### The Cumulative Distribution Function: Interpretation and Determination of Probabilities

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*}

## Question

Given 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

Solution

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

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

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