answer：Alright, so I have this problem about a public health official analyzing a new vaccine's effectiveness and public trust. It's divided into two parts. Let me try to tackle each part step by step.Starting with part 1: The vaccine has an efficacy rate of ( p ) percent. They conducted a study with 10,000 individuals, 7,500 of whom are vaccinated. They define a random variable ( X ) as the number of vaccinated individuals who do not contract the disease. It says the probability follows a binomial distribution, so I need to express ( P(X = k) ) and find the expected number. Then, see how it changes if ( p ) increases by 10%.Okay, binomial distribution. So, for a binomial distribution, the probability mass function is given by:[P(X = k) = C(n, k) times p^k times (1 - p)^{n - k}]Where ( n ) is the number of trials, ( k ) is the number of successes, ( p ) is the probability of success on a single trial.In this case, each vaccinated individual is a trial. So, ( n = 7500 ). A "success" here would be that the vaccine prevents the disease, so the individual does not contract it. So, the probability of success is ( p ) (but wait, is it ( p ) or ( 1 - p )?). Wait, the efficacy rate is ( p ), which is the probability that the vaccine prevents the disease. So, the probability that a vaccinated individual does not contract the disease is ( p ). So, yes, ( p ) is the probability of success.Therefore, the probability ( P(X = k) ) is:[P(X = k) = C(7500, k) times p^k times (1 - p)^{7500 - k}]Got that.Now, the expected number of vaccinated individuals who do not contract the disease. For a binomial distribution, the expectation is ( E[X] = n times p ). So, substituting the numbers:[E[X] = 7500 times p]That's straightforward.Then, the question is, how does this expected number change if the efficacy rate ( p ) increases by 10%? Hmm, so does that mean ( p ) becomes ( p + 0.10 ), or does it become ( 1.10p )?Wait, the wording is "increases by 10%". So, if ( p ) is, say, 80%, increasing by 10% would make it 90%, right? So, it's an absolute increase of 10 percentage points. But sometimes, "increases by 10%" can be interpreted as multiplying by 1.10. Hmm.Wait, the problem says "if the efficacy rate ( p ) increases by 10%". So, if ( p ) is a percentage, like 80%, then increasing by 10% would mean 80% + 10% = 90%. So, it's an absolute increase, not a multiplicative one. So, the new efficacy rate is ( p + 0.10 ).But wait, actually, ( p ) is given as a percentage, so if ( p = 80% ), increasing by 10% would be 90%, but if ( p ) is 50%, increasing by 10% would be 60%. So, yes, it's an absolute increase.But wait, let me confirm. If ( p ) is a probability, it's a value between 0 and 1. But the problem says "efficacy rate of ( p ) (as a percentage)", so ( p ) is a percentage, so for example, 80% efficacy would be ( p = 0.80 ). So, increasing by 10% would mean ( p + 0.10 ), but we have to ensure that it doesn't exceed 1. So, if ( p ) is 0.90, adding 0.10 would make it 1.00, which is acceptable.So, the new expected number would be ( 7500 times (p + 0.10) ). So, the change in expectation is ( 7500 times 0.10 = 750 ). So, the expected number increases by 750.Wait, but hold on. If ( p ) is a percentage, is it 0.10 or 10? Because if ( p ) is given as a percentage, like 80%, then 10% increase would be 80% + 10% = 90%, so 0.90. So, in terms of the variable ( p ), which is a percentage, the increase is 0.10 in decimal terms.Wait, actually, hold on. If ( p ) is a percentage, say 80%, that's 0.80 in decimal. So, increasing by 10% would be 0.80 + 0.10 = 0.90, which is 90%. So, yes, the increase is 0.10 in decimal terms, which is 10 percentage points.Alternatively, if it were a 10% increase relative to ( p ), it would be ( p times 1.10 ). But the wording is "increases by 10%", which is more likely an absolute increase, not a relative one.So, I think the correct interpretation is that the new efficacy rate is ( p + 0.10 ), so the expected number becomes ( 7500 times (p + 0.10) ). Therefore, the expected number increases by 750.Wait, but let me think again. If ( p ) is a percentage, like 80%, then 10% of that is 8%, so a 10% increase would be 88%. Hmm, but that's a different interpretation. So, the problem is a bit ambiguous.Wait, the problem says "the efficacy rate ( p ) (as a percentage)". So, ( p ) is a percentage, meaning it's a value between 0 and 100, not 0 and 1. So, for example, if ( p = 80 ), it's 80%. So, increasing ( p ) by 10% would mean 80 + 10 = 90, so 90% efficacy. So, in decimal terms, it's 0.90.Alternatively, if it's a 10% increase relative to ( p ), it would be ( p times 1.10 ). So, 80 * 1.10 = 88, so 88% efficacy.But the wording is "increases by 10%", which is more likely an absolute increase. So, adding 10 percentage points. So, 80% becomes 90%, 70% becomes 80%, etc.Therefore, I think the correct interpretation is that the new efficacy rate is ( p + 10 ) percentage points, which in decimal is ( p + 0.10 ). So, the expected number becomes ( 7500 times (p + 0.10) ), so an increase of 750.Therefore, the expected number changes from ( 7500p ) to ( 7500(p + 0.10) ), which is an increase of 750.So, that's part 1.Moving on to part 2: Assessing public trust levels, represented by a trust index ( T ), which is normally distributed with mean ( mu = 65 ) and standard deviation ( sigma = 10 ). The goal is to increase public trust so that at least 95% of the population has a trust index above 50. We need to determine the required increase in the mean ( mu ) to achieve this, keeping ( sigma ) constant.Alright, so currently, ( T sim N(65, 10^2) ). We need to find the new mean ( mu' ) such that ( P(T > 50) geq 0.95 ).Wait, actually, the wording is "at least 95% of the population has a trust index above 50". So, ( P(T > 50) geq 0.95 ).But wait, in a normal distribution, the probability that ( T > 50 ) is related to how far 50 is from the mean in terms of standard deviations.So, let's denote ( Z = frac{T - mu}{sigma} ), which is the standard normal variable.We need ( P(T > 50) geq 0.95 ). So, ( P(T > 50) = Pleft(Z > frac{50 - mu}{10}right) geq 0.95 ).We know that for the standard normal distribution, ( P(Z > z) = 0.05 ) when ( z = 1.645 ). Because the 95th percentile is at 1.645.Wait, actually, ( P(Z > 1.645) = 0.05 ), so ( P(Z leq 1.645) = 0.95 ). So, if we want ( P(T > 50) geq 0.95 ), that would mean ( P(T leq 50) leq 0.05 ). So, the 5th percentile is at 50.Therefore, we need:[frac{50 - mu'}{10} = -1.645]Wait, because if ( P(T leq 50) = 0.05 ), then ( Z = frac{50 - mu'}{10} = -1.645 ). Because the Z-score corresponding to 5% is -1.645.So, solving for ( mu' ):[frac{50 - mu'}{10} = -1.645 50 - mu' = -16.45 mu' = 50 + 16.45 mu' = 66.45]Wait, but hold on. Currently, the mean is 65. So, the required mean is 66.45. Therefore, the increase needed is ( 66.45 - 65 = 1.45 ).So, the mean needs to increase by approximately 1.45 to achieve that 95% of the population has a trust index above 50.Wait, let me double-check.We have ( T sim N(mu', 10^2) ). We want ( P(T > 50) geq 0.95 ). So, ( P(T leq 50) leq 0.05 ). So, 50 is the 5th percentile of the distribution.The Z-score for the 5th percentile is -1.645. So,[Z = frac{50 - mu'}{10} = -1.645 50 - mu' = -16.45 mu' = 50 + 16.45 = 66.45]Yes, that seems correct. So, the mean needs to be increased from 65 to 66.45, which is an increase of 1.45.But wait, is that the only consideration? Let me think.Alternatively, sometimes people might interpret "at least 95% above 50" as the 95th percentile being above 50. But in that case, it's a different calculation.Wait, if we want at least 95% above 50, that would mean that 50 is the 5th percentile, not the 95th. Because if 95% are above 50, then 5% are below 50.So, yes, that's correct. So, 50 is the 5th percentile, which corresponds to Z = -1.645.Therefore, the calculation is correct.So, the required increase in the mean is 1.45.But let me make sure about the direction. If the mean increases, the distribution shifts to the right, so more people will have higher trust indices. So, to have 95% above 50, we need the 5th percentile at 50, which requires shifting the mean to the right.Yes, that makes sense.So, summarizing part 2: The current mean is 65, and to have 95% above 50, the mean needs to be 66.45, so an increase of 1.45.Therefore, the required increase is approximately 1.45.But since the problem mentions the trust index is a whole number (from 0 to 100), but the mean can be a decimal. So, 1.45 is acceptable.Alternatively, if we need to round it, it would be approximately 1.45, which is about 1.5. But since the question doesn't specify rounding, we can keep it as 1.45.So, that's my thought process.**Final Answer**1. The probability is ( P(X = k) = binom{7500}{k} p^k (1 - p)^{7500 - k} ) and the expected number is ( boxed{7500p} ). If ( p ) increases by 10%, the expected number becomes ( boxed{7500(p + 0.10)} ).2. The required increase in the mean trust index is ( boxed{1.45} ).

answer：Okay, so I have this problem where a political science major and a software engineer are working together to analyze how social media influences public opinion during elections. They're using network theory and differential equations for their model. I need to tackle two parts here.Starting with part 1: They've given me a differential equation that models the opinion dynamics of each individual in the network. The equation is:[frac{dx_i(t)}{dt} = -alpha x_i(t) + beta sum_{j in N(i)} A_{ij} (x_j(t) - x_i(t))]Here, (x_i(t)) is the opinion of individual (i) at time (t), (alpha) and (beta) are constants, (A_{ij}) is the adjacency matrix, and (N(i)) is the set of neighbors of node (i). I need to analyze the stability of this system by finding the eigenvalues of the Laplacian matrix (L = D - A), where (D) is the degree matrix.Hmm, okay. So first, let me recall what the Laplacian matrix is. The Laplacian matrix (L) is defined as (D - A), where (D) is a diagonal matrix with the degrees of the nodes on the diagonal, and (A) is the adjacency matrix. The eigenvalues of the Laplacian matrix are important because they tell us about the connectivity and stability of the network.Now, the given differential equation seems to model how each individual's opinion changes over time. The term (-alpha x_i(t)) suggests some sort of decay or self-inhibition, while the second term (beta sum_{j in N(i)} A_{ij} (x_j(t) - x_i(t))) represents the influence from neighboring nodes. So, each node is influenced by its neighbors, and there's also a self-inhibition term.I think I can rewrite this differential equation in matrix form. Let me try that. If I consider the vector (x(t)) where each component is (x_i(t)), then the equation becomes:[frac{dx(t)}{dt} = -alpha x(t) + beta (A x(t) - D x(t))]Wait, because (sum_{j in N(i)} A_{ij} (x_j(t) - x_i(t))) can be written as (A x(t) - D x(t)), since (D x(t)) would be the diagonal matrix multiplied by (x(t)), effectively summing over the neighbors for each node.So, simplifying that, we get:[frac{dx(t)}{dt} = (-alpha I + beta (A - D)) x(t)]But (A - D) is just (-L), since (L = D - A). So substituting that in:[frac{dx(t)}{dt} = (-alpha I - beta L) x(t)]So, the system can be written as:[frac{dx(t)}{dt} = (-alpha I - beta L) x(t)]This is a linear system, and its stability is determined by the eigenvalues of the matrix (-alpha I - beta L). If all eigenvalues have negative real parts, the system is stable, and the opinions will converge to a fixed point.The eigenvalues of (-alpha I - beta L) can be found by noting that if (lambda) is an eigenvalue of (L), then (-alpha - beta lambda) is an eigenvalue of the system matrix. So, the eigenvalues of the system are (-alpha - beta lambda), where (lambda) are the eigenvalues of the Laplacian matrix (L).Now, the Laplacian matrix (L) is symmetric and positive semi-definite, so all its eigenvalues are real and non-negative. The smallest eigenvalue is 0, corresponding to the eigenvector where all components are equal (the steady state where everyone has the same opinion). The other eigenvalues are positive.So, for the system to be stable, we need all eigenvalues of (-alpha I - beta L) to have negative real parts. Since (lambda geq 0), then (-alpha - beta lambda) will be negative if (alpha > 0) and (beta > 0). Because both (alpha) and (beta) are positive constants, each eigenvalue (-alpha - beta lambda) will be negative, ensuring that the system is stable.Therefore, the system is stable, and the opinions will converge to a consensus or some fixed point depending on the initial conditions.Moving on to part 2: The network (G) is a small-world network with (n) nodes, where the probability of two nodes being connected is (p = frac{log(n)}{n}). I need to estimate the expected number of connections (edges) in the network and discuss how changes in (p) affect the spread of political opinions according to the differential equation.First, the expected number of edges in a network is given by (E = frac{n(n-1)}{2} p), since each pair of nodes can be connected with probability (p). For large (n), this approximates to (E approx frac{n^2}{2} p).Given (p = frac{log(n)}{n}), substituting that in:[E approx frac{n^2}{2} cdot frac{log(n)}{n} = frac{n log(n)}{2}]So, the expected number of edges is approximately (frac{n log(n)}{2}).Now, how does changing (p) affect the spread of opinions? The parameter (p) determines the density of the network. A higher (p) means more connections, making the network more densely connected.In the differential equation, the Laplacian matrix (L) plays a crucial role in the dynamics. The eigenvalues of (L) determine the convergence rate and stability. In a more connected network (higher (p)), the Laplacian tends to have larger eigenvalues, which would make the term (-beta lambda) more negative, potentially speeding up the convergence to the steady state.Additionally, in a small-world network, which typically has high clustering and short path lengths, increasing (p) can lead to a more efficient spread of opinions. This is because information can traverse the network more quickly through the many connections, leading to faster consensus or synchronization of opinions.On the other hand, if (p) is too low, the network becomes sparse, and the spread of opinions might be slower or even get stuck in local clusters without reaching a global consensus. The system might exhibit more variability or take longer to stabilize.So, in summary, increasing (p) leads to a denser network, which can facilitate faster and more efficient spread of political opinions, potentially leading to quicker consensus. Conversely, lower (p) results in a sparser network, which might hinder the spread and lead to slower convergence or fragmented opinions.I think that covers both parts. I need to make sure I didn't make any mistakes in the matrix manipulations or the reasoning about the eigenvalues. Let me double-check.In part 1, I rewrote the differential equation correctly in matrix form, identified the Laplacian, and correctly noted that the eigenvalues of the system matrix are (-alpha - beta lambda), which are all negative given positive (alpha) and (beta). So, the system is stable.In part 2, I calculated the expected number of edges correctly using the formula for expected edges in a random graph, substituted (p = frac{log(n)}{n}), and then discussed the implications of (p) on the network's connectivity and the spread of opinions. That seems solid.Yeah, I think that's all good.**Final Answer**1. The system is stable, and the eigenvalues of the Laplacian matrix (L) determine the convergence rate. The stability is ensured by the negative eigenvalues of the system matrix, leading to (boxed{text{Stable System}}).2. The expected number of edges is (boxed{frac{n log(n)}{2}}), and increasing (p) enhances the spread of opinions by increasing network connectivity.

answer：First, I need to determine the number of Pension Benefit (PB) cases, denoted as ( x ). According to the problem, the number of Disability Compensation (DC) cases is twice the number of PB cases, so DC cases are ( 2x ).Next, I'll calculate the number of successful cases for each type. For DC cases, the success rate is 75%, so successful DC cases are ( 0.75 times 2x = 1.5x ). For PB cases, the success rate is 60%, so successful PB cases are ( 0.60 times x = 0.6x ).The total number of successful cases is given as 300. Therefore, I can set up the equation:[1.5x + 0.6x = 300]Combining like terms:[2.1x = 300]Solving for ( x ):[x = frac{300}{2.1} = 142.857]Since the number of cases must be a whole number, I'll round ( x ) to 143. Thus, there are 143 PB cases and ( 2 times 143 = 286 ) DC cases.For part b, the success rates for both DC and PB cases will increase by 10%. The new success rates will be 85% for DC and 70% for PB. Using the same number of cases:- Successful DC cases: ( 0.85 times 286 = 243.1 )- Successful PB cases: ( 0.70 times 143 = 100.1 )Adding these together gives:[243.1 + 100.1 = 343.2]Rounding to the nearest whole number, the expected number of successful cases next year is 343.

answer：Okay, so I need to help my client understand their mortgage options. They’re looking at a 500,000 home and have two mortgage choices: a 30-year fixed-rate at 3.5% and a 15-year fixed-rate at 2.8%. Both require a 20% down payment. I need to calculate the monthly payments, total interest paid, and compare the two options.First, let me figure out the loan amount after the down payment. The down payment is 20% of 500,000. So, 20% of 500,000 is 0.2 * 500,000 = 100,000. That means the loan amount L is 500,000 - 100,000 = 400,000 for both mortgages.Next, I need to calculate the monthly payment for each option. The formula given is P = (L * r * (1 + r)^n) / ((1 + r)^n - 1). I remember that r is the monthly interest rate, so I need to convert the annual rate to monthly by dividing by 12. Also, n is the total number of payments, which is the term in years multiplied by 12.Starting with the 30-year mortgage at 3.5% annual interest. The monthly rate r is 3.5% / 12. Let me convert 3.5% to decimal first: 3.5 / 100 = 0.035. So, r = 0.035 / 12 ≈ 0.00291667. The number of payments n is 30 * 12 = 360.Plugging into the formula: P = (400,000 * 0.00291667 * (1 + 0.00291667)^360) / ((1 + 0.00291667)^360 - 1). Hmm, calculating (1 + 0.00291667)^360 might be a bit tricky. I think I can use a calculator for that. Let me compute that exponent first.Using a calculator, (1.00291667)^360 ≈ 2.42726. So, the numerator becomes 400,000 * 0.00291667 * 2.42726. Let me compute 400,000 * 0.00291667 first. That's approximately 400,000 * 0.00291667 ≈ 1,166.668. Then multiply by 2.42726: 1,166.668 * 2.42726 ≈ 2,830.52.The denominator is 2.42726 - 1 = 1.42726. So, P ≈ 2,830.52 / 1.42726 ≈ 1,982.07. So, the monthly payment for the 30-year mortgage is approximately 1,982.07.Now, for the 15-year mortgage at 2.8% annual interest. The monthly rate r is 2.8% / 12. Converting 2.8% to decimal: 0.028. So, r = 0.028 / 12 ≈ 0.00233333. The number of payments n is 15 * 12 = 180.Using the same formula: P = (400,000 * 0.00233333 * (1 + 0.00233333)^180) / ((1 + 0.00233333)^180 - 1). Let me compute (1.00233333)^180. Using a calculator, that's approximately 1.48985.Numerator: 400,000 * 0.00233333 ≈ 933.332. Then multiply by 1.48985: 933.332 * 1.48985 ≈ 1,390.00.Denominator: 1.48985 - 1 = 0.48985. So, P ≈ 1,390.00 / 0.48985 ≈ 2,838.00. Wait, that seems high. Let me double-check my calculations.Wait, 400,000 * 0.00233333 is indeed 933.332. Then 933.332 * 1.48985 is approximately 1,390.00. Then 1,390 / 0.48985 is approximately 2,838. Hmm, but that seems higher than I expected. Maybe I made a mistake in the exponent calculation.Let me recalculate (1.00233333)^180. Maybe using a more precise method. Alternatively, I can use the formula for monthly payment, but perhaps I can use an online calculator or a more accurate estimation. Alternatively, maybe I can use logarithms or another method.Alternatively, perhaps I can use the present value of an annuity formula. But I think my initial calculation might be correct. Let me check with another approach.Alternatively, I can use the formula step by step. Let me compute (1 + r)^n for the 15-year mortgage. r = 0.00233333, n = 180. So, (1.00233333)^180.Using natural logarithm: ln(1.00233333) ≈ 0.002328. Multiply by 180: 0.002328 * 180 ≈ 0.41904. Then exponentiate: e^0.41904 ≈ 1.519. Wait, that's different from my previous calculation of 1.48985. Hmm, maybe my initial exponent was off.Let me use a calculator for (1.00233333)^180. Let me compute it step by step. Alternatively, I can use the rule of 72 or another approximation, but perhaps it's better to use a calculator function.Assuming I have a calculator, let me compute (1.00233333)^180. Let me compute it as follows:First, compute ln(1.00233333) ≈ 0.002328. Multiply by 180: 0.002328 * 180 ≈ 0.41904. Then e^0.41904 ≈ 1.519. So, approximately 1.519.So, the numerator is 400,000 * 0.00233333 * 1.519 ≈ 400,000 * 0.00233333 ≈ 933.332, then 933.332 * 1.519 ≈ 1,416.67.Denominator: 1.519 - 1 = 0.519. So, P ≈ 1,416.67 / 0.519 ≈ 2,729.62.Wait, that's different from my previous calculation. I think I made a mistake earlier. Let me verify.Alternatively, perhaps I should use a more precise calculation for (1.00233333)^180. Let me use a calculator:1.00233333^180 ≈ e^(180 * ln(1.00233333)) ≈ e^(180 * 0.002328) ≈ e^0.41904 ≈ 1.519.So, the numerator is 400,000 * 0.00233333 * 1.519 ≈ 400,000 * 0.00233333 ≈ 933.332, then 933.332 * 1.519 ≈ 1,416.67.Denominator: 1.519 - 1 = 0.519. So, P ≈ 1,416.67 / 0.519 ≈ 2,729.62.Wait, that still seems high. Let me check with another method. Alternatively, perhaps I can use the formula for monthly payment more accurately.Alternatively, I can use the formula:P = L * [r(1 + r)^n] / [(1 + r)^n - 1]So, for the 15-year mortgage:r = 0.028 / 12 ≈ 0.00233333n = 180(1 + r)^n = (1.00233333)^180 ≈ 1.519So, numerator: 400,000 * 0.00233333 * 1.519 ≈ 400,000 * 0.0035444 ≈ 1,417.76Denominator: 1.519 - 1 = 0.519So, P ≈ 1,417.76 / 0.519 ≈ 2,729.62Wait, that seems consistent. So, the monthly payment for the 15-year mortgage is approximately 2,729.62.Wait, but earlier I thought it was around 2,838, but that was due to an incorrect exponent calculation. So, correcting that, it's approximately 2,729.62.Now, let me calculate the total interest paid for each option.For the 30-year mortgage:Monthly payment is approximately 1,982.07. Over 360 months, total payments are 1,982.07 * 360 ≈ 713,545.20.The loan amount is 400,000, so total interest paid is 713,545.20 - 400,000 = 313,545.20.For the 15-year mortgage:Monthly payment is approximately 2,729.62. Over 180 months, total payments are 2,729.62 * 180 ≈ 491,331.60.Total interest paid is 491,331.60 - 400,000 = 91,331.60.Comparing the two, the 15-year mortgage results in significantly less total interest paid. The difference is 313,545.20 - 91,331.60 = 222,213.60.So, the 15-year mortgage saves the client approximately 222,213.60 in interest over the life of the loan, but the monthly payments are higher: 2,729.62 vs. 1,982.07.I should present this information clearly to the client, explaining the trade-offs between lower monthly payments with the 30-year option versus significant interest savings with the 15-year option.

answer：Okay, so I have this problem where a nurse is analyzing patient data for a report to a politician. The data is about recovery times after a specific medical procedure. There are two parts to the problem. Let me try to work through each step carefully.First, the problem says that the recovery times follow a normal distribution with a mean of 12 days and a standard deviation of 3 days. So, for part 1, I need to calculate the probability that a randomly selected patient will recover in less than 10 days.Hmm, okay. Since it's a normal distribution, I can use the Z-score formula to standardize the value and then use the standard normal distribution table to find the probability. The Z-score formula is:Z = (X - μ) / σWhere X is the value we're interested in, which is 10 days. μ is the mean, 12 days, and σ is the standard deviation, 3 days.Let me plug in the numbers:Z = (10 - 12) / 3 = (-2) / 3 ≈ -0.6667So, the Z-score is approximately -0.6667. Now, I need to find the probability that Z is less than -0.6667. I can look this up in the standard normal distribution table or use a calculator.Looking at the Z-table, for Z = -0.67, the cumulative probability is about 0.2514. So, the probability that a patient recovers in less than 10 days is approximately 25.14%.Wait, let me double-check that. Sometimes, the Z-table gives the area to the left of the Z-score, which is exactly what we need here. So, yes, for Z = -0.67, it's about 0.2514. So, that seems correct.Okay, moving on to part 2. The nurse wants to test if the recovery time has decreased after improvements in medical education. They took a new sample of 50 patients with an average recovery time of 11 days and a standard deviation of 2.5 days. We need to conduct a hypothesis test at the 5% significance level.Alright, so this is a hypothesis test for the mean. Since the sample size is 50, which is greater than 30, we can use the Z-test. But wait, the population standard deviation is not given, only the sample standard deviation is provided. Hmm, so actually, should we use a t-test instead?Wait, the original distribution was normal, so even with a sample size of 50, if the population standard deviation is unknown, we should use a t-test. But sometimes, with large sample sizes, people approximate with the Z-test. Let me think.The problem says the original distribution is normal, so the sampling distribution of the mean will also be normal regardless of the sample size, especially since the original is normal. So, maybe we can use the Z-test here. Alternatively, since the sample size is 50, which is large, the t-test and Z-test will give similar results. But to be precise, since we don't know the population standard deviation after the improvement, we should use the sample standard deviation, which would make it a t-test.But wait, the original population had a standard deviation of 3, but after the improvement, the sample has a standard deviation of 2.5. So, is the population standard deviation known or not? The problem doesn't specify, so I think we have to assume it's unknown, hence a t-test.But let me check the question again. It says, "a new sample of 50 patients shows an average recovery time of 11 days with a standard deviation of 2.5 days." So, the standard deviation given is for the sample, not the population. Therefore, we should use a t-test.However, since the sample size is 50, which is large, the t-test and Z-test will be very similar. But to be accurate, I think we should use the t-test.Alright, so let's set up our hypotheses.Null hypothesis (H0): μ = 12 days (the mean recovery time has not decreased)Alternative hypothesis (H1): μ < 12 days (the mean recovery time has decreased)This is a one-tailed test because we're specifically testing if the mean has decreased.Now, the significance level is 5%, so α = 0.05.We need to calculate the test statistic. For a t-test, the formula is:t = (x̄ - μ) / (s / √n)Where x̄ is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.Plugging in the numbers:x̄ = 11 daysμ = 12 dayss = 2.5 daysn = 50So,t = (11 - 12) / (2.5 / √50) = (-1) / (2.5 / 7.0711) ≈ (-1) / (0.3536) ≈ -2.828So, the t-score is approximately -2.828.Now, we need to find the critical value for a one-tailed t-test with α = 0.05 and degrees of freedom (df) = n - 1 = 49.Looking up the t-table for df = 49 and α = 0.05 (one-tailed), the critical value is approximately -1.677. Wait, no, hold on. The critical value is the value that our test statistic needs to be less than (since it's a left-tailed test) to reject the null hypothesis.But actually, the critical value is the t-value that corresponds to 0.05 in the left tail. So, it's negative. Let me check the t-table.For df = 49, the critical value at α = 0.05 is approximately -1.677. So, if our calculated t-score is less than -1.677, we reject the null hypothesis.Our calculated t-score is -2.828, which is less than -1.677. Therefore, we reject the null hypothesis.Alternatively, we can calculate the p-value. The p-value is the probability of observing a t-score as extreme as -2.828 with 49 degrees of freedom.Using a t-table or calculator, for t = -2.828 and df = 49, the p-value is approximately 0.003 (since 2.828 is close to 2.81, which is around 0.005 for two-tailed, so one-tailed would be half that, around 0.0025). So, p ≈ 0.003.Since the p-value (0.003) is less than α (0.05), we reject the null hypothesis.Therefore, there is statistically significant evidence at the 5% significance level to conclude that the mean recovery time has decreased after the improvements in medical education.Wait, let me make sure I didn't make any calculation errors. So, t = (11 - 12)/(2.5/sqrt(50)).Calculating the denominator: 2.5 / sqrt(50) ≈ 2.5 / 7.0711 ≈ 0.3536.Then, t = -1 / 0.3536 ≈ -2.828. Yes, that's correct.Degrees of freedom is 50 - 1 = 49. Critical value is -1.677. Since -2.828 < -1.677, we reject H0.Alternatively, using the p-value approach, p ≈ 0.003, which is less than 0.05, so we reject H0.Therefore, the conclusion is that the mean recovery time has statistically significantly decreased.Wait, but just to be thorough, let me consider if using a Z-test would have given a different result. If we were to use the Z-test, the formula would be:Z = (x̄ - μ) / (σ / √n)But we don't have σ, only s. So, if we incorrectly used σ = 3, then:Z = (11 - 12)/(3 / sqrt(50)) ≈ (-1)/(0.4243) ≈ -2.357Then, the critical Z-value for α = 0.05 is -1.645. So, -2.357 < -1.645, so we would still reject H0.But since we don't know σ, it's more appropriate to use the t-test. However, both tests lead to the same conclusion in this case.So, either way, the result is significant.Therefore, the nurse can conclude that the mean recovery time has decreased after the improvements.**Final Answer**1. The probability is boxed{0.2514}.2. The mean recovery time has statistically significantly decreased, so we reject the null hypothesis. The final answer is boxed{text{Reject } H_0}.

answer：Alright, so I've got these two statistics problems to solve, and I need to figure them out step by step. Let me start with the first one.**Problem 1:** A social worker wants to find the probability that a randomly selected client will have a reduction of at least 2 points in their symptom severity score. The mean reduction is 2.5 points with a standard deviation of 0.8. The change in scores follows a normal distribution.Okay, so I remember that when dealing with probabilities in a normal distribution, we can use z-scores. The z-score formula is (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation.In this case, we want the probability that the reduction is at least 2 points. So, X is 2, μ is 2.5, and σ is 0.8. Let me plug those numbers into the formula.Z = (2 - 2.5) / 0.8 = (-0.5) / 0.8 = -0.625Hmm, so the z-score is -0.625. Now, I need to find the probability that Z is less than or equal to -0.625 because we want the probability that the reduction is at least 2 points. Wait, actually, no. If we're looking for the probability that the reduction is at least 2, that corresponds to the area to the right of 2 in the distribution. Since the z-score is negative, that area is actually the complement of the area to the left of -0.625.Let me recall how to find this probability. I can use a z-table or a calculator. Since I don't have a z-table here, I'll try to remember that a z-score of -0.625 is approximately between -0.6 and -0.7. Looking up -0.6 in the z-table gives about 0.2743, and -0.7 gives about 0.2420. Since -0.625 is closer to -0.6, maybe around 0.264? Wait, actually, I think it's better to use linear interpolation.The difference between -0.6 and -0.7 is 0.1 in z-score, which corresponds to a difference in probability of 0.2743 - 0.2420 = 0.0323. Since -0.625 is 0.025 away from -0.6, which is a quarter of the way from -0.6 to -0.7. So, 0.0323 * 0.25 = 0.008075. Therefore, the probability at -0.625 is approximately 0.2743 - 0.008075 = 0.2662.But wait, since we're looking for the probability that the reduction is at least 2, which is P(X ≥ 2), and since the z-score is negative, this corresponds to the area to the right of -0.625, which is 1 - 0.2662 = 0.7338. So, approximately 73.38% probability.But let me double-check. Alternatively, I can use the standard normal distribution function. If I have a calculator, I can compute the cumulative distribution function (CDF) at z = -0.625. The CDF gives P(Z ≤ z), so P(Z ≤ -0.625) is approximately 0.2660, so the probability we want is 1 - 0.2660 = 0.7340, which is about 73.4%.That seems reasonable. So, the probability is approximately 73.4%.**Problem 2:** Now, the social worker notices that clients with more than one co-occurring disorder have a different mean reduction. 40% of the clients have more than one co-occurring disorder with a mean reduction of 3 points and standard deviation 0.9. The remaining 60% have only one disorder with a mean reduction of 2 points and standard deviation 0.7. We need to calculate the overall expected mean reduction for the entire sample of 100 clients.Alright, so this is about calculating the weighted average of the mean reductions. Since 40% have a mean of 3 and 60% have a mean of 2, the overall mean is just 0.4*3 + 0.6*2.Let me compute that: 0.4*3 = 1.2, and 0.6*2 = 1.2. So, 1.2 + 1.2 = 2.4. Therefore, the overall expected mean reduction is 2.4 points.Wait, is that all? It seems straightforward. The standard deviations are given, but since we're only asked for the mean, we don't need them here. If we were asked for the overall standard deviation, that would be more complicated, but not in this case.So, yeah, the overall mean is 2.4.**Final Answer**1. The probability is boxed{0.734}.2. The overall expected mean reduction is boxed{2.4}.