Answer: 0.4375

## 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** $$ f_Y(y) = \int_{-\infty}^{\infty} f_{XY}(x,y)\, dx = \int_2^{10} \frac{1}{64}(10 - xy^2)\, dx $$ $$ f_Y(y) = \frac{1}{64} \left[10x - \frac{x^2 y^2}{2}\right]_2^{10} = \frac{1}{64}[80 - 48y^2] $$ So, $$ f_Y(y) = \begin{cases} \frac{1}{64}[80 - 48y^2], & 0 < y < 1 \\ 0, & \text{elsewhere} \end{cases} $$ 2. **Calculate the expectation E[Y]** $$ E(Y) = \int_{-\infty}^{\infty} y f_Y(y)\, dy = \int_0^1 y \cdot \left(\frac{1}{64}[80 - 48y^2]\right) dy $$ $$ E(Y) = \frac{1}{64} \int_0^1 (80y - 48y^3) dy = \frac{1}{64} \left[40y^2 - 12y^4\right]_0^1 $$ $$ E(Y) = \frac{1}{64}[40 - 12] = \frac{28}{64} = \frac{7}{16} = 0.4375 $$ Therefore, the expected length of time the contract has been in place is **0.4375**.

Author: Tanishq Prabhu