Question

Let X be a continuous rv with the following cdf. F(x) = 0 x ≤ 0 x 7 1 + ln 7 x 0 < x ≤ 7 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 P(X ≤ 2)? (Round your answer to three decimal places.) (b) What is P(2 ≤ X ≤ 3)? (Round your answer to three decimal places.) (c) What is the pdf of X?

Answer 1

P(X ≤ 2) is roughly 0.642, P(2 ≤ X ≤ 3) is approximately 0.351, and the pdf of X is f(x) = 1/7 + 1/(xln(7)) for the range 0 < x ≤ 7.

Step-by-step explanation:

Given the cumulative distribution function (CDF) for a continuous random variable (rv) X, we need to find the following probabilities:

1. P(X ≤ 2)
2. P(2 ≤ X ≤ 3)

We also need to determine the probability density function (pdf) of X.

The CDF of X provided in the given range is F(x) = x/7 + ln(x)/ln(7) for 0 < x ≤ 7.

We find P(X ≤ 2) by evaluating the CDF at x=2:

F(2) = 2/7 + ln(2)/ln(7) = 0.2857 + 0.3562 = 0.6419

Then, we round this value to three decimal places, obtaining 0.642.

To calculate P(2 ≤ X ≤ 3), we find F(3) and then subtract F(2) from it:

F(3) = 3/7 + ln(3)/ln(7)

P(2 ≤ X ≤ 3) = F(3) - F(2) = 0.4286 + 0.5646 - 0.6419 = 0.3513, which we round to 0.351.

For the pdf of X, we differentiate the CDF. The derivative of x/7 is 1/7, and the derivative of ln(x)/ln(7) is 1/(xln(7)). Combined, the pdf f(x) for 0 < x ≤ 7 is:

f(x) = 1/7 + 1/(xln(7))