Question

Let (x) denote the probability density function (p.d.f.) of a continuous variable, X. Let F(x) denote the cumulative distribution function (c.d.f.) of the p.d.f. Suppose... (x)=c(1 2x−1 8x2) o x ∈[0,2]

What value of c makes (x) a legitimate p.d.f.?

Answer 1

The constant c in the given pdf (x)=c(1+2x-¼x²) must be chosen such that the integral of this function from 0 to 2 equals 1. The integration process is done term-by-term, and the value of c is the inverse of the resulting computation.

Step-by-step explanation:

To make the function (x) = c(1 + 2x - ¼x2) a legitimate probability density function (pdf), the constant c must be chosen so that the total area under the curve (x) on the interval [0, 2] is equal to 1. This corresponds to the total probability for a continuous random variable X being 1, which is a fundamental property of probability distributions.

To find the correct value of c, we integrate the function (x) over the interval [0, 2]:

∫02 c(1 + 2x - ¼x2) dx = 1

The integration can be done term-by-term:

c∫ (dx) + c∫ (2x dx) - c∫ (¼x2 dx)

By solving the integral, we find the value of c that satisfies the equation. Assuming that students reading this response do not have a background in calculus, the integral would typically be solved using techniques from calculus

In this case, we would compute:

• 
  
• c(2 - 0)
• 
  
• + c(22 - 0)
• 
  
• - c(¼(23/3 - 0))

By solving this, we get c to be the inverse of the result obtained, ensuring that the total area under the curve is 1.