Cumulative Probability refers to the probability that a certain event occurs less than or equal to a certain number of times. It is obtained by adding up the probabilities of the random variable taking values less than or equal to the specified number.
Example:
Suppose that two coins are tossed. Here the possible events are {HH, HT, TH, TT}. Let X denote the number of heads obtained.
The probability of X=1, that is getting one heads is 2/4 = 0.5 since we have two favorable outcomes (HT, TH) out of four. So we obtain the probability distribution table as follows,
X = number of Heads | Probability P(X=x) |
0 | 1/4 = 0.25 |
1 | 2/4 = 0.5 |
2 | 1/4 = 0.25 |
We can obtain the cumulative probability for a particular value of X by adding up the probabilities of the current and previous values of X.
For example, P(X\leq1) = P(X=0) + P(X=1) = 0.25 + 0.5 = 0.75
In a similar manner we obtain the cumulative probability distribution as follows:
X = number of Heads | Probability P(X=x) | Cumulative Probability P(X\leqx) |
0 | 1/4 = 0.25 | 0.25 |
1 | 2/4 = 0.5 | 0.25+0.5 = 0.75 |
2 | 1/4 = 0.25 | 0.25+0.5+0.25 = 1 |
How do you calculate the cumulative probabilities for a continuous distribution?
As seen above we can calculate the cumulative probabilities for a discrete random variable by adding up all the previous probabilities.
\text{Cumulative Probability = }\sum_{-\infty}^{x} P(X=x)For a continuous random variable, given the probability distribution function (pdf) f(x) we can obtain the cumulative probability distribution function F(x) by integrating the pdf from minus infinity to X.
\text{Cumulative Probability = } P(X\leq x) = \int_{-\infty}^{x} f(x)dxExample: Suppose that a continuous random variable X has the pdf,
f(x) = 3x^2 \text{ where } 0\leq x \leq 1Find the cumulative probability distribution function.
Solution: F(x) = P(X\leq x) = \int_{0}^{x} f(x)dx = [x^3]_0^x = x^3 which is the required cumulative probability distribution function.