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.

## Formula

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.

## Example

**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).

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

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.**Solution:**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