Continuous Probability Distribution - Formula & Examples
Any variable can have two types of values. Either the values can be fix numbers which are also known as discrete values or a specified range that is known as continuous values. Based on these types of values a data set is defined as continuous or discrete. In continuous data type, the values can be lying anywhere within the range that is specified. 

For Example: the time required to drive back from office to home is always continuous.
So the probability distribution over a random variable "Y" where "Y" takes continuous values is termed as continuous probability distribution. 

There are three basic differences between a continuous and a discrete probability distribution:

i) The probability that a continuous variable will take a specific value is equal to zero.

ii) Because of this, we can never express continuous probability distribution in a tabular form.

iii) Thus we require an equation or a formula to describe such kind of distribution. Such equation is termed as probability density function.

Continuous Probability Distribution


We will discuss about probability distribution function here to understand the concept of continuous probability distribution:

A probability density function given over a rage a $\leq$ x $\leq$ b satisfies the following:

$f (x) \geq 0$ for all values of $‘x’$ lying between $a\ and\ b$.

The total area covered under the curve of the function lying in the range a and b is equal to 1.

So the probability of this continuous variable can now be found out by integrating this density function with respect to $‘x’$ over the interval $[a,\ b]$.

$P (a \leq x \leq b)$ = $\int_{a}^{b}$ $f (x) dx$

Common examples of continuous probability distributions are: uniform distribution, normal distribution, chi squared distribution etc.


Some solved examples on Continuous Probability Distribution are given below:

Example 1:  A continuous random variable say Y is following uniform distribution such that the probability between 4 and 9 is ‘r’. Find the value of ‘r’.

Solution: In a uniform distribution, the probability is same for all possible values of given variable between the specified range, so if ‘p’ is the probability of any value between say $a\ and\  b$ then p = $\frac{1}{(b - a)}$ since the sum of all probabilities is always equal to 1. 

Here, p = r, a = 4 and b = 9.

So r =$\frac{1}{(9 - 4)}$ = $\frac{1}{5}$.

Example 2: Write a short note on normal distribution.

Solution: A normal distribution function tells the probability of any real observation that is falling between two specified real limits (numbers) while the curve is reaching zero on either side.

Let $\mu$  be the expectation or mean of the normal distribution which is also equal to its mode and mean as well.  $\sigma$ denotes the standard deviation and thus $\sigma^2$ is the variation of the distribution. 

$f (x, \mu , \sigma )$ = $\frac{1}{\sigma  \sqrt (2\pi)}$ . $exp$ $(\frac{-(x - \mu)^2}{2\sigma^2})$

When $\sigma$ = 1 and $\mu$ = 0, then the normal distribution is said to be standard normal distribution and is denoted by N (0, 1).

Example 3: If the probability between a and b is equal to ‘r’. (a = 3 and b = 7). Find the expected value of the distribution.

It is given that expected value of a variable is following uniform distribution $\frac{(b – a)}{2}$ .
E (X) = $\frac{(b - a)}{2}$ = $\frac{(7 - 3)}{ 2}$ = $\frac{4}{2}$ = 2