2.4 – A Simple PDF Example

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Question

Let f(x) = k(3x2 + 1).

  1. Find the value of k that makes the given function a PDF on the interval 0 ≤ x ≤ 2.
  2. Let X be a continuous random variable whose PDF is f(x). Compute the probability that X is between 1 and 2.
  3. Find the distribution function of X.
  4. 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


source: http://wiki.ubc.ca/Science:MATH105_Probability/Lesson_2_CRV/2.06_A_Simple_PDF_Example

Previous: 2.3 – The Probability Density Function

Next: 2.5 – Some Common Continuous Distributions