A cumulative distribution offers a convenient tool for determining probabilities for a given random variable. As you will recall from a previous learning outcome statement, a cumulative distribution function, F(x), gives the probability that the random variable X is less than or equal to x, for every value x. It is usually expressed as:

$$ F(x) = P(X \le x) $$

**Example 1: Cumulative distribution function**

The random variable X has the following probability distribution function:

$$ \begin{matrix} P(x) = \frac { x }{ 150 } & \text{ for x} = 10, 20, 30, 40, 50 \\ 0 & \text{otherwise} \end{matrix} $$

Calculate and interpret F(20) and F(40), giving an interpretation for each.

**Solution**

As you will recall, we can determine the probability of each outcome for a random variable given the probability distribution function (pdf).

$$ \begin{align*} P(x) & =\cfrac {x}{150} \\ P(x) & = P(X = x) \\ \end{align*} $$

Thus,

$$ P(10) =\cfrac {10}{150} $$

Similarly,

$$ P(20) =\cfrac {20}{150} $$

$$ P(30) =\cfrac {30}{150} $$

$$ P(40) =\cfrac {40}{150} $$

And lastly,

$$ P(50) =\cfrac {50}{150} $$

*Note: We can prove that our pdf is correct by testing the first rule of probability distribution functions by adding all the probabilities.*

Now,

$$ F(x) = P(X \le x) $$

Therefore,

$$ \begin{align*} F(2) & = P(X \le 20) \\ & = P(X = 10) + P(X = 30) \\ & =\cfrac {10}{150} + \cfrac {20}{150} \\ &=\cfrac {30}{150} \text { or } \cfrac {1}{5} \\ \end{align*} $$

Interpretation: There is a 20% cumulative probability that outcomes 10 or 20 occur.

Similarly,

$$ \begin{align*} F(40) & = P(X \le 40) \\ & = P(X = 10) + P(X = 20) + P(X = 30) + P(X = 40) \\ & =\cfrac {10}{150} + \cfrac {20}{150} + \cfrac {30}{150} + \cfrac {40}{150} \\ & = \cfrac {100}{150} \text{ or } 66.67\% \\ \end{align*} $$

Interpretation: There is a 66.67% cumulative probability that outcomes 10, 20, 30, or 40 occur.

**Example 2: Cumulative distribution function**

Variable X can take the values 1, 2, 3, and 4. The cumulative probability distribution has been given below. Use it to calculate:

(a) P(X = 2)

(b) P(X = 4)

$$ \begin{array}{|c|c|c|c|c|} \hline \text{Outcome} & {1} & {2} & {3} & {4} \\ \hline \text{Cumulative Probability distribution} & {0.2} & {0.5} & {0.85} & {1} \\ \hline \end{array} $$

**Solution**

$$ F(x) = P(X \le x) $$

(a)

$$ \begin{align*} F(2) & = P(X \le 2) = 0.5 \\ 0.5 & = P(X = 1) + P(X = 2) \\ & = 0.2 + P(X = 2)\\ P(X = 2)& = 0.5 – 0.2 = 0.3 \\ \end{align*} $$

Note: a simpler, more direct approach can be:

$$ P(X = 2) = F(2) – F(3) $$

Therefore,

$$ P(X = 2) = 0.5 – 0.2 = 0.3 $$

(b)

$$ \begin{align*} P(X = 4) & = F(4) – F(3) \\ &= 1 – 0.85 = 0.15 \\ \end{align*} $$

QuestionGiven the following cumulative probability distribution, determine P(2).

$$ \begin{array}{|c|c|c|c|c|} \hline \text{Outcome} & {0} & {1} & {2} & {3} \\ \hline \text{Cumulative prob.} & {1/8} & {4/8} & {7/8} & {1} \\ \hline \end{array} $$

A. 7/8

B. 3/8

C. 1/8

SolutionThe correct answer is B:

$$ \begin{align*} P(2) & = P(X = 2) \\ & = F(2) – F(1) \\ &=\cfrac {7}{8} – \cfrac {4}{8} \\ & =\cfrac {3}{8} \\ \end{align*} $$

*Reading 9 LOS 9d:*

*Calculate and interpret probabilities for a random variable, given its cumulative distribution function.*