🧠:<think>Okay, so I have this problem about a young entrepreneur who took over their family's hotel and turned it into a boutique destination. The hotel has 50 rooms, and they've seen an increase in occupancy because of renovations. There are two parts to the problem, both involving maximizing revenue, but with different models.Starting with the first part: The revenue R(x) is modeled by a quadratic function, R(x) = -2x² + 40x + 120, where x is the number of rooms booked per day. I need to find the number of rooms that maximizes the daily revenue and then calculate that maximum revenue.Hmm, quadratic functions. I remember that a quadratic function in the form of ax² + bx + c has its vertex at x = -b/(2a). Since the coefficient of x² is negative (-2), the parabola opens downward, meaning the vertex is the maximum point. So, that should give me the x-value where the revenue is maximized.Let me write that down. For R(x) = -2x² + 40x + 120, a = -2, b = 40. So, the x-coordinate of the vertex is x = -b/(2a) = -40/(2*(-2)) = -40/(-4) = 10. So, x = 10 rooms booked per day would maximize the revenue.Wait, but the hotel has 50 rooms. So, 10 rooms is way below the capacity. That seems low. Maybe I made a mistake? Let me double-check. The formula is x = -b/(2a). Plugging in, it's -40 divided by (2*(-2)) which is -40/-4, which is indeed 10. So, mathematically, that's correct. But in reality, if the hotel has 50 rooms, why is the maximum revenue at 10 rooms? Maybe the pricing or other factors are such that beyond 10 rooms, the revenue starts decreasing because of the quadratic model.But let's not question the model; the problem says it's given. So, moving on, the maximum revenue would be R(10). Let me calculate that.R(10) = -2*(10)^2 + 40*(10) + 120 = -2*100 + 400 + 120 = -200 + 400 + 120 = 320. So, the maximum daily revenue is 320 when 10 rooms are booked.Wait, that seems low for a hotel with 50 rooms. Maybe the units are different? Or perhaps the model is simplified. Hmm, but the problem states it's a quadratic function, so I have to go with that.Moving on to the second part: The entrepreneur wants to introduce a dynamic pricing model. The price per room P(x) is given by P(x) = 200 - 3x, where x is the number of rooms booked per day. I need to calculate the number of rooms that should be booked per day to maximize the total revenue from room bookings, considering this dynamic pricing model.Alright, so in this case, the revenue isn't given directly. Instead, we have the price per room as a function of x. So, total revenue would be the number of rooms booked multiplied by the price per room. So, R(x) = x * P(x) = x*(200 - 3x) = 200x - 3x².So, R(x) = -3x² + 200x. That's another quadratic function, and since the coefficient of x² is negative (-3), it opens downward, so the vertex will give the maximum revenue.Again, using the vertex formula x = -b/(2a). Here, a = -3, b = 200. So, x = -200/(2*(-3)) = -200/(-6) ≈ 33.333. Since the number of rooms booked has to be an integer, we can check x = 33 and x = 34 to see which gives a higher revenue.But let me compute it exactly first. x = 200/(6) = 100/3 ≈ 33.333. So, approximately 33.33 rooms. Since you can't book a fraction of a room, we need to check both 33 and 34.Calculating R(33): R(33) = -3*(33)^2 + 200*(33) = -3*1089 + 6600 = -3267 + 6600 = 3333.Calculating R(34): R(34) = -3*(34)^2 + 200*(34) = -3*1156 + 6800 = -3468 + 6800 = 3332.So, R(33) is 3333 and R(34) is 3332. So, 33 rooms give a slightly higher revenue. Therefore, the optimal number of rooms to book per day is 33.Wait, but the hotel has 50 rooms. 33 is within capacity, so that's feasible. So, that seems reasonable.But let me think again. The first model gave 10 rooms for maximum revenue, but that seems too low. Maybe the models are different because in the first part, the revenue function is given, whereas in the second part, it's derived from the price function.So, in the first part, the revenue is already given as a quadratic, so we just find its maximum. In the second part, we have to derive the revenue function from the price function and then maximize it.So, in the first part, the maximum is at x=10, which is quite low, but perhaps that's because the model includes other factors. Maybe the cost structure or something else is affecting the revenue. But in the second part, with the dynamic pricing, the maximum is at x≈33, which is more in line with expectations.So, putting it all together, for part 1, x=10 rooms booked per day, maximum revenue 320. For part 2, x=33 rooms booked per day, maximum revenue 3333.Wait, but in the first part, the revenue is only 320? That seems really low for a hotel. Maybe the units are different? Or perhaps it's per room revenue? Wait, no, the function is R(x), which is total revenue, so it's in dollars. So, 320 per day? That seems too low for a hotel with 50 rooms.Wait, maybe the function is in hundreds of dollars? Or perhaps it's a different currency? The problem doesn't specify, so I have to go with the given function. So, if R(x) = -2x² +40x +120, then at x=10, R=320. So, maybe it's in a different unit or perhaps it's a simplified model.Anyway, the problem didn't specify units beyond dollars, so I have to assume it's correct.So, to recap:1. For the quadratic revenue function, maximum at x=10, R=320.2. For the dynamic pricing model, revenue function is R(x)= -3x² +200x, maximum at x≈33.33, so x=33 gives higher revenue.Therefore, the answers are 10 rooms and 320 for part 1, and 33 rooms for part 2.But wait, the second part asks for the number of rooms to maximize total revenue, so just 33. It doesn't ask for the revenue amount, so maybe I don't need to calculate that.But in the first part, it asks for both the number of rooms and the maximum revenue.So, to make sure, let me write down the steps clearly.For part 1:Given R(x) = -2x² +40x +120.To find maximum, find vertex.x = -b/(2a) = -40/(2*(-2)) = 10.Then R(10) = -2*(10)^2 +40*10 +120 = -200 +400 +120 = 320.So, 10 rooms, 320.For part 2:Price per room P(x) =200 -3x.Total revenue R(x) = x*P(x) = x*(200 -3x) = 200x -3x².This is a quadratic function, R(x) = -3x² +200x.To find maximum, vertex at x = -b/(2a) = -200/(2*(-3)) = 200/6 ≈33.333.Since x must be integer, check x=33 and x=34.R(33)= -3*(33)^2 +200*33 = -3*1089 +6600= -3267+6600=3333.R(34)= -3*(34)^2 +200*34= -3*1156 +6800= -3468+6800=3332.So, x=33 gives higher revenue.Therefore, the answers are:1. 10 rooms, 320.2. 33 rooms.I think that's it. I don't see any mistakes in the calculations. The first part seems counterintuitive because 10 rooms is low, but given the quadratic model, that's where the maximum is. The second part makes more sense with 33 rooms being a reasonable number for a 50-room hotel.Final Answer1. The number of rooms booked per day that maximizes the daily revenue is boxed{10}, and the maximum daily revenue is boxed{320} dollars.2. The number of rooms that should be booked per day to maximize the total revenue from room bookings is boxed{33}.</think>

🧠:<think>Okay, so I have this problem about John analyzing two interventions for his wife's health: an exercise regimen and a nutritional plan. There are two parts to this problem. Let me try to tackle them one by one.Starting with the first part: They give me two functions, E(t) and N(t). E(t) is the energy level from the exercise regimen, which is modeled by E(t) = 5 sin(2πt) + 10. And N(t) is the energy level from the nutritional plan, which is N(t) = kt², where k is a constant. After 5 weeks, the total energy level from both interventions combined is 50 units. I need to find the value of k.Alright, so let's break this down. The total energy after t weeks is E(t) + N(t). So, when t = 5, E(5) + N(5) = 50.First, let me compute E(5). E(t) = 5 sin(2πt) + 10. Plugging in t = 5:E(5) = 5 sin(2π * 5) + 10.Hmm, sin(2π * 5) is sin(10π). I remember that sin(nπ) where n is an integer is zero. So sin(10π) = 0. Therefore, E(5) = 5*0 + 10 = 10.Okay, so E(5) is 10. Then N(5) is k*(5)^2 = 25k.So, total energy at t=5 is E(5) + N(5) = 10 + 25k = 50.So, 10 + 25k = 50. Let me solve for k.Subtract 10 from both sides: 25k = 40.Divide both sides by 25: k = 40 / 25 = 8/5 = 1.6.Wait, 40 divided by 25 is 1.6? Let me check that. 25*1.6 is 40, yes. So k = 1.6. Hmm, 1.6 is 8/5, so maybe better to write it as a fraction.So, k = 8/5.Alright, that seems straightforward. Let me just recap to make sure I didn't make a mistake. E(t) is 5 sin(2πt) + 10. At t=5, sin(10π) is zero, so E(5)=10. N(5)=25k. Total is 10 + 25k = 50, so 25k=40, k=40/25=8/5. Yep, that seems right.Moving on to the second part: John is monitoring the family's well-being, W(t), which is defined as the integral from 0 to t of (E(τ) + N(τ) - S(τ)) dτ. S(t) is given as 3t + 2. We need to calculate W(t) after 5 weeks, so W(5).So, W(t) = ∫₀ᵗ [E(τ) + N(τ) - S(τ)] dτ.Let me write out what E(τ), N(τ), and S(τ) are:E(τ) = 5 sin(2πτ) + 10N(τ) = k τ², and we found k = 8/5.S(τ) = 3τ + 2.So, substituting these into the integral:W(t) = ∫₀ᵗ [5 sin(2πτ) + 10 + (8/5)τ² - (3τ + 2)] dτ.Let me simplify the integrand first:First, expand the terms:5 sin(2πτ) + 10 + (8/5)τ² - 3τ - 2.Combine like terms:10 - 2 = 8.So, the integrand becomes:5 sin(2πτ) + (8/5)τ² - 3τ + 8.Therefore, W(t) = ∫₀ᵗ [5 sin(2πτ) + (8/5)τ² - 3τ + 8] dτ.Now, let's compute this integral term by term.First, the integral of 5 sin(2πτ) dτ.The integral of sin(aτ) dτ is (-1/a) cos(aτ) + C. So, for 5 sin(2πτ), the integral is 5 * (-1/(2π)) cos(2πτ) + C = (-5/(2π)) cos(2πτ) + C.Second term: integral of (8/5)τ² dτ.Integral of τ² is (τ³)/3, so (8/5)*(τ³/3) = (8/15) τ³.Third term: integral of -3τ dτ.Integral of τ is (τ²)/2, so -3*(τ²/2) = (-3/2) τ².Fourth term: integral of 8 dτ.Integral of 8 is 8τ.Putting it all together, the integral from 0 to t is:[ (-5/(2π)) cos(2πτ) + (8/15)τ³ - (3/2)τ² + 8τ ] evaluated from 0 to t.So, W(t) = [ (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t ] - [ (-5/(2π)) cos(0) + (8/15)(0)³ - (3/2)(0)² + 8*0 ].Simplify the expression at τ=0:cos(0) = 1, so the first term is (-5/(2π))*1 = -5/(2π).The rest of the terms at τ=0 are zero.So, W(t) = [ (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t ] - [ -5/(2π) + 0 - 0 + 0 ].Which simplifies to:(-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t + 5/(2π).So, W(t) = (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t + 5/(2π).Now, we need to compute W(5). Let's plug t=5 into this expression.First, let's compute each term step by step.1. (-5/(2π)) cos(2π*5):cos(10π) again, which is cos(0) because cosine has a period of 2π, so cos(10π) = cos(0) = 1.So, this term becomes (-5/(2π)) * 1 = -5/(2π).2. (8/15)*(5)^3:5^3 is 125, so (8/15)*125 = (8*125)/15.125 divided by 15 is approximately 8.333..., but let me compute it exactly.125 ÷ 15 = 25/3, so 8*(25/3) = 200/3 ≈ 66.666...But let's keep it as a fraction: 200/3.3. -(3/2)*(5)^2:5^2 is 25, so -(3/2)*25 = -75/2 = -37.5.4. 8*5 = 40.5. The last term is +5/(2π).So, putting all these together:W(5) = (-5/(2π)) + (200/3) - 37.5 + 40 + 5/(2π).Let me simplify term by term.First, let's combine the terms with π:-5/(2π) + 5/(2π) = 0. So those cancel out.Now, the remaining terms:200/3 - 37.5 + 40.Let me convert all to fractions to make it easier.200/3 is approximately 66.666...37.5 is 75/2.40 is 40/1.So, 200/3 - 75/2 + 40.To combine these, find a common denominator. The denominators are 3, 2, and 1. The least common denominator is 6.Convert each term:200/3 = (200*2)/(3*2) = 400/675/2 = (75*3)/(2*3) = 225/640 = 40*6/6 = 240/6So, now:400/6 - 225/6 + 240/6 = (400 - 225 + 240)/6.Compute numerator:400 - 225 = 175175 + 240 = 415So, 415/6.Convert that back to a mixed number or decimal if needed.415 divided by 6 is 69 with a remainder of 1, so 69 1/6, which is approximately 69.166...So, W(5) = 415/6 ≈ 69.1667.But let me double-check my calculations to make sure I didn't make a mistake.Wait, let's go back to the integral:W(t) = ∫₀ᵗ [5 sin(2πτ) + (8/5)τ² - 3τ + 8] dτ.Wait, hold on. When I simplified the integrand, I had:5 sin(2πτ) + (8/5)τ² - 3τ + 8.But when I set up the integral, I think I missed the negative sign on the 3τ term. Wait, no, in the original expression, it's E(τ) + N(τ) - S(τ). So, E(τ) is 5 sin(2πτ) +10, N(τ) is (8/5)τ², and S(τ) is 3τ + 2. So, E + N - S is 5 sin(2πτ) +10 + (8/5)τ² - 3τ - 2.Which simplifies to 5 sin(2πτ) + (8/5)τ² - 3τ + 8. So that part was correct.Then, integrating term by term:Integral of 5 sin(2πτ) is (-5/(2π)) cos(2πτ).Integral of (8/5)τ² is (8/15)τ³.Integral of -3τ is (-3/2)τ².Integral of 8 is 8τ.So, the antiderivative is correct.Then, evaluating from 0 to t:At t, we have (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t.At 0, we have (-5/(2π)) cos(0) + 0 - 0 + 0 = -5/(2π).So, subtracting, we get:[ (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t ] - [ -5/(2π) ].Which is (-5/(2π)) cos(2πt) + (8/15)t³ - (3/2)t² + 8t + 5/(2π).So, that expression is correct.Then, plugging in t=5:First term: (-5/(2π)) cos(10π) = (-5/(2π))*1 = -5/(2π).Second term: (8/15)*(125) = 200/3.Third term: -(3/2)*(25) = -75/2.Fourth term: 8*5 = 40.Fifth term: +5/(2π).So, combining:-5/(2π) + 200/3 - 75/2 + 40 + 5/(2π).As before, the -5/(2π) and +5/(2π) cancel out.So, left with 200/3 - 75/2 + 40.Convert to sixths:200/3 = 400/675/2 = 225/640 = 240/6So, 400/6 - 225/6 + 240/6 = (400 - 225 + 240)/6 = (175 + 240)/6 = 415/6 ≈ 69.1667.So, 415/6 is the exact value, which is approximately 69.1667.But let me check the arithmetic again:200/3 is approximately 66.6667-75/2 is -37.5+40 is +40.So, 66.6667 - 37.5 = 29.166729.1667 + 40 = 69.1667.Yes, that matches.So, W(5) = 415/6, which is approximately 69.1667.But since the problem didn't specify whether to leave it as a fraction or decimal, I think it's better to present it as an exact fraction, 415/6.Wait, but let me make sure that 415 and 6 have no common factors. 415 divided by 5 is 83, and 6 is 2*3. So, no common factors. So, 415/6 is the simplest form.So, summarizing:1. k = 8/52. W(5) = 415/6I think that's it. Let me just go through once more to ensure I didn't make any mistakes.First part: E(5) = 10, N(5)=25k, total 50. So, 10 +25k=50, 25k=40, k=40/25=8/5. Correct.Second part: Integral of E + N - S from 0 to 5.E = 5 sin(2πτ) +10N = (8/5)τ²S = 3τ +2So, E + N - S = 5 sin(2πτ) +10 + (8/5)τ² -3τ -2 = 5 sin(2πτ) + (8/5)τ² -3τ +8.Integrate term by term:Integral of 5 sin(2πτ) is (-5/(2π)) cos(2πτ)Integral of (8/5)τ² is (8/15)τ³Integral of -3τ is (-3/2)τ²Integral of 8 is 8τSo, antiderivative is (-5/(2π)) cos(2πτ) + (8/15)τ³ - (3/2)τ² +8τ.Evaluate from 0 to 5:At 5: (-5/(2π)) cos(10π) + (8/15)*125 - (3/2)*25 + 8*5= (-5/(2π))*1 + 200/3 - 75/2 +40At 0: (-5/(2π)) cos(0) +0 -0 +0 = -5/(2π)So, subtracting:[ (-5/(2π)) + 200/3 -75/2 +40 ] - [ -5/(2π) ] = (-5/(2π) +200/3 -75/2 +40) +5/(2π) = 200/3 -75/2 +40.Convert to sixths:200/3 = 400/675/2 = 225/640 = 240/6So, 400/6 -225/6 +240/6 = (400 -225 +240)/6 = 415/6.Yes, that's correct.So, I think my answers are correct.Final Answer1. The value of ( k ) is boxed{dfrac{8}{5}}.2. The family's well-being after 5 weeks is boxed{dfrac{415}{6}}.</think>

🧠:<think>Okay, so I need to calculate the complexity score ( C ) for a document using the given formula ( C = alpha H + beta D ). The coefficients ( alpha ) and ( beta ) are given as 0.7 and 0.3 respectively. To find ( C ), I first need to compute the Shannon entropy ( H ) and the average dependency parse tree depth ( D ).Starting with the Shannon entropy ( H ). The document has a vocabulary of 50 unique words, and the probabilities ( p_i ) follow Zipf's law with parameter ( s = 1.2 ). Zipf's law states that the probability of the ( i )-th word is proportional to ( i^{-s} ). So, the probability ( p_i ) can be calculated as ( p_i = frac{i^{-s}}{sum_{k=1}^{n} k^{-s}} ), where ( n = 50 ).First, I need to compute the denominator, which is the sum of ( k^{-s} ) from ( k = 1 ) to ( 50 ). Let me denote this sum as ( Z ). So, ( Z = sum_{k=1}^{50} k^{-1.2} ). Calculating this sum might be a bit tedious, but I can approximate it or use a calculator. Alternatively, I might remember that for large ( n ), the sum ( sum_{k=1}^{n} k^{-s} ) approximates the Riemann zeta function ( zeta(s) ), but since ( n = 50 ) isn't extremely large, the approximation might not be perfect. However, for the sake of this problem, I might need to compute it step by step.Let me try to compute ( Z ) manually. Starting with ( k = 1 ) to ( k = 50 ), each term is ( k^{-1.2} ). So, for ( k = 1 ), it's ( 1^{-1.2} = 1 ). For ( k = 2 ), it's ( 2^{-1.2} approx 0.5743 ). For ( k = 3 ), ( 3^{-1.2} approx 0.3833 ). Continuing this way up to ( k = 50 ) would take a lot of time, but maybe I can find a pattern or use a formula. Alternatively, perhaps I can use the fact that the sum converges and use an approximation. However, since the exact value is needed for ( H ), I might need to compute it more accurately.Alternatively, maybe I can use a known approximation for the sum ( sum_{k=1}^{n} k^{-s} ). For ( s = 1.2 ), the sum can be approximated using the integral test. The integral from 1 to ( n ) of ( x^{-s} dx ) is ( frac{n^{1 - s} - 1}{1 - s} ). Plugging in ( s = 1.2 ), we get ( frac{n^{-0.2} - 1}{-0.2} = frac{1 - n^{-0.2}}{0.2} ). For ( n = 50 ), ( n^{-0.2} = 50^{-0.2} approx 0.398 ). So, the integral approximation gives ( frac{1 - 0.398}{0.2} = frac{0.602}{0.2} = 3.01 ). However, this is just an approximation, and the actual sum might be slightly different. But since I don't have a calculator here, maybe I can accept this approximation for now, but I should note that it might not be very accurate.Wait, actually, the integral test gives an approximation of the sum, but it's usually for large ( n ). Since ( n = 50 ) is not extremely large, the approximation might not be precise. Maybe I can compute the sum in parts. Let's compute the first few terms exactly and then approximate the rest.Compute the first 10 terms:1: 12: 0.57433: 0.38334: 4^{-1.2} ≈ 0.25^{0.6} ≈ 0.25^0.6 ≈ e^{0.6 ln 0.25} ≈ e^{0.6*(-1.3863)} ≈ e^{-0.8318} ≈ 0.436Wait, actually, 4^{-1.2} = (4^1)^{-1.2} = 4^{-1.2} = (2^2)^{-1.2} = 2^{-2.4} ≈ 0.25^{1.2} ≈ 0.25*0.25^{0.2} ≈ 0.25*0.8409 ≈ 0.2102Wait, I think I made a mistake earlier. Let me recalculate:For ( k = 4 ), ( 4^{-1.2} = (2^2)^{-1.2} = 2^{-2.4} ). Now, ( 2^{-2} = 0.25 ), and ( 2^{-0.4} ≈ 0.7579 ). So, ( 2^{-2.4} = 2^{-2} * 2^{-0.4} ≈ 0.25 * 0.7579 ≈ 0.1895 ).Similarly, for ( k = 5 ), ( 5^{-1.2} ≈ e^{-1.2 ln 5} ≈ e^{-1.2*1.6094} ≈ e^{-1.9313} ≈ 0.1455 ).Continuing:k=6: 6^{-1.2} ≈ e^{-1.2 ln6} ≈ e^{-1.2*1.7918} ≈ e^{-2.1502} ≈ 0.1165k=7: 7^{-1.2} ≈ e^{-1.2 ln7} ≈ e^{-1.2*1.9459} ≈ e^{-2.3351} ≈ 0.0967k=8: 8^{-1.2} = (2^3)^{-1.2} = 2^{-3.6} ≈ 0.125 * 2^{-0.6} ≈ 0.125 * 0.6598 ≈ 0.0825k=9: 9^{-1.2} = (3^2)^{-1.2} = 3^{-2.4} ≈ (1/9)^{1.2} ≈ (0.1111)^{1.2} ≈ 0.0631k=10: 10^{-1.2} ≈ e^{-1.2 ln10} ≈ e^{-1.2*2.3026} ≈ e^{-2.7631} ≈ 0.0631Wait, let me check that. 10^{-1.2} is 10^(-1 -0.2) = 10^{-1} * 10^{-0.2} ≈ 0.1 * 0.6309 ≈ 0.0631. Yes, that's correct.So, summing up the first 10 terms:1 + 0.5743 + 0.3833 + 0.1895 + 0.1455 + 0.1165 + 0.0967 + 0.0825 + 0.0631 + 0.0631 ≈Let me add them step by step:1 + 0.5743 = 1.57431.5743 + 0.3833 ≈ 1.95761.9576 + 0.1895 ≈ 2.14712.1471 + 0.1455 ≈ 2.29262.2926 + 0.1165 ≈ 2.40912.4091 + 0.0967 ≈ 2.50582.5058 + 0.0825 ≈ 2.58832.5883 + 0.0631 ≈ 2.65142.6514 + 0.0631 ≈ 2.7145So, the first 10 terms sum to approximately 2.7145.Now, for the remaining terms from k=11 to k=50, I need to estimate their sum. Since each term decreases as k increases, the sum from k=11 to 50 will be significantly smaller than the first 10 terms. Maybe I can approximate this sum using an integral.The sum ( sum_{k=11}^{50} k^{-1.2} ) can be approximated by the integral ( int_{10.5}^{50.5} x^{-1.2} dx ). The integral of ( x^{-1.2} ) is ( frac{x^{-0.2}}{-0.2} ). So, evaluating from 10.5 to 50.5:( frac{(50.5)^{-0.2} - (10.5)^{-0.2}}{-0.2} )First, compute ( (50.5)^{-0.2} ). Since 50.5 is approximately 50, and 50^{-0.2} ≈ 0.398 as I calculated earlier. Similarly, 10.5^{-0.2} ≈ (10)^{-0.2} ≈ 0.6309.So, plugging in:( frac{0.398 - 0.6309}{-0.2} = frac{-0.2329}{-0.2} ≈ 1.1645 )So, the integral approximation for the sum from k=11 to 50 is approximately 1.1645.Therefore, the total sum ( Z ≈ 2.7145 + 1.1645 ≈ 3.879 ).Wait, but earlier I approximated the integral from 1 to 50 as 3.01, but when I broke it down, the first 10 terms are 2.7145 and the rest are 1.1645, totaling 3.879. Hmm, that seems inconsistent. Maybe my integral approximation was too rough.Alternatively, perhaps I should use the exact sum for the first 10 terms and then approximate the rest using the integral from 11 to 50. Let's proceed with that.So, total sum ( Z ≈ 2.7145 + 1.1645 ≈ 3.879 ).Now, each ( p_i = frac{i^{-1.2}}{Z} ≈ frac{i^{-1.2}}{3.879} ).Now, to compute the Shannon entropy ( H = -sum_{i=1}^{50} p_i log_2 p_i ).This requires calculating each term ( -p_i log_2 p_i ) and summing them up. Since this is a bit tedious, maybe I can find a pattern or use an approximation.Alternatively, since the probabilities follow Zipf's law, the entropy can be approximated using known results. However, I'm not sure about the exact formula, so I might need to compute it numerically.But since I don't have a calculator here, maybe I can approximate the sum by recognizing that the entropy for Zipf's distribution can be approximated as ( H ≈ log_2 n - frac{zeta'(s)}{zeta(s)} ), but I'm not sure if that's accurate. Alternatively, perhaps I can use the fact that for large ( n ), the entropy approaches ( log_2 n - frac{zeta'(s)}{zeta(s)} ), but since ( n = 50 ) is not extremely large, this might not be precise.Alternatively, maybe I can use the fact that for Zipf's law with parameter ( s ), the entropy can be approximated as ( H ≈ log_2 n - frac{zeta'(s)}{zeta(s)} ), but I need to verify this.Wait, actually, the entropy for Zipf's distribution is given by ( H = log_2 Z + frac{zeta'(s)}{zeta(s)} ), where ( Z = sum_{k=1}^n k^{-s} ). But I'm not sure if that's correct. Alternatively, perhaps it's ( H = log_2 Z - frac{zeta'(s)}{zeta(s)} ). I need to check.Wait, let's recall that for a probability distribution ( p_i = frac{i^{-s}}{Z} ), the entropy is ( H = -sum p_i log_2 p_i = sum p_i log_2 Z - sum p_i log_2 i^{-s} = log_2 Z - s sum p_i log_2 i ).So, ( H = log_2 Z - s sum p_i log_2 i ).Now, ( sum p_i log_2 i = sum frac{i^{-s}}{Z} log_2 i = frac{1}{Z} sum_{i=1}^{n} i^{-s} log_2 i ).This sum can be approximated or computed numerically. However, without a calculator, it's challenging. Alternatively, perhaps I can approximate it using integrals.Let me denote ( S = sum_{i=1}^{n} i^{-s} log_2 i ). Then, ( S ) can be approximated by the integral ( int_{1}^{n} x^{-s} log_2 x dx ).Let me compute this integral. Let ( u = log_2 x ), so ( du = frac{1}{x ln 2} dx ). Let me make a substitution:Let ( t = x^{-s} ), then ( dt = -s x^{-s -1} dx ). Hmm, maybe integration by parts is better.Let me set ( u = log_2 x ), ( dv = x^{-s} dx ). Then, ( du = frac{1}{x ln 2} dx ), and ( v = frac{x^{-s +1}}{-s +1} ).So, integration by parts gives:( int u dv = uv - int v du = log_2 x cdot frac{x^{-s +1}}{-s +1} - int frac{x^{-s +1}}{-s +1} cdot frac{1}{x ln 2} dx ).Simplify:= ( frac{log_2 x cdot x^{-s +1}}{-s +1} - frac{1}{(-s +1) ln 2} int x^{-s} dx )= ( frac{log_2 x cdot x^{-s +1}}{-s +1} - frac{1}{(-s +1)^2 ln 2} x^{-s +1} ) + CNow, evaluating from 1 to n:= ( left[ frac{log_2 n cdot n^{-s +1}}{-s +1} - frac{n^{-s +1}}{(-s +1)^2 ln 2} right] - left[ frac{log_2 1 cdot 1^{-s +1}}{-s +1} - frac{1^{-s +1}}{(-s +1)^2 ln 2} right] )Since ( log_2 1 = 0 ) and ( 1^{-s +1} = 1 ), this simplifies to:= ( frac{log_2 n cdot n^{-s +1}}{-s +1} - frac{n^{-s +1}}{(-s +1)^2 ln 2} + frac{1}{(-s +1)^2 ln 2} )Now, plugging in ( s = 1.2 ):= ( frac{log_2 n cdot n^{-0.2}}{-0.2} - frac{n^{-0.2}}{(-0.2)^2 ln 2} + frac{1}{(-0.2)^2 ln 2} )Simplify:= ( frac{log_2 n cdot n^{-0.2}}{-0.2} - frac{n^{-0.2}}{0.04 ln 2} + frac{1}{0.04 ln 2} )= ( -5 log_2 n cdot n^{-0.2} - frac{n^{-0.2}}{0.04 ln 2} + frac{1}{0.04 ln 2} )Now, ( n = 50 ), so ( n^{-0.2} ≈ 0.398 ), ( log_2 50 ≈ 5.6439 ), and ( ln 2 ≈ 0.6931 ).Plugging in:= ( -5 * 5.6439 * 0.398 - frac{0.398}{0.04 * 0.6931} + frac{1}{0.04 * 0.6931} )Compute each term:First term: ( -5 * 5.6439 * 0.398 ≈ -5 * 2.246 ≈ -11.23 )Second term: ( frac{0.398}{0.027724} ≈ 14.35 )Third term: ( frac{1}{0.027724} ≈ 36.08 )So, total integral approximation:≈ -11.23 -14.35 + 36.08 ≈ (-11.23 -14.35) + 36.08 ≈ (-25.58) + 36.08 ≈ 10.5So, the integral approximation for ( S ≈ 10.5 ).But remember, ( S = sum p_i log_2 i = frac{1}{Z} sum i^{-s} log_2 i ≈ frac{1}{3.879} * 10.5 ≈ 2.706 ).Therefore, ( H = log_2 Z - s * S ≈ log_2(3.879) - 1.2 * 2.706 ).Compute ( log_2(3.879) ). Since ( 2^2 = 4 ), so ( log_2(3.879) ≈ 1.95 ).Then, ( 1.2 * 2.706 ≈ 3.247 ).So, ( H ≈ 1.95 - 3.247 ≈ -1.297 ). Wait, that can't be right because entropy can't be negative. I must have made a mistake in the approximation.Wait, no, actually, the formula is ( H = log_2 Z - s * sum p_i log_2 i ). So, ( H = log_2 Z - s * S ).But ( S = sum p_i log_2 i ≈ 2.706 ), so ( H = log_2(3.879) - 1.2 * 2.706 ≈ 1.95 - 3.247 ≈ -1.297 ). This is negative, which is impossible because entropy is always non-negative. Therefore, my approximation must be incorrect.Perhaps the integral approximation for ( S ) is not accurate enough. Alternatively, maybe I made a mistake in the integration by parts.Alternatively, maybe I should compute ( S ) numerically by summing the first few terms and approximating the rest.Let me try that. Compute ( S = sum_{i=1}^{50} frac{i^{-1.2}}{Z} log_2 i ). Since ( Z ≈ 3.879 ), each term is ( frac{i^{-1.2}}{3.879} log_2 i ).Compute the first few terms:i=1: ( frac{1^{-1.2}}{3.879} log_2 1 = 0 )i=2: ( frac{2^{-1.2}}{3.879} log_2 2 ≈ frac{0.5743}{3.879} * 1 ≈ 0.148 )i=3: ( frac{3^{-1.2}}{3.879} log_2 3 ≈ frac{0.3833}{3.879} * 1.58496 ≈ 0.152 )i=4: ( frac{4^{-1.2}}{3.879} log_2 4 ≈ frac{0.1895}{3.879} * 2 ≈ 0.097 )i=5: ( frac{5^{-1.2}}{3.879} log_2 5 ≈ frac{0.1455}{3.879} * 2.3219 ≈ 0.093 )i=6: ( frac{6^{-1.2}}{3.879} log_2 6 ≈ frac{0.1165}{3.879} * 2.58496 ≈ 0.078 )i=7: ( frac{7^{-1.2}}{3.879} log_2 7 ≈ frac{0.0967}{3.879} * 2.8074 ≈ 0.070 )i=8: ( frac{8^{-1.2}}{3.879} log_2 8 ≈ frac{0.0825}{3.879} * 3 ≈ 0.063 )i=9: ( frac{9^{-1.2}}{3.879} log_2 9 ≈ frac{0.0631}{3.879} * 3.1699 ≈ 0.051 )i=10: ( frac{10^{-1.2}}{3.879} log_2 10 ≈ frac{0.0631}{3.879} * 3.3219 ≈ 0.054 )Adding these up:0 + 0.148 + 0.152 + 0.097 + 0.093 + 0.078 + 0.070 + 0.063 + 0.051 + 0.054 ≈0.148 + 0.152 = 0.30.3 + 0.097 = 0.3970.397 + 0.093 = 0.490.49 + 0.078 = 0.5680.568 + 0.070 = 0.6380.638 + 0.063 = 0.7010.701 + 0.051 = 0.7520.752 + 0.054 = 0.806So, the first 10 terms sum to approximately 0.806.Now, for the remaining terms from i=11 to 50, I need to estimate their sum. Since each term is decreasing, maybe I can approximate the sum using an integral.Let me denote ( S' = sum_{i=11}^{50} frac{i^{-1.2}}{Z} log_2 i ). Then, ( S' ≈ frac{1}{Z} int_{10.5}^{50.5} x^{-1.2} log_2 x dx ).Using the same integration by parts as before, but now from 10.5 to 50.5.From earlier, the integral ( int x^{-1.2} log_2 x dx ) evaluated from a to b is:( frac{log_2 b cdot b^{-0.2}}{-0.2} - frac{b^{-0.2}}{(-0.2)^2 ln 2} + frac{log_2 a cdot a^{-0.2}}{-0.2} - frac{a^{-0.2}}{(-0.2)^2 ln 2} )Wait, actually, from the earlier result, the integral from a to b is:( left[ frac{log_2 x cdot x^{-0.2}}{-0.2} - frac{x^{-0.2}}{(-0.2)^2 ln 2} right]_{a}^{b} )So, plugging in a=10.5 and b=50.5:= ( left( frac{log_2 50.5 cdot (50.5)^{-0.2}}{-0.2} - frac{(50.5)^{-0.2}}{(-0.2)^2 ln 2} right) - left( frac{log_2 10.5 cdot (10.5)^{-0.2}}{-0.2} - frac{(10.5)^{-0.2}}{(-0.2)^2 ln 2} right) )Compute each part:First, compute ( log_2 50.5 ≈ log_2 50 ≈ 5.6439 ), ( (50.5)^{-0.2} ≈ 0.398 ).Similarly, ( log_2 10.5 ≈ log_2 10 ≈ 3.3219 ), ( (10.5)^{-0.2} ≈ 0.6309 ).Now, compute each term:First term: ( frac{5.6439 * 0.398}{-0.2} ≈ frac{2.246}{-0.2} ≈ -11.23 )Second term: ( frac{0.398}{0.04 * 0.6931} ≈ frac{0.398}{0.027724} ≈ 14.35 )Third term: ( frac{3.3219 * 0.6309}{-0.2} ≈ frac{2.092}{-0.2} ≈ -10.46 )Fourth term: ( frac{0.6309}{0.04 * 0.6931} ≈ frac{0.6309}{0.027724} ≈ 22.76 )Putting it all together:= [ -11.23 - 14.35 ] - [ -10.46 - 22.76 ] ≈ (-25.58) - (-33.22) ≈ 7.64So, the integral approximation for ( S' ≈ frac{1}{3.879} * 7.64 ≈ 1.97 ).Therefore, the total ( S ≈ 0.806 + 1.97 ≈ 2.776 ).Now, going back to the entropy formula:( H = log_2 Z - s * S ≈ log_2(3.879) - 1.2 * 2.776 ).Compute ( log_2(3.879) ≈ 1.95 ) as before.Compute ( 1.2 * 2.776 ≈ 3.331 ).So, ( H ≈ 1.95 - 3.331 ≈ -1.381 ). Again, negative, which is impossible. Clearly, my approximation is flawed.Perhaps I should abandon the integral approximation and instead use the exact sum for ( S ) by computing more terms.Alternatively, maybe I can use the fact that for Zipf's law, the entropy can be approximated as ( H ≈ log_2 n - frac{zeta'(s)}{zeta(s)} ), but I need to verify this.Wait, actually, the entropy for a Zipf distribution is given by ( H = log_2 Z + frac{zeta'(s)}{zeta(s)} ), where ( Z = sum_{k=1}^n k^{-s} ). But I'm not sure if that's correct. Alternatively, perhaps it's ( H = log_2 Z - frac{zeta'(s)}{zeta(s)} ).Wait, let's think about it. The entropy is ( H = -sum p_i log_2 p_i = sum p_i log_2 Z - sum p_i log_2 i^{-s} = log_2 Z + s sum p_i log_2 i ).Wait, that's different from what I had earlier. So, ( H = log_2 Z + s sum p_i log_2 i ).So, in that case, ( H = log_2 Z + s * S ), where ( S = sum p_i log_2 i ).Given that, and with ( S ≈ 2.776 ), then ( H ≈ 1.95 + 1.2 * 2.776 ≈ 1.95 + 3.331 ≈ 5.281 ).That makes more sense, as entropy is positive.Wait, so earlier I had a sign error. The correct formula is ( H = log_2 Z + s * S ), not minus. So, that explains why it was negative before. So, with the correct formula, ( H ≈ 5.281 ).But let's verify this with the exact sum.Alternatively, perhaps I can use the known result that for Zipf's distribution, the entropy is approximately ( H ≈ log_2 n - frac{zeta'(s)}{zeta(s)} ). For ( s = 1.2 ), ( zeta(s) ≈ 1.497 ), and ( zeta'(s) ≈ -0.058 ). So, ( H ≈ log_2 50 - (-0.058 / 1.497) ≈ 5.6439 + 0.0387 ≈ 5.6826 ). This is close to my earlier approximation of 5.281, but not exact. However, since ( n = 50 ) is finite, the approximation might not be perfect.Alternatively, perhaps the exact value is around 5.28 to 5.68.Given that, maybe I can take an average or use the approximation from the sum.Wait, earlier, when I computed ( S ≈ 2.776 ), then ( H ≈ 1.95 + 1.2 * 2.776 ≈ 1.95 + 3.331 ≈ 5.281 ).Alternatively, using the known approximation, ( H ≈ 5.68 ).Given the discrepancy, perhaps I should proceed with the value obtained from the sum, which is approximately 5.28.But to be more accurate, maybe I can compute more terms of ( S ).Compute terms from i=11 to i=20:i=11: ( frac{11^{-1.2}}{3.879} log_2 11 ≈ frac{0.0513}{3.879} * 3.4594 ≈ 0.045 )i=12: ( frac{12^{-1.2}}{3.879} log_2 12 ≈ frac{0.0443}{3.879} * 3.58496 ≈ 0.041 )i=13: ( frac{13^{-1.2}}{3.879} log_2 13 ≈ frac{0.0387}{3.879} * 3.7004 ≈ 0.037 )i=14: ( frac{14^{-1.2}}{3.879} log_2 14 ≈ frac{0.0341}{3.879} * 3.8074 ≈ 0.033 )i=15: ( frac{15^{-1.2}}{3.879} log_2 15 ≈ frac{0.0304}{3.879} * 3.9069 ≈ 0.030 )i=16: ( frac{16^{-1.2}}{3.879} log_2 16 ≈ frac{0.0272}{3.879} * 4 ≈ 0.028 )i=17: ( frac{17^{-1.2}}{3.879} log_2 17 ≈ frac{0.0245}{3.879} * 4.09 ≈ 0.026 )i=18: ( frac{18^{-1.2}}{3.879} log_2 18 ≈ frac{0.0222}{3.879} * 4.17 ≈ 0.023 )i=19: ( frac{19^{-1.2}}{3.879} log_2 19 ≈ frac{0.0202}{3.879} * 4.2479 ≈ 0.021 )i=20: ( frac{20^{-1.2}}{3.879} log_2 20 ≈ frac{0.0186}{3.879} * 4.3219 ≈ 0.020 )Adding these up:0.045 + 0.041 = 0.0860.086 + 0.037 = 0.1230.123 + 0.033 = 0.1560.156 + 0.030 = 0.1860.186 + 0.028 = 0.2140.214 + 0.026 = 0.2400.240 + 0.023 = 0.2630.263 + 0.021 = 0.2840.284 + 0.020 = 0.304So, terms from i=11 to 20 sum to approximately 0.304.Similarly, terms from i=21 to 30 will be smaller. Let's approximate them.Assuming each term is about half of the previous set, but actually, they decrease as i increases. Let's say the sum from i=21 to 30 is approximately 0.15.Similarly, from i=31 to 40: 0.075From i=41 to 50: 0.0375So, total sum from i=11 to 50: 0.304 + 0.15 + 0.075 + 0.0375 ≈ 0.5665Therefore, total ( S ≈ 0.806 + 0.5665 ≈ 1.3725 ).Wait, but earlier I had an integral approximation of 1.97, which is higher. This discrepancy suggests that my manual summation might be underestimating the sum.Alternatively, perhaps I should accept that without a calculator, it's difficult to compute this accurately, and proceed with the approximation that ( H ≈ 5.28 ).Alternatively, perhaps I can use the known result that for Zipf's law with ( s = 1 ), the entropy is ( log_2 n - 1 ), but for ( s = 1.2 ), it's slightly higher.Given the time constraints, I'll proceed with ( H ≈ 5.28 ).Now, moving on to calculating the average dependency parse tree depth ( D ).Given that there are 100 sentences, and each sentence's depth ( d_j = 3 + sinleft(frac{2 pi j}{100}right) ) for ( j = 1, 2, ldots, 100 ).So, ( D = frac{1}{100} sum_{j=1}^{100} d_j = frac{1}{100} sum_{j=1}^{100} left(3 + sinleft(frac{2 pi j}{100}right)right) ).This can be split into two sums:( D = frac{1}{100} left( sum_{j=1}^{100} 3 + sum_{j=1}^{100} sinleft(frac{2 pi j}{100}right) right) ).The first sum is straightforward: ( sum_{j=1}^{100} 3 = 3 * 100 = 300 ).The second sum is ( sum_{j=1}^{100} sinleft(frac{2 pi j}{100}right) ). This is the sum of sine function over a full period, which is known to be zero because the sine function is symmetric and positive and negative areas cancel out.Therefore, ( sum_{j=1}^{100} sinleft(frac{2 pi j}{100}right) = 0 ).Thus, ( D = frac{1}{100} (300 + 0) = 3 ).So, the average depth ( D = 3 ).Now, combining ( H ≈ 5.28 ) and ( D = 3 ), with ( alpha = 0.7 ) and ( beta = 0.3 ), the complexity score ( C ) is:( C = 0.7 * 5.28 + 0.3 * 3 ≈ 3.696 + 0.9 ≈ 4.596 ).Rounding to a reasonable number of decimal places, say two, ( C ≈ 4.60 ).However, considering that my approximation for ( H ) might be a bit off, perhaps the exact value is slightly different. But given the steps, this is a reasonable estimate.Alternatively, if I use the known approximation for ( H ≈ 5.68 ), then ( C = 0.7 * 5.68 + 0.3 * 3 ≈ 3.976 + 0.9 ≈ 4.876 ), which rounds to 4.88.But since my manual calculation gave ( H ≈ 5.28 ), I'll stick with ( C ≈ 4.60 ).However, to be more precise, perhaps I should compute ( H ) more accurately.Wait, another approach: since the sum ( sum_{i=1}^{n} i^{-s} ) for ( s = 1.2 ) and ( n = 50 ) can be computed using the Hurwitz zeta function, but without a calculator, it's difficult. Alternatively, perhaps I can use the fact that ( Z = sum_{k=1}^{50} k^{-1.2} ≈ zeta(1.2) - zeta(1.2, 51) ), where ( zeta(s, q) ) is the Hurwitz zeta function. But I don't have the exact values for these.Alternatively, perhaps I can use the approximation ( Z ≈ zeta(1.2) ) since ( n = 50 ) is large enough. The value of ( zeta(1.2) ≈ 6.847 ). But earlier, my manual sum was around 3.879, which is much lower. So, perhaps the approximation isn't valid for ( n = 50 ).Wait, actually, the Riemann zeta function ( zeta(s) ) is the sum from ( k=1 ) to infinity of ( k^{-s} ). For ( s = 1.2 ), ( zeta(1.2) ≈ 6.847 ). Therefore, the sum from ( k=1 ) to 50 is less than that, which aligns with my manual sum of approximately 3.879.Therefore, ( Z ≈ 3.879 ).Now, going back to the entropy formula:( H = log_2 Z + s * S ≈ log_2(3.879) + 1.2 * 2.776 ≈ 1.95 + 3.331 ≈ 5.281 ).So, ( H ≈ 5.28 ).Therefore, the complexity score ( C = 0.7 * 5.28 + 0.3 * 3 ≈ 3.696 + 0.9 ≈ 4.596 ), which is approximately 4.60.Thus, the final complexity score ( C ) is approximately 4.60.</think>

🧠:<think>Alright, so I have this problem where I need to help a college guidance counselor analyze the academic performance of Asian and Pacific Island American (APIA) students in three STEM fields: Computer Science, Engineering, and Biology. The data provided includes the GPA distributions for each field, which are all normal distributions with given means and variances.First, let me break down the problem into its two sub-problems.Sub-problem 1: I need to find the probability that a randomly selected APIA student has a GPA higher than 3.8 in each of the three fields. That means I have to calculate P(GPA > 3.8) for Computer Science, Engineering, and Biology separately.Sub-problem 2: I need to determine the GPA threshold that corresponds to the top 10% of students in each field. In other words, I need to find the 90th percentile for each of the three distributions.Okay, starting with Sub-problem 1.For each field, the GPA is normally distributed with specific means (μ) and standard deviations (σ). The formula for the normal distribution is N(μ, σ²). So, for each field:- Computer Science: N(3.5, 0.4²)- Engineering: N(3.2, 0.5²)- Biology: N(3.7, 0.3²)To find P(GPA > 3.8), I can use the Z-score formula to standardize the GPA and then use the standard normal distribution table or a calculator to find the probability.The Z-score formula is:Z = (X - μ) / σWhere X is the GPA we're interested in, which is 3.8.Once I calculate the Z-score, I can find the probability that a Z-score is greater than that value. Since standard normal tables usually give the probability that Z is less than a certain value, I'll subtract the table value from 1 to get the probability that Z is greater than that value.Let me go through each field one by one.1. Computer Science:μ = 3.5, σ = 0.4X = 3.8Z = (3.8 - 3.5) / 0.4 = 0.3 / 0.4 = 0.75So, Z = 0.75Looking up Z = 0.75 in the standard normal distribution table, the cumulative probability (P(Z < 0.75)) is approximately 0.7734.Therefore, P(Z > 0.75) = 1 - 0.7734 = 0.2266So, approximately 22.66% chance that a Computer Science student has a GPA higher than 3.8.Wait, that seems a bit high. Let me double-check.Yes, the mean is 3.5, so 3.8 is only 0.3 above the mean, which is 0.75 standard deviations. Since the normal distribution is symmetric, the area beyond 0.75σ is about 22.66%, which is correct.2. Engineering:μ = 3.2, σ = 0.5X = 3.8Z = (3.8 - 3.2) / 0.5 = 0.6 / 0.5 = 1.2So, Z = 1.2Looking up Z = 1.2, the cumulative probability is approximately 0.8849.Therefore, P(Z > 1.2) = 1 - 0.8849 = 0.1151So, approximately 11.51% chance that an Engineering student has a GPA higher than 3.8.That makes sense because 3.8 is 1.2 standard deviations above the mean of 3.2, which is further out than Computer Science, so the probability is lower.3. Biology:μ = 3.7, σ = 0.3X = 3.8Z = (3.8 - 3.7) / 0.3 = 0.1 / 0.3 ≈ 0.3333So, Z ≈ 0.33Looking up Z = 0.33, the cumulative probability is approximately 0.6293.Therefore, P(Z > 0.33) = 1 - 0.6293 = 0.3707So, approximately 37.07% chance that a Biology student has a GPA higher than 3.8.Wait, that seems high. Let me think. The mean is 3.7, so 3.8 is only 0.1 above the mean, which is about 0.33 standard deviations. Since the distribution is normal, the area beyond 0.33σ is about 37%, which is correct.So, summarizing Sub-problem 1:- Computer Science: ~22.66%- Engineering: ~11.51%- Biology: ~37.07%Now, moving on to Sub-problem 2.I need to find the GPA threshold that corresponds to the top 10% of students in each field. That is, the 90th percentile. This means that 90% of students have a GPA below this threshold, and 10% have a GPA above it.To find the 90th percentile, I need to find the GPA value such that P(GPA < X) = 0.90.Again, using the Z-score formula, but this time, I need to find the Z-score that corresponds to the 90th percentile and then convert it back to the original GPA scale.In the standard normal distribution, the Z-score corresponding to the 90th percentile is approximately 1.28. This is because P(Z < 1.28) ≈ 0.8997, which is roughly 0.90.So, Z = 1.28Now, using the formula:X = μ + Z * σLet me calculate this for each field.1. Computer Science:μ = 3.5, σ = 0.4X = 3.5 + 1.28 * 0.4 = 3.5 + 0.512 = 4.012So, approximately 4.01 GPA.Wait, that's a 4.01, which is above the typical 4.0 scale. Is that possible? Well, technically, the normal distribution extends beyond the mean by several standard deviations, but in reality, GPAs are capped at 4.0. However, since the problem states that the distribution is normal, I think we should proceed with the calculation as is, even if it exceeds 4.0.So, the threshold is approximately 4.01.2. Engineering:μ = 3.2, σ = 0.5X = 3.2 + 1.28 * 0.5 = 3.2 + 0.64 = 3.84So, approximately 3.84 GPA.3. Biology:μ = 3.7, σ = 0.3X = 3.7 + 1.28 * 0.3 = 3.7 + 0.384 = 4.084Again, similar to Computer Science, this is above 4.0, but we'll go with the calculation.So, approximately 4.08 GPA.Wait, let me verify the Z-score for the 90th percentile.Yes, the Z-score for 0.90 is indeed approximately 1.28. Sometimes it's approximated as 1.28, sometimes as 1.2815 or 1.2816. For simplicity, 1.28 is acceptable.Alternatively, using more precise tables or a calculator, the exact Z-score for 0.90 is about 1.28155. So, using 1.28155 would give a slightly more accurate result.Let me recalculate with Z = 1.28155.1. Computer Science:X = 3.5 + 1.28155 * 0.4 = 3.5 + 0.51262 ≈ 4.0126So, approximately 4.0126, which is about 4.01.2. Engineering:X = 3.2 + 1.28155 * 0.5 = 3.2 + 0.640775 ≈ 3.8408So, approximately 3.8408, which is about 3.84.3. Biology:X = 3.7 + 1.28155 * 0.3 = 3.7 + 0.384465 ≈ 4.0845So, approximately 4.0845, which is about 4.08.Therefore, the thresholds are approximately:- Computer Science: 4.01- Engineering: 3.84- Biology: 4.08But since GPAs are typically reported to two decimal places, we can round these to:- Computer Science: 4.01- Engineering: 3.84- Biology: 4.08However, in reality, GPAs can't exceed 4.0, so for Computer Science and Biology, the 90th percentile would be at the maximum GPA of 4.0. But since the problem states the distributions as normal, I think we should proceed with the calculated values, even if they exceed 4.0.Alternatively, if we consider that GPAs can't go beyond 4.0, then for Computer Science and Biology, the 90th percentile would be 4.0. But since the problem doesn't specify this constraint, I think it's safer to provide the calculated values.So, summarizing Sub-problem 2:- Computer Science: ~4.01- Engineering: ~3.84- Biology: ~4.08Wait, but let me think again. If the distribution is normal, it's possible for the 90th percentile to be above 4.0, but in reality, GPAs are capped. So, perhaps the counselor should be aware that in fields where the calculated threshold exceeds 4.0, the actual threshold would be 4.0.But since the problem doesn't specify this, I think we should proceed with the calculated values.Therefore, the final answers for Sub-problem 2 are approximately 4.01, 3.84, and 4.08 for Computer Science, Engineering, and Biology respectively.Let me just recap the steps to ensure I didn't make any mistakes.For Sub-problem 1:1. For each field, calculate Z = (3.8 - μ) / σ2. Use Z-table to find P(Z < z)3. Subtract from 1 to get P(Z > z)For Sub-problem 2:1. Find Z corresponding to 90th percentile (Z ≈ 1.28)2. Calculate X = μ + Z * σ for each fieldYes, that seems correct.Just to double-check the Z-scores:- For Computer Science: Z = 0.75, which is correct.- For Engineering: Z = 1.2, correct.- For Biology: Z ≈ 0.33, correct.And for the percentiles:- Computer Science: X ≈ 4.01- Engineering: X ≈ 3.84- Biology: X ≈ 4.08Yes, all calculations seem correct.One thing to note is that in Biology, the mean is already 3.7, so the 90th percentile is quite high, almost 4.08, which is beyond 4.0. Similarly, Computer Science's 90th percentile is just over 4.0. Engineering is more reasonable at 3.84.I think that's all. I don't see any errors in my calculations.</think>