# Science:MATH105 Probability/Lesson 2 CRV/2.06 A Simple PDF Example

## Question

Let *f*(*x*) = *k*(3*x*^{2} + 1).

- Find the value of
*k*that makes the given function a PDF on the interval 0 ≤*x*≤ 2. - Let
*X*be a continuous random variable whose PDF is*f*(*x*). Compute the probability that*X*is between 1 and 2. - Find the distribution function of
*X*. - Find the probability that
*X*is*exactly*equal to 1.

## Solution

### Part 1)

Therefore, *k* = 1/10.

Notice that *f*(*x*) ≥ 0 for all *x*. Also notice that we can rewrite this PDF in the obvious way so that it is defined for all real numbers:

### Part 2)

Using our value for *k* from Part 1:

Therefore, Pr(1 ≤ *X* ≤ 2) is 4/5.

### Part 3)

Using the Fundamental Theorem of Calculus, the CDF of *X* at *x* in [0,2] is

We can also easily verify that *F*(*x*) = 0 for all *x* < 0 and that *F*(*x*) = 1 for all *x* > 2.

### Part 4)

Since *X* is a continuous random variable, we immediately know that the probability that it equals any one particular value must be zero. More directly, we compute