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Cumulative Distribution Function (CDF)

A cumulative distribution function F(x)F(x) assigns a "running total" probability for the each value of XX.

For discrete random variables, we use a cumulative distribution table.
    x       F(x)=P(Xx)  x1P(x1)x2P(x1)+P(x2)1\begin{array}{c|c} \ \ \ \ x\ \ \ \ \ &\ \ F(x)=P(X\le x)\ \ \\ \hline x_1&P(x_1)\\ x_2&P(x_1)+P(x_2)\\ \vdots&\vdots\\ &1 \end{array}

Wize Tip
  • The column xx must contain all possible values of XX
  • The first F(x)F\left(x\right) value in the CDF is the same as in the PDF
  • The F(xn)F\left(x_n\right) values must increase as we go down the column
  • The last F(x)F\left(x\right) value in the CDF must equal 1
  • Given a PDF you can always get a CDF (vice versa)
  • Don't forget about calculating probabilities using the complement rule!

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Example: Cumulative Distribution Function (CDF)

Let XX be a discrete random variable with values all possible even integers. The following cumulative distribution information is given:


a) Find P(X>4)P\left(X>4\right)

P(X>4)=P(X=6,8,10,12,14,)P\left(X>4\right)=P\left(X=6,8,10,12,14,\dots\right)

Since we don’t have information for X>10X > 10, we will use the complement

P(X>4)=1P(X4)=10.25=0.75P\left(X>4\right)=1−P\left(X\le4\right)=1−0.25=0.75


b) Find P(2X8)P\left(2\le X\le8\right)


P(2X8)=P(X=2,4,6,8)=P(𝑋8)P(𝑋0)=0.590.05=0.54P\left(2\le X\le8\right)=P\left(X=2,4,6,8\right)=P\left(𝑋\le8\right)−P\left(𝑋\le0\right)=0.59−0.05=0.54



c) Find P(X=6)P\left(X=6\right)

The answer is not 0.39!

P(X=6)=F(6)F(4)=0.390.25=0.14P\left(X=6\right)=F\left(6\right)-F\left(4\right)=0.39-0.25=0.14

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Or, you could have created the probability distribution function first, and then answered the questions.


The cumulative distribution function (cdf) of a discrete random variable that only takes on integer values is given by F(x)F(x).

Suppose F(1)=0.61F(-1)=0.61, F(2)=0.75F(2)=0.75, F(3)=0.80F(3)=0.80,F(4)=0.87F(4)=0.87 and F(5)=0.95F(5)=0.95

Find P(2<X3)+P(X4)P\left(2<X\le3\right)+P\left(X\ge4\right)