The predicted value of the student's score on test A with the smallest MSE is 72, and the value of this minimum MSE is 100.
Given a student's score on test B is 100, the predicted value of her score on test A with the smallest mean squared error (MSE) is equal to the mean score of test A. This is because the mean of test A minimizes the sum of squared deviations from the predicted value, which is the MSE.
To calculate the MSE for the mean of test A, we need the variance of test A. The variance of test A is given by:
σ^2_A = (1/n) * Σ(x_i - μ_A)^2
where:
σ^2_A is the variance of test A
n is the number of students
x_i is the score of the i-th student on test A
μ_A is the mean of test A
Using the data from Exercise 2, we can calculate the mean and variance of test A:
μ_A = (1/10) * Σx_i = 72
σ^2_A = (1/10) * Σ(x_i - 72)^2 = 100
Therefore, the predicted value of the student's score on test A with the smallest MSE is 72, and the value of this minimum MSE is 100.