a. The marginal density function for y2 is y2^2, defined for 0 <= y2 <= 1.
b. The conditional density function f(y1 | y2) is defined for all values of y2 within the range 0 <= y2 <= 1.
c. The probability that more than half a tank is sold given that three-fourths of a tank is stocked is 0.5787.
a. Finding the marginal density function for y2:
The marginal density function for y2 is given by:
f_y2(y2) = ∫ f_joint(y1, y2) dy1
where f_joint(y1, y2) is the joint density function of y1 and y2.
Substituting the given joint density function, we get:
f_y2(y2) = ∫ 2 * y1 * y2 dy1
Evaluating the integral, we get:
f_y2(y2) = y2^2
Therefore, the marginal density function for y2 is y2^2, defined for 0 <= y2 <= 1.
b. Determining the values of y2 for which the conditional density f(y1 | y2) is defined:
The conditional density function f(y1 | y2) is defined for all values of y2 within the range 0 <= y2 <= 1. This is because the joint density function f_joint(y1, y2) is positive for all values of y1 and y2 within this range.
c. Calculating the probability that more than half a tank is sold given that three-fourths of a tank is stocked:
Let y1 represent the proportion of the tank sold and y2 represent the proportion of the tank stocked. We are given that y2 = 0.75. We want to find the probability that y1 > 0.5, given that y2 = 0.75.
Using the conditional density function f(y1 | y2), we can calculate the probability that y1 > 0.5 as follows:
P(y1 > 0.5 | y2 = 0.75) = ∫ f(y1 | y2) dy1, where y1 > 0.5
Substituting the given conditional density function, we get:
P(y1 > 0.5 | y2 = 0.75) = ∫ 2 * y1 * 0.75 dy1, where y1 > 0.5
Evaluating the integral, we get:
P(y1 > 0.5 | y2 = 0.75) = 0.5787
Therefore, the probability that more than half a tank is sold given that three-fourths of a tank is stocked is 0.5787.