex, so a_i > 0.So, theta_j is expressed in terms of lambda. Now, we can substitute this into the constraint equation.Sum_{i=1}^n theta_i = kSubstituting theta_i:sum_{i=1}^n [ (-b_i - lambda) / (2 a_i) ] = kLet me factor out 1/2:(1/2) sum_{i=1}^n [ (-b_i - lambda) / a_i ] = kMultiply both sides by 2:sum_{i=1}^n [ (-b_i - lambda) / a_i ] = 2kLet me write this as:sum_{i=1}^n (-b_i / a_i) - lambda sum_{i=1}^n (1 / a_i) = 2kLet me denote S = sum_{i=1}^n (1 / a_i)Then, the equation becomes:- sum_{i=1}^n (b_i / a_i) - lambda S = 2kSo, solving for lambda:- lambda S = 2k + sum_{i=1}^n (b_i / a_i)Multiply both sides by -1:lambda S = -2k - sum_{i=1}^n (b_i / a_i)Therefore,lambda = [ -2k - sum_{i=1}^n (b_i / a_i) ] / SBut S is sum_{i=1}^n (1 / a_i), so:lambda = [ -2k - sum_{i=1}^n (b_i / a_i) ] / [ sum_{i=1}^n (1 / a_i) ]Once we have lambda, we can substitute back into theta_j:theta_j = (-b_j - lambda) / (2 a_j)So, plugging lambda:theta_j = [ -b_j - ( [ -2k - sum_{i=1}^n (b_i / a_i) ] / [ sum_{i=1}^n (1 / a_i) ] ) ] / (2 a_j)Simplify numerator:= [ -b_j + (2k + sum_{i=1}^n (b_i / a_i)) / sum_{i=1}^n (1 / a_i) ) ] / (2 a_j)Let me write this as:theta_j = [ (-b_j * sum_{i=1}^n (1 / a_i) ) + 2k + sum_{i=1}^n (b_i / a_i) ) ] / (2 a_j * sum_{i=1}^n (1 / a_i) )Wait, let me check that step again.Wait, when I have:theta_j = [ -b_j + (2k + sum(b_i / a_i)) / sum(1/a_i) ] / (2 a_j)Yes, that's correct.So, to make it clearer:theta_j = [ -b_j + (2k + sum_{i=1}^n (b_i / a_i)) / sum_{i=1}^n (1 / a_i) ) ] / (2 a_j)Alternatively, factor out 1 / sum(1/a_i):theta_j = [ (-b_j * sum(1/a_i) + 2k + sum(b_i / a_i) ) ] / (2 a_j * sum(1/a_i))Which can be written as:theta_j = [ 2k + sum_{i=1}^n (b_i / a_i) - b_j * sum_{i=1}^n (1 / a_i) ) ] / (2 a_j * sum_{i=1}^n (1 / a_i))Alternatively, factor out 1 / sum(1/a_i):theta_j = [2k + sum_{i=1}^n (b_i / a_i) - b_j * sum_{i=1}^n (1 / a_i) ] / (2 a_j sum_{i=1}^n (1 / a_i))Alternatively, we can write:theta_j = [2k + sum_{i=1}^n (b_i / a_i) ] / (2 a_j sum_{i=1}^n (1 / a_i)) - [b_j / (2 a_j)]Wait, that might not be necessary. Maybe it's better to leave it as:theta_j = [ -b_j + (2k + sum_{i=1}^n (b_i / a_i)) / sum_{i=1}^n (1 / a_i) ) ] / (2 a_j)Alternatively, factor out 1 / sum(1/a_i):theta_j = [ (2k + sum_{i=1}^n (b_i / a_i) - b_j sum_{i=1}^n (1 / a_i)) ] / (2 a_j sum_{i=1}^n (1 / a_i))Wait, that seems a bit messy. Maybe it's better to just express theta_j in terms of lambda.But regardless, the key point is that each theta_j is determined by this expression involving lambda, which itself is determined by the constraint.So, summarizing, the optimal theta_j is:theta_j = (-b_j - lambda) / (2 a_j)where lambda is:lambda = [ -2k - sum_{i=1}^n (b_i / a_i) ] / [ sum_{i=1}^n (1 / a_i) ]Alternatively, we can write lambda as:lambda = [ -2k - sum_{i=1}^n (b_i / a_i) ] / S, where S = sum_{i=1}^n (1 / a_i)So, that's part 1.Moving on to part 2: After optimizing the parameters, the developer wants to estimate the sensitivity of the prediction error to small changes in the input data x. The prediction error function is now E(theta, x) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i + sum_{j=1}^m d_{ij} x_j ]So, E(theta, x) is a function of both theta and x. The developer wants the gradient of E with respect to x, which is the vector of partial derivatives of E with respect to each x_j.So, let's compute the gradient ∇_x E(theta, x).First, E(theta, x) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i + sum_{j=1}^m d_{ij} x_j ]So, E(theta, x) can be written as:E(theta, x) = sum_{i=1}^n [a_i theta_i^2 + b_i theta_i] + sum_{i=1}^n sum_{j=1}^m d_{ij} x_jBut the first part is just the original E(theta), which we've already optimized. The second part is a linear term in x.So, to compute the gradient with respect to x, we can focus on the second part.The second part is sum_{i=1}^n sum_{j=1}^m d_{ij} x_jWhich can be rewritten as sum_{j=1}^m [ sum_{i=1}^n d_{ij} ] x_jSo, for each x_j, the coefficient is sum_{i=1}^n d_{ij}Therefore, the partial derivative of E with respect to x_j is sum_{i=1}^n d_{ij}Hence, the gradient ∇_x E(theta, x) is the vector where each component j is sum_{i=1}^n d_{ij}So, ∇_x E(theta, x) = [ sum_{i=1}^n d_{i1}, sum_{i=1}^n d_{i2}, ..., sum_{i=1}^n d_{im} ]^TIn other words, each element of the gradient is the sum of the corresponding column in the matrix D, where D is the matrix with entries d_{ij}.Now, interpreting this in the context of improving the algorithm's robustness to data variability: The gradient ∇_x E(theta, x) tells us the sensitivity of the prediction error to changes in each input data point x_j. Specifically, the j-th component of the gradient indicates how much the prediction error E changes per unit change in x_j, holding theta constant.Therefore, if a particular x_j has a high absolute value in the gradient, it means that small changes in x_j can lead to较大的变化 in the prediction error. This suggests that the algorithm is sensitive to variations in that input feature x_j. To improve robustness, the developer might want to consider normalizing or standardizing the input data, especially features with high sensitivity, to reduce their impact on prediction variability. Alternatively, they could look into regularization techniques or robust optimization methods to make the algorithm less sensitive to such changes.So, in summary, the gradient provides a measure of how each input feature affects the prediction error, guiding the developer on which features to focus on for improving the model's stability and robustness.Final Answer1. The optimal parameter values are given by: [ theta_j = frac{-b_j - lambda}{2a_j} ] where ( lambda ) is determined by the constraint and is: [ lambda = frac{-2k - sum_{i=1}^n frac{b_i}{a_i}}{sum_{i=1}^n frac{1}{a_i}} ] Thus, the minimizing ( theta ) is: [ boxed{theta_j = frac{-b_j - lambda}{2a_j}} ]2. The gradient of the prediction error with respect to the input data ( x ) is: [ nabla_x E(theta, x) = left( sum_{i=1}^n d_{i1}, sum_{i=1}^n d_{i2}, ldots, sum_{i=1}^n d_{im} right)^T ] This gradient indicates the sensitivity of the prediction error to changes in each input feature, which is crucial for enhancing the algorithm's robustness. [ boxed{nabla_x E(theta, x) = left( sum_{i=1}^n d_{i1}, sum_{i=1}^n d_{i2}, ldots, sum_{i=1}^n d_{im} right)^T} ]</think>

🧠:<think>Okay, so I have this problem about an IT technician who helps users reset their passwords. Each user has a probability of forgetting their password on any given day, and this follows a Poisson distribution with a mean rate λ. There are two parts to the problem.Starting with the first part: If the technician helps k users reset their passwords on a particular day, what is the probability distribution of k? I need to express this in terms of n and λ.Hmm, let me recall what a Poisson distribution is. It's a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The formula is P(k) = (λ^k * e^(-λ)) / k!But wait, in this case, each user has their own Poisson process. So, each user has a probability of forgetting their password, which is Poisson with mean λ. So, for each user, the number of times they forget their password in a day is Poisson(λ). But actually, the problem says the probability that any given user forgets their password on any given day follows a Poisson distribution. Hmm, that might mean that the number of forgotten passwords per user is Poisson distributed with mean λ.But actually, when they say "the probability that any given user forgets their password on any given day follows a Poisson distribution," that might be a bit confusing. Because Poisson distribution is for counts, not probabilities. So, maybe they mean that the number of times a user forgets their password in a day is Poisson distributed with mean λ. So, each user can forget their password multiple times in a day, but each time, the technician has to reset it. So, the total number of resets per user is Poisson(λ), and since there are n users, the total number of resets k is the sum of n independent Poisson(λ) random variables.I remember that the sum of independent Poisson random variables is also Poisson, with the parameter being the sum of the individual parameters. So, if each user contributes a Poisson(λ) number of resets, then the total number k is Poisson(nλ). So, the probability distribution of k is Poisson with parameter nλ.Wait, let me think again. If each user has a Poisson(λ) number of resets, then the total across n users is Poisson(nλ). So, yes, the distribution of k is Poisson(nλ). Therefore, the probability P(k) = ( (nλ)^k * e^(-nλ) ) / k!.That seems right. So, part 1 is answered.Moving on to part 2: Assuming the IT technician can handle a maximum of m password resets per day before becoming overloaded, what is the probability that the technician becomes overloaded on any given day?So, becoming overloaded means that the number of resets k exceeds m. So, the probability that k > m.Given that k follows a Poisson(nλ) distribution, the probability that k > m is equal to 1 minus the probability that k ≤ m.So, P(k > m) = 1 - P(k ≤ m) = 1 - Σ (from i=0 to m) [ ( (nλ)^i * e^(-nλ) ) / i! ]So, that's the expression for the probability of being overloaded.Alternatively, sometimes people use the complement of the cumulative distribution function, which is exactly what this is.But just to make sure, let me think if there's another way to model this. Since each user's number of resets is Poisson(λ), and the total is Poisson(nλ), then yes, the probability that the total exceeds m is 1 minus the sum from 0 to m of the Poisson probabilities.Alternatively, if n is large and λ is small, such that nλ is moderate, the Poisson distribution is a good approximation. But in this case, since we're given that each user follows Poisson(λ), and we're summing n of them, it's exactly Poisson(nλ), so we don't need any approximation.So, summarizing:1. The distribution of k is Poisson(nλ), so P(k) = ( (nλ)^k e^{-nλ} ) / k!2. The probability of overload is 1 - Σ_{i=0}^m [ ( (nλ)^i e^{-nλ} ) / i! ]I think that's correct.Final Answer1. The probability distribution of ( k ) is given by ( boxed{P(k) = frac{(nlambda)^k e^{-nlambda}}{k!}} ).2. The probability that the technician becomes overloaded is ( boxed{1 - sum_{i=0}^{m} frac{(nlambda)^i e^{-nlambda}}{i!}} ).</think>