# Continuous Probability Distribution Examples

Continuous variables does not take discrete values. They do take continuous values such as 1.1, 2.33, and so on. To find a random value in uniform continuous distribution, the probability of finding a value in an interval [a, b] will be,
$p$ = $\frac{1}{b - a}$
If a probability density function is given as $f(x)$ then the probability of finding the continuous variable in the range $[a,\ b]$ can be obtained by the formula:
$P(a\leq x\leq b)= \int_{a}^{b}f(x)dx$
Example: Find the probability of getting a random number from the interval $[2,\ 7]$.

Solution: Probability of getting a number in $[2,\ 7]$ = $\frac{1}{7 - 2}$ = $\frac{1}{5}$

## Word Problems

Problem 1:

Find the probability of finding a random variable between $[34,\ 88]$.

Solution:

To find the probability, the formula will be $\frac{1}{b - a}$.

$b\ =\ 88,\ a\ =\ 34$

$Probability$ = $\frac{1}{88 - 34}$ = $\frac{1}{54}$
Problem 2:

For a given probability function $f(x)\ =\ x^{-2}$, if $x\ \geq\ 1$ find $P(X\ \leq\ 2)$.

Solution:

The interval given is $[1, 2]$.

$P(X \leq 2)$ = $\int_{1}^{2}x^{-2}$

= $\frac{x^{-1}}{-1}_{1}^{2}$

= $\frac{-1}{x}_{1}^{2}$ = $\frac{-1}{2}$ - $\frac{-1}{1}$

= $\frac{1}{2}$ = $0.5$

Hence, Probability is $0.5.$
Problem 3:

Find $P(2\ \leq\ P\ \leq\ 4)$ for probability function $f(x)$ = $2x^{-3}$ for $x\ \geq\ 1$.

Solution:

The given interval is $[2,\ 4]$.

$P(2\ \leq\ P\ \leq\ 4)$ = $\int_{2}^{4}2x^{-3}dx$

= $[\frac{2x^{-3+1}}{-3+1}]_{2}^{4}$ = $[-x^{-2}]_{2}^{4}$

= $-[\frac{1}{16}$ -$\frac{1}{4}]$ = $\frac{3}{16}$

The probability is $\frac{3}{16}$.
Problem 4:

Find the expected value over the interval $[5,\ 11]$ for a uniform distribution.

Solution:

The expected value in uniform distribution in an interval $[a,\ b]$ can be found by $\frac{b - a}{2}$.

$E(X)$ = $\frac{11 - 5}{2}$ = $\frac{6}{2}$ = $3$.