Question

Let X be a continuous rv with the following cdf. XSO ****) + In OCXS7 1 X > 7 (This type of cdf is suggested in the article "Variability in Measured Bedload-Transport Rates": as a model for a certain hydrologic variable.] (a) What is PIX S 2)? (Round your answer to three decimal places.) 0.6436 (b) What is P(2 S XS 5)? (Round your answer to three decimal places.) (C) What is the pdf of X? 31 (5) X

Answer 1

The cdf of the continuous random variable X is given. The probabilities P(X ≤ 2) and P(2 ≤ X ≤ 5) are calculated using the cdf. The pdf of X is obtained by differentiating the cdf.

Step-by-step explanation:

The given cdf for the continuous random variable X can be written as:

(a) To find P(X ≤ 2), we can substitute x = 2 into the cdf expression:

F(2) = 0 for x ≤ 7

F(2) = 1 for x > 7

Therefore, P(X ≤ 2) = 0.6436 (rounded to three decimal places).

(b) To find P(2 ≤ X ≤ 5), we can subtract the cdf values at x = 2 and x = 5:

P(2 ≤ X ≤ 5) = F(5) - F(2)

P(2 ≤ X ≤ 5) = 1 - 0.6436 = 0.3564 (rounded to three decimal places).

(c) To find the pdf of X, we can differentiate the cdf:

f(x) = d/dx F(x)

Since the cdf has a jump at x = 7, the pdf f(x) will have a Dirac delta function at x = 7, denoted as δ(x - 7). The pdf will be zero for all other values of x. Therefore, the pdf of X is:

f(x) = δ(x - 7)