Final answer:
To solve the given problem, we need to find the value of c that makes the density function integrate to 1. Then, we can find the cumulative distribution function F(x) by integrating the density function from negative infinity to x. Finally, we can find the mean and variance of X using the formulas.
Step-by-step explanation:
To solve the given problem, we need to follow these steps:
a) To show that c=12, we need to find the value of c that makes the density function integrate to 1. Since the density function is given by f(x) = c(x^2 - 2x + 3) for 1≤x≤3 and f(x) = 0 for x<1 or x>3, we can integrate this function over the interval [1,3] and set it equal to 1 to solve for c:
1 = ∫[1,3] c(x^2 - 2x + 3) dx
1 = c[(1/3)x^3 - x^2 + 3x] from 1 to 3
c = 12
b) To find the cumulative distribution function F(x), we need to integrate the density function from negative infinity to x. Since the density function is 0 for x<1, ∫[1,x] f(t) dt = ∫[1,x] 12(t^2 - 2t + 3) dt = 12(∫[1,x] t^2 - 2t + 3) dt = 12[(1/3)x^3 - x^2 + 3x] - 12[(1/3) - 1 + 3] = 4x^3 - 12x^2 + 36x - 20
c) To find the probabilities:
d) To find the mean and variance of X, we need to use the formulas:
Mean (E[X]) = ∫x f(x) dx = ∫[1,3] x * 12(x^2 - 2x + 3) dx = ... (calculations)
Variance (Var(X)) = E[X^2] - (E[X])^2 = ... (calculations)