Solution Explanation

To find the expected length of time the contract has been in place (E[Y]), we need to:

1. Find the marginal distribution of Y by integrating the joint density function over all possible values of X: f_Y(y) = \int_{2}^{10} \frac{1}{64}(10 - xy^2)\,dx $`$2`. **Compute the integral**: f_Y(y) = \frac{1}{64}\int_{2}^{10} (10 - xy^2),dx = \frac{1}{64}\left[10x - \frac{x^2y^2}{2}\right]_{2}^{10} $$$$ = \frac{1}{64}\left[(100 - 50y^2) - (20 - 2y^2)\right] = \frac{1}{64}[80 - 48y^2] $$So,$$ f_Y(y) = \begin{cases} \frac{1}{64}[80 - 48y^2], & 0 < y < 1 \ 0, & \text{elsewhere} \end{cases} $`$3`. Calculate the expected value of Y: $$E[Y] = \int_{0}^{1} y \cdot f_Y(y)\,dy = \int_{0}^{1} y \cdot \frac{1}{64}[80 - 48y^2]\,dy$$ $$= \frac{1}{64}\int_{0}^{1} (80y - 48y^3)\,dy = \frac{1}{64}\left[40y^2 - 12y^4\right]_{0}^{1}$$ $$= \frac{1}{64}[40 - 12] = \frac{28}{64} = \frac{7}{16} = 0.4375$$

Key Concepts:

• Marginal distribution from joint distribution
• Expected value of a continuous random variable
• Integration of polynomial functions

Verification: The marginal distribution integrates to 1 over its support:

$$\int_{0}^{1} \frac{1}{64}[80 - 48y^2]\,dy = \frac{1}{64}\left[80y - 16y^3\right]_{0}^{1} = \frac{1}{64}[80 - 16] = \frac{64}{64} = 1$$

This confirms our marginal distribution is properly normalized.