Suppose X is a random variable whose CDF is given by

F(x) = 0, X<0
X^3, 0<X<1;
1, 1<X

Then the mean for this random variable is:

I don't understand how to do CDF (cumulitive distribution function) or how to get the mean of it. Thank You!

The average or mean of a distribution F(x) is the integral of F(x)dx

(which in this case = 1/4 when the limits are 0 and 1) divided by the width of the interval (which in this case = 1). So the answer is 1/4.

To find the mean of a random variable X, you need to calculate the expected value, which is denoted as E(X).

In order to find E(X), we can use the formula:

E(X) = ∫x * f(x) dx

Where f(x) is the probability density function (PDF) of the random variable.

In this case, we are given the cumulative distribution function (CDF), denoted as F(x), instead of the PDF. Hence, we need to find the PDF first.

To find the PDF from the given CDF, we can differentiate it. The PDF is the derivative of the CDF.

Let's find the PDF for the given CDF:

F(x) = 0, X < 0
X^3, 0 < X < 1
1, X > 1

Since the CDF is constant (equal to 0) for X < 0 and X > 1, the PDF will be 0 for X < 0 and X > 1.

For 0 < X < 1, the CDF is X^3. To find the PDF in this interval, we differentiate the CDF with respect to X:

f(x) = d/dx (X^3) = 3X^2

Now that we have the PDF, we can find the mean:

E(X) = ∫x * f(x) dx

We need to calculate this integral over the range 0 to 1, as the PDF is defined only in this range.

So, the mean for this random variable will be:

E(X) = ∫x * 3X^2 dx (from 0 to 1)

E(X) = 3 * ∫X^3 dx (from 0 to 1)

E(X) = 3 * [(1/4) * X^4] (from 0 to 1)

E(X) = 3 * [(1/4) * (1^4 - 0^4)]

E(X) = 3 * (1/4)

E(X) = 3/4

So, the mean for this random variable is 3/4.