answer:To find the optimal boundary that separates "Good" performers from "Bad" performers using discriminant analysis, we first need to calculate the mean vectors, covariance matrices, and the within-class scatter matrix for both classes. Then, we can use these values to compute the linear discriminant function and find the optimal boundary. Step 1: Calculate the mean vectors for both classes. Good performers: Age mean: (34 + 42 + 39 + 35 + 48 + 38) / 6 = 236 / 6 = 39.33 Experience mean: (8 + 14 + 10 + 9 + 20 + 12) / 6 = 73 / 6 = 12.17 Mean vector for Good performers: (39.33, 12.17) Bad performers: Age mean: (28 + 24 + 29 + 31) / 4 = 112 / 4 = 28 Experience mean: (4 + 2 + 6 + 7) / 4 = 19 / 4 = 4.75 Mean vector for Bad performers: (28, 4.75) Step 2: Calculate the covariance matrices for both classes. Good performers: Covariance matrix = [Σ(x_i - mean_x)(x_i - mean_x)^T] / (n - 1) For simplicity, we'll calculate the covariance matrix for age and experience separately and then combine them. Covariance_age = Σ(age_i - mean_age)^2 / (n - 1) = [(34-39.33)^2 + (42-39.33)^2 + (39-39.33)^2 + (35-39.33)^2 + (48-39.33)^2 + (38-39.33)^2] / 5 = 52.47 Covariance_experience = Σ(experience_i - mean_experience)^2 / (n - 1) = [(8-12.17)^2 + (14-12.17)^2 + (10-12.17)^2 + (9-12.17)^2 + (20-12.17)^2 + (12-12.17)^2] / 5 = 34.47 Covariance matrix for Good performers: [52.47, 0] [0, 34.47] Bad performers: Covariance matrix = [Σ(x_i - mean_x)(x_i - mean_x)^T] / (n - 1) Covariance_age = Σ(age_i - mean_age)^2 / (n - 1) = [(28-28)^2 + (24-28)^2 + (29-28)^2 + (31-28)^2] / 3 = 8.67 Covariance_experience = Σ(experience_i - mean_experience)^2 / (n - 1) = [(4-4.75)^2 + (2-4.75)^2 + (6-4.75)^2 + (7-4.75)^2] / 3 = 5.25 Covariance matrix for Bad performers: [8.67, 0] [0, 5.25] Step 3: Calculate the within-class scatter matrix. Within-class scatter matrix = Covariance matrix for Good performers * (n_good - 1) + Covariance matrix for Bad performers * (n_bad - 1) Within-class scatter matrix: [52.47 * 5 + 8.67 * 3, 0] [0, 34.47 * 5 + 5.25 * 3] Within-class scatter matrix: [314.35, 0] [0, 186.6] Step 4: Compute the linear discriminant function. The linear discriminant function is given by: w = inv(S_w) * (mean_good - mean_bad) w = [1 / 314.35, 0] [0, 1 / 186.6] * [(39.33 - 28), (12.17 - 4.75)] w = [0.00318, 0] [0, 0.00536] * [11.33, 7.42] w = [0.036, 0.0398] Step 5: Find the optimal boundary. The optimal boundary can be found by calculating the midpoint between the mean vectors of both classes and then projecting it onto the linear discriminant function. Midpoint = (mean_good + mean_bad) / 2 Midpoint = (39.33 + 28, 12.17 + 4.75) / 2 Midpoint = (33.665, 8.46) Optimal boundary = w^T * Midpoint Optimal boundary = [0.036, 0.0398] * [33.665, 8.46] Optimal boundary = 0.036 * 33.665 + 0.0398 * 8.46 Optimal boundary = 1.21034 + 0.33668 Optimal boundary = 1.54702 The optimal boundary that separates "Good" performers from "Bad" performers based on the age and experience of employees is approximately 1.547.

answer:To predict the class for a new data point (5, 5) using discriminant analysis, we need to compute the discriminant function for each class and compare their values. The discriminant function is given by: g(x) = -0.5 * (x - μ).T * Σ^(-1) * (x - μ) - 0.5 * ln(det(Σ)) + ln(P(ω)) where x is the data point, μ is the mean, Σ is the covariance matrix, and P(ω) is the prior probability of the class. Assuming equal prior probabilities for both classes, we can ignore the last term. For Class 0: μ0 = [3, 4] Σ0 = [[2, 1], [1, 2]] For Class 1: μ1 = [7, 6] Σ1 = [[3, 1], [1, 3]] Let's compute the discriminant function for each class: 1. Calculate the inverse and determinant of the covariance matrices: Σ0_inv = [[1.5, -0.5], [-0.5, 1.5]] det(Σ0) = 3 Σ1_inv = [[1, -0.3333], [-0.3333, 1]] det(Σ1) = 8 2. Calculate the difference between the data point and the mean for each class: x - μ0 = [2, 1] x - μ1 = [-2, -1] 3. Calculate the discriminant function for each class: g0(x) = -0.5 * [2, 1] * Σ0_inv * [2, 1].T - 0.5 * ln(3) g0(x) = -0.5 * [2, 1] * [4.5, -1.5] - 0.5 * ln(3) g0(x) = -0.5 * (6) - 0.5 * ln(3) g0(x) = -3 - 0.5 * ln(3) g1(x) = -0.5 * [-2, -1] * Σ1_inv * [-2, -1].T - 0.5 * ln(8) g1(x) = -0.5 * [-2, -1] * [1.6667, 0.3333] - 0.5 * ln(8) g1(x) = -0.5 * (-1) - 0.5 * ln(8) g1(x) = 0.5 - 0.5 * ln(8) Now, compare the values of g0(x) and g1(x): g0(x) = -3 - 0.5 * ln(3) g1(x) = 0.5 - 0.5 * ln(8) Since g1(x) > g0(x), the predicted class for the new data point (5, 5) is Class 1.

answer:To solve this problem, we will use the discriminant analysis method, which is a statistical technique used to classify data points into two or more groups based on their characteristics. In this case, we have two sets of data: Set A and Set B. We want to find the boundary value that separates these two sets with a classification accuracy of at least 90%. First, let's define the variables: Set A: - n1 = 10 (number of data points) - μ1 = 25 (mean) - σ1 = 5 (standard deviation) Set B: - n2 = 12 (number of data points) - μ2 = 20 (mean) - σ2 = 3 (standard deviation) To find the boundary value, we will use the following discriminant function: D(x) = (x - μ1)^2 / σ1^2 - (x - μ2)^2 / σ2^2 We want to find the value of x that satisfies the condition D(x) = 0, which indicates the boundary between the two sets. To do this, we will set D(x) = 0 and solve for x: (x - μ1)^2 / σ1^2 - (x - μ2)^2 / σ2^2 = 0 Now, let's plug in the values for μ1, σ1, μ2, and σ2: (x - 25)^2 / 5^2 - (x - 20)^2 / 3^2 = 0 Solve for x: (1/25)(x^2 - 50x + 625) - (1/9)(x^2 - 40x + 400) = 0 Multiply both sides by 225 (the least common multiple of 25 and 9) to eliminate the fractions: 9(x^2 - 50x + 625) - 25(x^2 - 40x + 400) = 0 Expand and simplify: 9x^2 - 450x + 5625 - 25x^2 + 1000x - 10000 = 0 Combine like terms: -16x^2 + 550x - 4375 = 0 Now, we have a quadratic equation in the form of ax^2 + bx + c = 0. We can use the quadratic formula to solve for x: x = (-b ± √(b^2 - 4ac)) / 2a Plug in the values for a, b, and c: x = (-550 ± √(550^2 - 4(-16)(-4375))) / (2(-16)) x = (-550 ± √(302500 - 280000)) / (-32) x = (-550 ± √(22500)) / (-32) x = (-550 ± 150) / (-32) There are two possible solutions for x: x1 = (-550 + 150) / (-32) = -400 / (-32) = 12.5 x2 = (-550 - 150) / (-32) = -700 / (-32) = 21.875 Since we are looking for the boundary value that separates the two sets, we should choose the value that is between the means of the two sets (25 and 20). Therefore, the boundary value is x = 21.875. To achieve a classification accuracy of at least 90%, the boundary value of 21.875 should be used to separate Set A and Set B.

answer:To calculate the discriminant function, we first need to find the inverse of the covariance matrix. The covariance matrix is given as: | | GPA | Entrance exam | | --- | --- | --- | | GPA | 0.1 | 8 | | Entrance exam | 8 | 100 | The determinant of the covariance matrix is: det = (0.1 * 100) - (8 * 8) = 10 - 64 = -54 The inverse of the covariance matrix is (1/det) * adjoint(matrix), where adjoint(matrix) is the matrix obtained by swapping the diagonal elements and changing the sign of the off-diagonal elements. So, the inverse covariance matrix is: | | GPA | Entrance exam | | --- | --- | --- | | GPA | -100/54 | 8/54 | | Entrance exam | 8/54 | -0.1/54 | Now, let's calculate the difference in means for GPA and entrance exam scores between the two schools: ΔGPA = 3.6 - 3.2 = 0.4 ΔEntranceExam = 170 - 150 = 20 Next, we multiply the inverse covariance matrix by the difference in means: | | GPA | Entrance exam | | --- | --- | --- | | GPA | -100/54 * 0.4 + 8/54 * 20 | -0.1/54 * 0.4 + 8/54 * 20 | This results in: | | GPA | Entrance exam | | --- | --- | --- | | GPA | -40/54 + 160/54 | -0.4/54 + 160/54 | Simplifying the matrix: | | GPA | Entrance exam | | --- | --- | --- | | GPA | 120/54 | 159.6/54 | Now, we can calculate the discriminant function value (D) for each school: D_A = (120/54) * 3.2 + (159.6/54) * 150 D_B = (120/54) * 3.6 + (159.6/54) * 170 D_A = 2.222 * 3.2 + 2.956 * 150 ≈ 7.111 + 443.333 ≈ 450.444 D_B = 2.222 * 3.6 + 2.956 * 170 ≈ 8.000 + 500.556 ≈ 508.556 Since the cutoff score is set at a discriminant value of 0, we can see that both schools' applicants have discriminant values greater than 0. However, School B's applicants have a higher discriminant value (508.556) compared to School A's applicants (450.444). Therefore, applicants from School B are more likely to be admitted.