Question

Suppose that X AND Y are random variables whose joint density is given by:

f(XY) (x,y) = 1/π √3 - 2/3 ( x² + y² -xy) for (x,y) in R².
Calculate the value of the expression (2X+Y).

Answer 1

Given normal distributions X~ N(5, 6) and Y~ N(2, 1), and provided values x = 17 and y = 4, the z-scores are calculated respectively. For Y, the z-score is found to be 2 by employing the formula Z = (Y - μ) / σ.

Step-by-step explanation:

The question involves calculating a value based on the joint density function of two random variables, X and Y. However, the latter part of the question appears to perturb to the calculations involving normal distributions and z-scores. Given normal distributions X~ N(5, 6) and Y~ N(2, 1), if x = 17, then z for X is 2. If y = 4, to find the z-score for Y, we use the formula:

Z = (Y - μ) / σ

Where μ is the mean and σ is the standard deviation of the normal distribution. Substituting the mean and standard deviation for Y into the equation:

Z = (4 - 2) / 1 = 2

Thus, if y = 4, the z-score for Y is also 2.