Final answer:
The expected value of Z is 3 and the variance of Z is 51. To find P(0 ≤ Z ≤ 1.5), we standardize the values using the mean and standard deviation of Z.
Step-by-step explanation:
To determine the expected value (E(Z)) and variance (V(Z)) of Z, we can use the properties of expectations and variances. Since X and Y are independent variables, E(Z) = E(3X) - E(2Y) + E(1) and V(Z) = V(3X) + V(2Y) + 0 (since the variance of a constant is 0).
Since X ~ N(2,5), E(Z) = 3 * E(X) - 2 * E(Y) + 1 = 3 * 2 - 2 * 3 + 1 = 3.
Similarly, V(Z) = 3^2 * V(X) + (-2)^2 * V(Y) = 9 * 5 + 4 * 2 = 51.
Therefore, the expected value of Z is 3 and the variance of Z is 51.
To find P(0 ≤ Z ≤ 1.5), we can standardize the values using the mean and standard deviation of Z. Since Z = 3X - 2Y + 1 and X ~ N(2,5) and Y ~ N(3,4), we have Z ~ N(3E(X) - 2E(Y) + E(1), 3^2V(X) + (-2)^2V(Y)), which simplifies to Z ~ N(3(2) - 2(3) + 1, 51).
To standardize the values, we use the formula: Z = (X - μ) / σ, where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. Plugging in the values, we get Z = (1.5 - 3) / sqrt(51) = -1.5 / sqrt(51).
Using a standard normal table or a calculator, we can find that P(Z ≤ -1.5 / sqrt(51)) ≈ 0.066, which is approximately equal to P(Z ≥ -1.5 / sqrt(51)) ≈ 1 - 0.066 = 0.934.