Cumulative distribution function are the function talking about the probability of an an event over a given interval. The cumulative distribution function can be obtained by integrating the probability density function. The probability can be obtained by integrating the cumulative distribution function over a definite interval. A cumulative distribution function will always be a non-decreasing function.

## Word Problems

**Problem 1: **

Life expectancy of a certain bacteria having the density function,

$p(x)$ = $\frac{1}{x^3}$ if $x\ \geq\ 1$,

$p(x)$ = $0$ if $x\ <\ 1$

Find the probability of bacteria living from $2$ to $5$ days.

**Solution: **

The density function of bacteria living from two to ten days,

$p(x)$ = $\frac{1}{x^3}$

To get the probability this function has to integrated from $2$ to $10$.

$\int_{2}^{10}p(x)\ dx$ = $\int_{2}^{10}p(x)\ dx$ = $\int_{2}^{10}$ $\frac{1}{x^3}$ $dx$ = -$\frac{1}{2}$ $[x^{-2}]_{2}^{10}$ = $0.12$

**Problem 2:**

If probability density function of a random variable $X$ is given to be $f(x)$ = $\frac{1}{2x^2}$ for $0\ <\ x\ <\ 2$ then find its cumulative distribution function.

**Solution:**

To find the cumulative distribution function the probability density function is integrated.

The cumulative distribution function, $F(x)$ = $\int_{0}^{x}f(x)dx$

$F(x)$ = $\int_{0}^{x}$ $\frac{1}{2x^2}$ $dx$ = -$\frac{1}{2}$ $[x^{-1}]_{0}^{x}$ = $\frac{-1}{2x}$

Hence, cumulative distribution function is $\frac{-1}{2x}$ for $0\ <\ x\ <\ 2$.

**Problem 3: **

A chemical can have life on the basis of given density function.

$p(x)$ = $\frac{1}{3x^5}$ if $x\ \geq\ 1$,

$p(x)$ = $0$ if $x\ <\ 1$

Find the probability that the chemical a life from $0$ to $5$ days.

**Solution: **

The probability can be obtained by integrating the function over the given interval.

$P(x)$ = $\int_{0}^{5}p(x)dx$

Hence,

$P(x)$ = $\int_{0}^{1}0\ dx + \int_{1}^{5}$ $\frac{1}{3x^{5}}$ $dx$

= $[\frac{-1}{12}$ $x^{-4}]_{1}^{5}$

= $0.0832$