Calculating Probabilities from Cumulative Distribution Function

A cumulative distribution offers a convenient tool for determining probabilities for a given random variable. As seen above, the cumulative distribution function, \(F(x)\), gives the probability that the random variable \(X\) is less than or equal to \(x\) for every \(x\) value. It is usually expressed as:

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

Example: 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 \(F(40)\).

Solution

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

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

Therefore,

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

Similarly,

$$\begin{align} P(20) &=\cfrac {20}{150} \\ P(30) &=\cfrac {30}{150}\\ P(40) &=\cfrac {40}{150}\end{align}$$

And lastly,

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

Note to candidates: 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: Calculating Probabilities Given Cumulative Distribution Function

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

(a) P(X = 2).

(b) P(X = 4).

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

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 to candidates: 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 \(P(X=2)\).

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

1. \(\frac{7}{8}\).
2. \(\frac{3}{8}\).
3. \(\frac{1}{8}\).

Solution

The correct answer is B.

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