Question

Let x ∼ unif[−4, 4], and y = (x 1)2. (a)find f y( 9 4 ) and f y(16), where f y(·) is the probability density function of y. separate your answers with a comma. (b)find p{y < 16}.

Answer 1

Without the specific probability density function fy(y), we cannot determine fy(9/4) and fy(16), but we calculated that P(y < 16) is 7/8 by finding the corresponding x-values for which y < 16 and using the properties of the uniform distribution.

Step-by-step explanation:

The question at hand involves the continuous uniform distribution, transformation of random variables, and computing probabilities for a given distribution. We begin with a random variable x that follows a uniform distribution on the interval [-4, 4], and we define a new variable y = (x + 1)^2. For part (a), we need to find the probability density function (fy(y)) of y at two specific points, 9/4 and 16, then for part (b) we calculate the cumulative probability P(y < 16). Since fy is not provided, we cannot determine fy(9/4) and fy(16).

However, we can approach the cumulative probability P(y < 16) by first finding the corresponding values for x that make y < 16. In this case, we solve for (x + 1)^2 < 16, giving us -5 < x < 3, remembering that x cannot be smaller than -4 (lower bound of the original distribution). Then, the cumulative probability is the area under the distribution of x from -4 to 3, which is easily calculated as the difference in x values divided by the total span of the distribution, 8. Thus, P(y < 16) = (3 - (-4)) / 8 = 7/8.