🧠:<think>Okay, so I have this problem where a journalist is looking into how the number of resources allocated to criminal cases affects the time it takes to solve them. There are 10 cases with data on days taken and resources used. The first part is to calculate the Pearson correlation coefficient, and the second part is to do a linear regression to model the relationship and make a prediction.Starting with the first part, Pearson correlation. I remember that Pearson's r measures the linear relationship between two variables. It ranges from -1 to 1, where -1 is a perfect negative correlation, 0 is no correlation, and 1 is a perfect positive correlation.So, I need to calculate r using the given data. The formula for Pearson's r is:r = [nΣ(xy) - ΣxΣy] / sqrt([nΣx² - (Σx)²][nΣy² - (Σy)²])Where n is the number of data points, which is 10 here.First, I need to compute the sums of x, y, xy, x², and y².Let me list out the data:Case 1: x=5, y=30Case 2: x=7, y=45Case 3: x=4, y=20Case 4: x=6, y=35Case 5: x=8, y=50Case 6: x=5, y=25Case 7: x=7, y=40Case 8: x=9, y=55Case 9: x=6, y=30Case 10: x=10, y=60So, let me tabulate these:x: 5,7,4,6,8,5,7,9,6,10y:30,45,20,35,50,25,40,55,30,60Now, I need to compute Σx, Σy, Σxy, Σx², Σy².Let me compute each step by step.First, Σx:5 + 7 + 4 + 6 + 8 + 5 + 7 + 9 + 6 + 10Let me add them:5 + 7 = 1212 + 4 = 1616 + 6 = 2222 + 8 = 3030 + 5 = 3535 + 7 = 4242 + 9 = 5151 + 6 = 5757 + 10 = 67So, Σx = 67Next, Σy:30 + 45 + 20 + 35 + 50 + 25 + 40 + 55 + 30 + 60Adding step by step:30 + 45 = 7575 + 20 = 9595 + 35 = 130130 + 50 = 180180 + 25 = 205205 + 40 = 245245 + 55 = 300300 + 30 = 330330 + 60 = 390So, Σy = 390Now, Σxy:Multiply each x and y pair and sum them up.Case 1: 5*30=150Case 2:7*45=315Case3:4*20=80Case4:6*35=210Case5:8*50=400Case6:5*25=125Case7:7*40=280Case8:9*55=495Case9:6*30=180Case10:10*60=600Now, sum these products:150 + 315 = 465465 + 80 = 545545 + 210 = 755755 + 400 = 11551155 + 125 = 12801280 + 280 = 15601560 + 495 = 20552055 + 180 = 22352235 + 600 = 2835So, Σxy = 2835Next, Σx²:Compute each x squared and sum.5²=257²=494²=166²=368²=645²=257²=499²=816²=3610²=100Now, sum these:25 + 49 = 7474 + 16 = 9090 + 36 = 126126 + 64 = 190190 + 25 = 215215 + 49 = 264264 + 81 = 345345 + 36 = 381381 + 100 = 481So, Σx² = 481Similarly, Σy²:Compute each y squared and sum.30²=90045²=202520²=40035²=122550²=250025²=62540²=160055²=302530²=90060²=3600Now, sum these:900 + 2025 = 29252925 + 400 = 33253325 + 1225 = 45504550 + 2500 = 70507050 + 625 = 76757675 + 1600 = 92759275 + 3025 = 1230012300 + 900 = 1320013200 + 3600 = 16800So, Σy² = 16800Now, plug these into the Pearson formula.First, compute numerator:nΣxy - ΣxΣy = 10*2835 - 67*390Compute 10*2835: 28350Compute 67*390:Let me compute 67*390:First, 60*390 = 23,4007*390 = 2,730So, total is 23,400 + 2,730 = 26,130So, numerator = 28,350 - 26,130 = 2,220Now, compute denominator:sqrt([nΣx² - (Σx)²][nΣy² - (Σy)²])First, compute [nΣx² - (Σx)²]:nΣx² = 10*481 = 4,810(Σx)² = 67² = 4,489So, 4,810 - 4,489 = 321Next, compute [nΣy² - (Σy)²]:nΣy² = 10*16,800 = 168,000(Σy)² = 390² = 152,100So, 168,000 - 152,100 = 15,900Now, multiply these two results: 321 * 15,900Compute 321 * 15,900:First, 300*15,900 = 4,770,00021*15,900 = 333,900So, total is 4,770,000 + 333,900 = 5,103,900So, denominator = sqrt(5,103,900)Compute sqrt(5,103,900). Let me see:First, note that 2,259² = 5,103, because 2,259*2,259 is approximately 5,103,000.Wait, let me compute 2,259²:2,259 * 2,259:Compute 2,000² = 4,000,0002*2,000*259 = 2*2,000*259 = 4,000*259 = 1,036,000259² = 67,081So, total is 4,000,000 + 1,036,000 + 67,081 = 5,103,081Which is very close to 5,103,900. So, sqrt(5,103,900) ≈ 2,259. So, approximately 2,259.But let me check 2,259² = 5,103,081, which is 819 less than 5,103,900. So, 5,103,900 - 5,103,081 = 819.So, 2,259 + x squared ≈ 5,103,900.(2,259 + x)^2 ≈ 5,103,900Approximate x:2,259² + 2*2,259*x + x² ≈ 5,103,9005,103,081 + 4,518x + x² ≈ 5,103,900So, 4,518x ≈ 5,103,900 - 5,103,081 = 819So, x ≈ 819 / 4,518 ≈ 0.181So, sqrt ≈ 2,259 + 0.181 ≈ 2,259.18So, approximately 2,259.18.So, denominator ≈ 2,259.18Therefore, Pearson's r = 2,220 / 2,259.18 ≈ 0.9826So, approximately 0.983.So, the Pearson correlation coefficient is about 0.983.Interpretation: This is a very strong positive correlation. It suggests that as the number of resources allocated increases, the time taken to solve the cases also increases. Wait, that seems counterintuitive. More resources should mean cases are solved faster, right? So, a positive correlation here implies that more resources are associated with longer times, which is unexpected.But let me double-check my calculations because that seems odd.Wait, let me go back through the steps.First, Σx=67, Σy=390, Σxy=2835, Σx²=481, Σy²=16800.Numerator: 10*2835 - 67*390 = 28,350 - 26,130 = 2,220. That seems correct.Denominator: sqrt[(10*481 - 67²)(10*16800 - 390²)] = sqrt[(4810 - 4489)(168000 - 152100)] = sqrt[321 * 15900]321*15900=5,103,900. sqrt(5,103,900)= approximately 2,259.18.So, 2,220 / 2,259.18 ≈ 0.9826.So, r≈0.983.So, that's correct. So, a very strong positive correlation.But the journalist is looking into whether more resources lead to quicker solving. So, if more resources lead to longer times, that's an issue. But perhaps there's an error in the data or in the interpretation.Wait, let me check the data again.Looking at the data:Case 1: 5 resources, 30 daysCase 2:7,45Case3:4,20Case4:6,35Case5:8,50Case6:5,25Case7:7,40Case8:9,55Case9:6,30Case10:10,60Looking at this, as resources increase, time also increases. For example, case 10 has the most resources (10) and the longest time (60 days). Similarly, case 8 has 9 resources and 55 days. So, it does seem that higher resources are associated with higher times.But that's counterintuitive. Maybe the data is flawed, or perhaps more resources are allocated to more complex cases which take longer, so it's not necessarily that more resources cause longer times, but that more complex cases require more resources and take longer.So, the correlation is positive, but it might not imply causation. It could be that the cases requiring more resources are inherently more time-consuming.But in terms of the Pearson coefficient, it's a strong positive correlation.So, moving on to the second part: linear regression.We need to find the equation of the best-fit line T = aR + b.The formula for the slope a is:a = [nΣxy - ΣxΣy] / [nΣx² - (Σx)²]Which is the numerator we already computed: 2,220Denominator: 321So, a = 2,220 / 321 ≈ 6.916So, a ≈ 6.916Then, the intercept b is:b = (Σy - aΣx)/nCompute Σy = 390, Σx = 67, n=10So, b = (390 - 6.916*67)/10First, compute 6.916*67:6*67=4020.916*67≈61.492So, total ≈402 + 61.492≈463.492So, 390 - 463.492 ≈ -73.492Divide by 10: -7.3492So, b ≈ -7.35Therefore, the regression equation is T = 6.916R - 7.35Now, predict the time for 11 resources:T = 6.916*11 - 7.35Compute 6.916*11:6*11=660.916*11≈10.076Total≈66 + 10.076≈76.076Subtract 7.35: 76.076 - 7.35≈68.726So, approximately 68.73 days.But let me do this more accurately.Compute a more precisely:a = 2,220 / 321Divide 2,220 by 321:321*6=1,9262,220 - 1,926=294321*0.916≈294 (since 321*0.9=288.9, 321*0.016≈5.136, total≈294.036)So, a≈6.916Similarly, b:b = (390 - 6.916*67)/10Compute 6.916*67:Let me compute 6*67=4020.916*67:0.9*67=60.30.016*67≈1.072So, 60.3 +1.072≈61.372Total 402 +61.372≈463.372So, 390 - 463.372≈-73.372Divide by 10: -7.3372So, b≈-7.337Thus, the equation is T = 6.916R -7.337So, for R=11:T=6.916*11 -7.3376.916*11:6*11=660.916*11=10.076Total=66+10.076=76.07676.076 -7.337≈68.739So, approximately 68.74 days.So, the predicted time is about 68.74 days.But let me check if I did the calculations correctly.Wait, in the regression formula, T = aR + b, where a is the slope and b is the intercept.We have a≈6.916, b≈-7.337So, plugging R=11:6.916*11 = 76.07676.076 -7.337=68.739Yes, that's correct.So, the predicted time is approximately 68.74 days.But let me think about this. The correlation is positive, so as R increases, T increases. So, the regression line reflects that. So, with 11 resources, the time is predicted to be about 68.74 days.But wait, looking at the data, when R=10, T=60. So, with R=11, it's predicting longer time, which is consistent with the positive correlation.But again, this seems counterintuitive because more resources should lead to faster solving. So, perhaps the relationship is not causal, or there are other factors at play.But in terms of the mathematical model, this is the result.So, summarizing:1. Pearson correlation coefficient is approximately 0.983, indicating a very strong positive correlation between resources and time to solve.2. The regression equation is T = 6.916R -7.337, and for R=11, T≈68.74 days.But I should present the answers more precisely.For Pearson's r, let me compute it more accurately.We had numerator=2,220, denominator≈2,259.18So, r=2,220 / 2,259.18≈0.9826So, r≈0.983For the regression slope a=2,220 / 321≈6.916And intercept b=(390 -6.916*67)/10≈(390 -463.372)/10≈-73.372/10≈-7.337So, the equation is T=6.916R -7.337For R=11:T=6.916*11 -7.337≈76.076 -7.337≈68.739≈68.74 daysSo, rounding to two decimal places, 68.74 days.Alternatively, if we want to present it as a whole number, maybe 69 days, but since the data is in whole days, perhaps 69 days.But the question says "predict the time it would take", so it's fine to have a decimal.So, final answers:1. Pearson's r≈0.983, indicating a very strong positive correlation.2. Regression equation: T=6.916R -7.337, and for R=11, T≈68.74 days.But let me check if I made any calculation errors.Wait, in the Pearson formula, I used n=10, which is correct.Σx=67, Σy=390, Σxy=2835, Σx²=481, Σy²=16800.Numerator:10*2835=28,350; 67*390=26,130; 28,350-26,130=2,220.Denominator: sqrt[(10*481 -67²)(10*16800 -390²)] = sqrt[(4810-4489)(168000-152100)] = sqrt[321*15900]=sqrt[5,103,900]≈2,259.18So, r=2,220 /2,259.18≈0.9826≈0.983.Yes, correct.For regression:a=2,220 /321≈6.916b=(390 -6.916*67)/10≈(390 -463.372)/10≈-7.337Yes.So, all calculations seem correct.Therefore, the answers are:1. Pearson correlation coefficient is approximately 0.983, indicating a very strong positive relationship.2. The regression equation is T=6.916R -7.337, and for 11 resources, the predicted time is approximately 68.74 days.</think>

🧠:<think>Okay, so I have this problem about optimizing matrix multiplications in a deep neural network using TensorFlow's XLA. The first part is about finding the optimal way to parenthesize the matrix chain to minimize FLOPs. Hmm, I remember that matrix chain multiplication is a classic dynamic programming problem. The goal is to find the order of multiplying matrices such that the total number of scalar multiplications is minimized.Each layer has a weight matrix W_i of size m_i x n_i. The computation involves multiplying these matrices in sequence, and each multiplication has a cost of 2*m*k*n FLOPs. So, for a chain of matrices, the total cost depends on the order in which we perform the multiplications.Let me think about how dynamic programming applies here. The standard approach is to define a table where dp[i][j] represents the minimum cost to multiply matrices from i to j. Then, for each possible split point k between i and j, we calculate the cost of multiplying i to k and k+1 to j, plus the cost of multiplying the resulting matrices. The minimum over all k gives the optimal cost for dp[i][j].So, for part 1, I can structure the algorithm as follows:1. Initialize a 2D array dp where dp[i][j] will store the minimum cost for multiplying matrices i through j.2. For each length l from 1 to k (where k is the number of matrices), compute the cost for all possible sequences of length l.3. For each sequence starting at i and ending at j = i + l - 1, try all possible k from i to j-1.4. For each k, compute the cost as dp[i][k] + dp[k+1][j] + 2 * m_i * k * n_j, where m_i is the number of rows of matrix i, and n_j is the number of columns of matrix j.5. The minimum of these costs is stored in dp[i][j].This should give the optimal parenthesization. The time complexity is O(k^3) because we have three nested loops: over the length l, over the starting index i, and over the split point k.Now, for part 2, we have to consider parallel execution with p processors. Each processor can perform up to q FLOPs per second. The goal is to minimize both the total FLOPs and the wall-clock time. Since communication costs are negligible, we can focus on distributing the FLOPs across processors.Wait, but how does parallelism affect the parenthesization? In the sequential case, the order affects the total FLOPs, but in parallel, maybe we can compute independent parts simultaneously. However, matrix multiplications are not entirely independent because each multiplication depends on the result of the previous one. So, the chain is inherently sequential in terms of dependencies, but perhaps we can find a way to parallelize within each multiplication step.Alternatively, maybe the parenthesization can be chosen such that certain multiplications can be done in parallel. For example, if the chain can be split into independent subchains, those can be processed in parallel. But in a linear chain, each multiplication depends on the previous, so maybe the only way to parallelize is within each individual matrix multiplication.But the problem says each matrix multiplication can be distributed across multiple processors. So, for each multiplication, we can split the computation across p processors, each handling a part of the FLOPs. The total FLOPs remain the same, but the time per multiplication is reduced by a factor of p, assuming perfect distribution.Wait, but the total FLOPs are fixed by the parenthesization. So, the wall-clock time would be the sum of the times for each multiplication step, where each step's time is (FLOPs for that step) / (p * q). However, since the multiplications are sequential, the total time is the sum of each step's time.But if we can find a parenthesization that allows some multiplications to be done in parallel, that could reduce the total time. However, in a linear chain, each multiplication is dependent on the previous, so they can't be done in parallel. So, the only way to speed up is to parallelize each individual multiplication.Therefore, the optimal parenthesization in terms of FLOPs remains the same as in part 1, but the wall-clock time is the sum of (cost of each multiplication step) / (p * q). Since each multiplication step's cost is fixed by the parenthesization, the total time is the sum of (2 * m_i * k * n_j) / (p * q) for each multiplication step.Wait, but maybe the way we parenthesize affects how much we can parallelize within each multiplication. For example, larger matrices might allow for better parallelization efficiency. But the problem states that communication costs are negligible, so we can assume that the FLOPs can be perfectly distributed.Therefore, the optimal strategy is still to minimize the total FLOPs as in part 1, and then the wall-clock time is simply the total FLOPs divided by (p * q). But that can't be right because the multiplications are sequential, so the total time is the sum of each step's time, not the total FLOPs divided by p*q.Wait, let me clarify. Each matrix multiplication can be distributed across p processors, so each multiplication's time is (FLOPs for that multiplication) / (p * q). Since the multiplications are sequential, the total time is the sum of these individual times.Therefore, to minimize the total time, we need to minimize the sum of (FLOPs for each multiplication) / (p * q). But since p and q are constants, minimizing the sum is equivalent to minimizing the total FLOPs. So, the optimal parenthesization remains the same as in part 1.However, if we consider that some parenthesizations might allow for more parallelism within the multiplications, but since each multiplication is a single step, the only way to parallelize is within each step. So, the total time is determined by the sum of each step's time, which is proportional to the total FLOPs divided by p*q.Wait, no. The total FLOPs are fixed by the parenthesization, but the wall-clock time is the sum of each step's time. Each step's time is (FLOPs for that step) / (p * q). So, the total time is (total FLOPs) / (p * q). Therefore, the total time is inversely proportional to p*q. So, the speedup is p*q times compared to a single processor.But that can't be right because the multiplications are sequential. If you have p processors, you can only parallelize within each multiplication, not across multiplications. So, each multiplication's time is reduced by a factor of p, but since they are sequential, the total time is the sum of each multiplication's time, each divided by p.Therefore, the total time is (sum of FLOPs for each multiplication) / (p * q). The sum of FLOPs is the total FLOPs, which is minimized by the optimal parenthesization. So, the total time is (C_total) / (p * q), where C_total is the minimal total FLOPs from part 1.Thus, the speedup compared to a single processor is p times, assuming q remains the same. Because with p processors, each multiplication is done p times faster, so the total time is reduced by p.Wait, but if each multiplication is done in parallel across p processors, then each multiplication's time is (FLOPs) / (p * q). So, the total time is sum over all multiplications of (FLOPs_i) / (p * q) = (sum FLOPs_i) / (p * q) = C_total / (p * q). Without parallelism, it would be C_total / q. So, the speedup is p times.But actually, in the sequential case, each multiplication is done on one processor, so the time is sum (FLOPs_i) / q. With p processors, it's sum (FLOPs_i) / (p * q). So, the speedup is p times.However, this assumes that each multiplication can be perfectly split across p processors, which might not always be the case, but the problem states to assume communication costs are negligible, so we can assume perfect distribution.Therefore, the optimal parenthesization remains the same as in part 1, and the total time is reduced by a factor of p. So, the speedup is p times.Wait, but in reality, the number of multiplications is fixed by the number of layers minus one. So, if we have k layers, we have k-1 multiplications. Each multiplication can be parallelized across p processors. So, the total time is (sum of FLOPs for each multiplication) / (p * q). Since the sum is minimized by the optimal parenthesization, the total time is minimized accordingly.So, in conclusion, for part 2, the algorithm remains the same as in part 1 because the parenthesization only affects the total FLOPs, and the wall-clock time is directly proportional to the total FLOPs divided by p*q. Therefore, minimizing the total FLOPs also minimizes the wall-clock time in the parallel model.But wait, is there a way to change the parenthesization to allow more parallelism across multiplications? For example, if we can group some multiplications to be done in parallel, but in a linear chain, each multiplication depends on the previous, so they can't be done in parallel. Therefore, the only parallelism possible is within each individual multiplication step.Hence, the optimal parenthesization for minimizing FLOPs is still the best approach, and the wall-clock time is the total FLOPs divided by p*q. The speedup is p times compared to a single processor.I think that's the reasoning. Now, let me structure this into a proper answer.</think>

🧠:<think>Alright, so I have this problem about a banker analyzing the impact of a financial crisis on a community of artists. There are 100 artists, each with different incomes and financial resilience. The problem is split into two parts. Let me try to tackle each part step by step.Starting with part 1: We have ( I_i ) representing the annual income of the ( i )-th artist before the crisis. It's normally distributed with a mean of 50,000 and a standard deviation of 15,000. Then, the financial crisis causes a drop in income, modeled by ( D_i ), which is also normally distributed with a mean of 10,000 and a standard deviation of 5,000. Importantly, ( D_i ) is independent of ( I_i ). We need to compute the expected value and variance of the artists' income after the crisis.Okay, so after the crisis, the income of each artist would be ( I_i - D_i ). Since both ( I_i ) and ( D_i ) are normal variables, their difference should also be normal. I remember that for normal distributions, the mean of the difference is the difference of the means, and the variance is the sum of the variances if they're independent. Since ( I_i ) and ( D_i ) are independent, that should hold.Let me write that down:- The expected value (mean) of ( I_i - D_i ) is ( E[I_i] - E[D_i] ).- The variance of ( I_i - D_i ) is ( Var(I_i) + Var(D_i) ).Calculating the mean first:( E[I_i] = 50,000 )( E[D_i] = 10,000 )So, ( E[I_i - D_i] = 50,000 - 10,000 = 40,000 ).That makes sense; on average, each artist's income drops by 10,000.Now, the variance:( Var(I_i) = (15,000)^2 = 225,000,000 )( Var(D_i) = (5,000)^2 = 25,000,000 )So, ( Var(I_i - D_i) = 225,000,000 + 25,000,000 = 250,000,000 ).Therefore, the standard deviation after the crisis would be the square root of 250,000,000, which is approximately 15,811.39. But since the question only asks for variance, we can leave it at 250,000,000.Wait, hold on. Is that correct? Let me double-check. Since variance is additive for independent variables, regardless of whether we're adding or subtracting, right? So, yes, subtracting ( D_i ) is the same as adding a negative ( D_i ), and since variance doesn't depend on the sign, it's just adding the variances. So, 225,000,000 + 25,000,000 is indeed 250,000,000.Okay, so part 1 seems straightforward. The expected income after the crisis is 40,000, and the variance is 250,000,000.Moving on to part 2: The banker wants to know how many artists will fall below the poverty line of 20,000 after the crisis. So, we need to find the probability that a randomly selected artist's income is below 20,000. Using the combined distribution from part 1, which is normal with mean 40,000 and variance 250,000,000 (or standard deviation ~15,811.39).So, essentially, we have a normal distribution ( N(40,000, 250,000,000) ). We need to find ( P(I_i - D_i < 20,000) ).To find this probability, we can standardize the variable. Let me denote ( X = I_i - D_i ). Then, ( X sim N(40,000, 250,000,000) ). We need ( P(X < 20,000) ).First, compute the z-score:( z = frac{20,000 - 40,000}{sqrt{250,000,000}} )Calculating the numerator: 20,000 - 40,000 = -20,000.Denominator: sqrt(250,000,000). Let me compute that.250,000,000 is 250 million. The square root of 250,000,000 is the same as sqrt(250) * 1000. Since sqrt(250) is approximately 15.811388, so sqrt(250,000,000) is approximately 15,811.39.So, z = (-20,000) / 15,811.39 ≈ -1.2649.So, the z-score is approximately -1.2649.Now, we need to find the probability that a standard normal variable is less than -1.2649. That is, ( P(Z < -1.2649) ).Looking at standard normal distribution tables or using a calculator, the cumulative probability for z = -1.26 is approximately 0.1038, and for z = -1.27, it's approximately 0.1020. Since -1.2649 is between -1.26 and -1.27, we can interpolate.Alternatively, using a calculator or precise z-table, let me recall that the exact value for z = -1.2649 can be found using the error function or a calculator.But since I don't have a calculator here, I can approximate it. Alternatively, I remember that the cumulative distribution function (CDF) for z = -1.26 is about 0.1038, and for z = -1.27, it's about 0.1020. The difference between z = -1.26 and z = -1.27 is 0.01 in z, and the difference in probabilities is about 0.1038 - 0.1020 = 0.0018.Our z is -1.2649, which is 0.0049 above -1.26. So, the fraction is 0.0049 / 0.01 = 0.49. So, we can approximate the probability as 0.1038 - (0.49 * 0.0018) ≈ 0.1038 - 0.000882 ≈ 0.1029.Alternatively, using linear approximation:The change in z is 0.0049 from -1.26 to -1.2649. The change in probability is approximately the derivative of the CDF at z = -1.26 times the change in z.The derivative of the CDF is the PDF, which is ( phi(z) = frac{1}{sqrt{2pi}} e^{-z^2 / 2} ).At z = -1.26, ( phi(-1.26) = frac{1}{sqrt{2pi}} e^{-(1.26)^2 / 2} ).Calculating ( (1.26)^2 = 1.5876 ), so ( e^{-1.5876 / 2} = e^{-0.7938} ≈ 0.4523 ).Thus, ( phi(-1.26) ≈ 0.4523 / 2.5066 ≈ 0.1804 ).So, the change in probability is approximately 0.1804 * 0.0049 ≈ 0.000884.Therefore, the probability at z = -1.2649 is approximately 0.1038 - 0.000884 ≈ 0.1029.So, approximately 10.29% probability.Alternatively, if I use a calculator or precise computation, let me recall that for z = -1.2649, the exact probability can be found using:( P(Z < z) = frac{1}{2} left(1 + text{erf}left( frac{z}{sqrt{2}} right)right) )But without a calculator, it's hard. Alternatively, using the approximation:For z between -3 and 3, the CDF can be approximated with the formula:( P(Z < z) ≈ frac{1}{2} left(1 + text{sign}(z) sqrt{1 - e^{- (z^2 / 4.236)}} right) )But I might be misremembering. Alternatively, perhaps it's better to stick with the linear approximation.Given that, I think 10.29% is a reasonable approximation. So, about 10.3% chance that an artist's income falls below 20,000.Therefore, in a community of 100 artists, we can expect approximately 10.3 artists to fall below the poverty line. Since we can't have a fraction of a person, we might round it to 10 or 11 artists.But the question asks for the probability, not the number. So, the probability is approximately 10.3%.Wait, but let me cross-verify. Maybe using another method.Alternatively, using the z-score of approximately -1.2649, we can use the standard normal table.Looking up z = -1.26: 0.1038z = -1.27: 0.1020So, the difference is 0.0018 over 0.01 z-units.Our z is -1.2649, which is 0.0049 beyond -1.26 towards -1.27.So, the proportion is 0.0049 / 0.01 = 0.49.So, the probability is 0.1038 - 0.49*(0.1038 - 0.1020) = 0.1038 - 0.49*0.0018 ≈ 0.1038 - 0.000882 ≈ 0.1029, as before.So, 0.1029, which is approximately 10.29%.Therefore, the probability is approximately 10.3%.Alternatively, using more precise methods, perhaps with a calculator, the exact value is about 0.1029, so 10.29%.Hence, the probability that a randomly selected artist will have an income below 20,000 after the crisis is approximately 10.3%.Wait, but let me think again. Is the distribution of ( X = I_i - D_i ) correctly modeled?Yes, because both ( I_i ) and ( D_i ) are normal, independent, so their difference is normal with mean 40,000 and variance 250,000,000.Therefore, the calculations seem correct.So, summarizing:1. Expected income after crisis: 40,000Variance: 250,000,0002. Probability of income below 20,000: approximately 10.3%Therefore, the banker can expect about 10-11 artists out of 100 to fall below the poverty line.Final Answer1. The expected value is boxed{40000} and the variance is boxed{250000000}.2. The probability is approximately boxed{0.103}.</think>

🧠:<think>Okay, so I have this problem about the impact of deregulation on a petroleum company's profits and the market equilibrium. Let me try to break it down step by step.Starting with part 1: The company has a profit function under regulation, which is P(x) = 50x - 0.5x². After deregulation, the profit function changes to P_dereg(x) = 60x - 0.4x². I need to find the maximum profit in both cases and then calculate the percentage increase due to deregulation.Hmm, profit functions are quadratic, so they should have a maximum point. Since the coefficient of x² is negative in both cases, the parabola opens downward, meaning the vertex is the maximum point. The x-coordinate of the vertex is at -b/(2a). Let me apply that.For the regulated market:P(x) = 50x - 0.5x²Here, a = -0.5 and b = 50.So, x = -50/(2*(-0.5)) = -50/(-1) = 50.Wait, x is in millions of barrels, so 50 million barrels. Now, plug this back into P(x) to find the maximum profit.P(50) = 50*50 - 0.5*(50)² = 2500 - 0.5*2500 = 2500 - 1250 = 1250.So, maximum profit under regulation is 1,250 million.Now, for the deregulated market:P_dereg(x) = 60x - 0.4x²Here, a = -0.4 and b = 60.So, x = -60/(2*(-0.4)) = -60/(-0.8) = 75.So, x is 75 million barrels. Plugging back into P_dereg(x):P_dereg(75) = 60*75 - 0.4*(75)² = 4500 - 0.4*5625 = 4500 - 2250 = 2250.Maximum profit after deregulation is 2,250 million.Now, percentage increase in maximum profit. The increase is 2250 - 1250 = 1000. So, percentage increase is (1000/1250)*100% = 80%.Alright, that seems straightforward. Let me move on to part 2.Part 2: The demand function changes from D(x) = 100 - 2x to D_dereg(x) = 120 - 1.5x. The supply function remains S(x) = 20 + 0.5x. I need to find the equilibrium quantity and price before and after deregulation, and then calculate the change in consumer surplus.First, equilibrium occurs where demand equals supply. So, for the regulated market:Set D(x) = S(x):100 - 2x = 20 + 0.5xLet me solve for x.100 - 20 = 2x + 0.5x80 = 2.5xx = 80 / 2.5 = 32.So, equilibrium quantity is 32 million barrels. Now, plug this back into either D(x) or S(x) to find the price.Using D(x): P = 100 - 2*32 = 100 - 64 = 36.Alternatively, using S(x): P = 20 + 0.5*32 = 20 + 16 = 36. Same result.Now, for the deregulated market:Set D_dereg(x) = S(x):120 - 1.5x = 20 + 0.5xSolving for x:120 - 20 = 1.5x + 0.5x100 = 2xx = 50.So, equilibrium quantity is 50 million barrels. Plugging back into D_dereg(x):P = 120 - 1.5*50 = 120 - 75 = 45.Alternatively, S(x): P = 20 + 0.5*50 = 20 + 25 = 45. Same.Now, I need to find the change in consumer surplus. Consumer surplus is the area under the demand curve and above the equilibrium price, up to the equilibrium quantity.First, for the regulated market:Demand curve is D(x) = 100 - 2x.At equilibrium, price is 36. So, consumer surplus is the integral from 0 to 32 of (100 - 2x - 36) dx.Simplify: (64 - 2x) dx.Integral of 64 is 64x, integral of -2x is -x². So, evaluated from 0 to 32:[64*32 - (32)²] - [0 - 0] = (2048 - 1024) = 1024.So, consumer surplus in regulated market is 1024.For the deregulated market:Demand curve is D_dereg(x) = 120 - 1.5x.Equilibrium price is 45. So, consumer surplus is integral from 0 to 50 of (120 - 1.5x - 45) dx.Simplify: (75 - 1.5x) dx.Integral of 75 is 75x, integral of -1.5x is -0.75x². Evaluated from 0 to 50:[75*50 - 0.75*(50)²] - [0 - 0] = (3750 - 0.75*2500) = 3750 - 1875 = 1875.So, consumer surplus in deregulated market is 1875.Wait, hold on. That can't be right. Because when price increases, consumer surplus should decrease, but here it's increasing. Hmm, maybe I made a mistake.Wait, no. Let me think again. In the regulated market, equilibrium price was 36, and in deregulated, it's 45. So, price increased, which should decrease consumer surplus. But according to my calculations, it increased from 1024 to 1875. That seems contradictory.Wait, maybe I messed up the demand functions. Let me double-check.In the regulated market, D(x) = 100 - 2x. At equilibrium, x=32, P=36.So, consumer surplus is the area under D(x) from 0 to 32, minus the area under P=36.Which is indeed the integral of (100 - 2x - 36) dx from 0 to32, which is (64 - 2x) dx. That integral is correct.Similarly, in deregulated, D_dereg(x) = 120 - 1.5x. At equilibrium, x=50, P=45.So, consumer surplus is integral of (120 - 1.5x - 45) dx from 0 to50, which is (75 - 1.5x) dx. That integral is correct.But why is consumer surplus increasing when price increased? Because the demand curve shifted to the right, so even though price went up, the quantity increased more, leading to a larger area.Wait, let's visualize it. In the regulated market, the demand is steeper (slope -2) and supply is flatter (slope 0.5). After deregulation, demand becomes less steep (slope -1.5) and supply remains the same. So, the equilibrium quantity increases, but the price also increases.But the consumer surplus is the area under the demand curve above the price. So, even though the price is higher, the quantity is much higher, so the area might actually be larger.Wait, let me calculate the actual values again.Regulated consumer surplus: 1024.Deregulated: 1875.So, the change is 1875 - 1024 = 851. So, consumer surplus increased by 851.But wait, that seems counterintuitive because price went up. Maybe because the demand curve is more elastic, the increase in quantity outweighs the increase in price.Alternatively, perhaps I should have calculated the change as the difference between the two consumer surpluses.But the question says, "What is the change in consumer surplus due to deregulation?" So, it's 1875 - 1024 = 851. So, consumer surplus increased by 851.But wait, let me think again. In the regulated market, the equilibrium was at x=32, P=36. In the deregulated, x=50, P=45.So, the price increased, but the quantity increased as well. The demand curve shifted to the right, so the area under the new demand curve is larger, even though the price is higher.But let me think about the formula for consumer surplus. It's the integral from 0 to x of (D(x) - P) dx.In the regulated case, D(x) = 100 - 2x, P=36, x=32.So, integral of (100 - 2x - 36) dx from 0 to32 = integral of (64 - 2x) dx = [64x - x²] from 0 to32 = (64*32 - 32²) = 2048 - 1024 = 1024.In the deregulated case, D(x)=120 -1.5x, P=45, x=50.Integral of (120 -1.5x -45) dx from 0 to50 = integral of (75 -1.5x) dx = [75x - 0.75x²] from 0 to50 = (75*50 - 0.75*2500) = 3750 - 1875 = 1875.So, the change is indeed +851. So, consumer surplus increased by 851.But wait, that seems odd because usually, when prices go up, consumer surplus decreases. But in this case, the demand curve shifted right, so the area under the new demand curve is larger, even though the price is higher. So, the net effect is an increase in consumer surplus.Alternatively, maybe I should have considered the change in consumer surplus as the difference between the two areas. So, 1875 - 1024 = 851. So, the change is an increase of 851.Alternatively, maybe I should have calculated the consumer surplus in both cases and then subtracted. So, yes, 1875 - 1024 = 851.Wait, but let me think about it in terms of the formula. Consumer surplus is the area under the demand curve above the price. So, if the demand curve shifts right and the price increases, the area could either increase or decrease depending on the relative changes.In this case, the demand curve shifted right (from 100 -2x to 120 -1.5x), which is both an increase in the intercept and a decrease in slope. So, the demand is more elastic and higher. So, even though the price went up, the quantity increased enough to make the consumer surplus larger.So, the change in consumer surplus is +851.Wait, but let me check the calculations again.Regulated:Integral of (64 - 2x) from 0 to32:At x=32: 64*32 = 2048, 2x² = 2*(32)^2 = 2048, so 2048 - 2048 = 0? Wait, no, wait, the integral is 64x - x². So, at x=32, it's 64*32 - (32)^2 = 2048 - 1024 = 1024. Correct.Deregulated:Integral of (75 -1.5x) from 0 to50:At x=50: 75*50 = 3750, 0.75*(50)^2 = 0.75*2500=1875, so 3750 - 1875 = 1875. Correct.So, the change is indeed +851.So, summarizing:1. Maximum profit under regulation: 1,250 million. After deregulation: 2,250 million. Percentage increase: 80%.2. Equilibrium quantity before: 32 million barrels, price: 36. After: 50 million barrels, price: 45. Change in consumer surplus: +851.Wait, but the question says "change in consumer surplus due to deregulation." So, it's 1875 - 1024 = 851. So, an increase of 851.But let me think again: when the price increases, consumer surplus should decrease, but in this case, the demand curve shifted right, so the area under the new demand curve is larger. So, even though the price is higher, the quantity is much higher, leading to a larger consumer surplus.Alternatively, maybe I should have considered the consumer surplus as the area above the price and below the demand curve. So, in the regulated case, it's 1024, and in the deregulated, it's 1875, so the change is +851.Alternatively, maybe the question expects the change in consumer surplus as the difference between the two, regardless of direction. So, 851.Alternatively, perhaps I should have calculated the consumer surplus in both cases and then subtracted the regulated from the deregulated.Yes, that's what I did.So, the change is +851.Wait, but let me think about the formula again. Consumer surplus is the integral from 0 to x of (D(x) - P) dx.In the regulated case, D(x) = 100 - 2x, P=36, x=32.So, integral of (100 - 2x -36) dx from 0 to32 = integral of (64 - 2x) dx = [64x - x²] from 0 to32 = 64*32 - 32² = 2048 - 1024 = 1024.In the deregulated case, D(x)=120 -1.5x, P=45, x=50.Integral of (120 -1.5x -45) dx from 0 to50 = integral of (75 -1.5x) dx = [75x - 0.75x²] from 0 to50 = 75*50 - 0.75*2500 = 3750 - 1875 = 1875.So, the change is 1875 - 1024 = 851.Therefore, the change in consumer surplus is an increase of 851.Wait, but let me think about the units. The problem didn't specify units, but since x is in millions of barrels, and P is in dollars per barrel, the consumer surplus would be in dollars. So, 851 million dollars.But let me check if I did the integrals correctly.For the regulated market:Integral of (64 - 2x) dx from 0 to32.Antiderivative is 64x - x².At x=32: 64*32 = 2048, x²=1024. So, 2048 - 1024 = 1024.Correct.For the deregulated market:Integral of (75 -1.5x) dx from 0 to50.Antiderivative is 75x - 0.75x².At x=50: 75*50=3750, 0.75*(50)^2=0.75*2500=1875. So, 3750 - 1875=1875.Correct.So, the change is indeed 1875 - 1024 = 851.Therefore, the change in consumer surplus due to deregulation is an increase of 851 million dollars.Wait, but let me think again: when the price increases, consumer surplus should decrease, but in this case, the demand curve shifted right, so the area under the new demand curve is larger, leading to a larger consumer surplus despite the higher price.Yes, that makes sense because the quantity demanded increased significantly, which can offset the higher price.So, in conclusion:1. Maximum profit under regulation: 1,250 million. After deregulation: 2,250 million. Percentage increase: 80%.2. Equilibrium quantity before: 32 million barrels, price: 36. After: 50 million barrels, price: 45. Change in consumer surplus: +851 million dollars.I think that's it.</think>