Probability density functions and probability of X in an interval

Probability density functions must meet two specific criteria

Probability density refers to the probability that a continuous random variable $X$ will exist within a set of conditions.

It follows that using the probability density equations will tell us the likelihood of an $X$ existing in the interval $[a,b]$.

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A probability density function $f(x)$ must meet these conditions:

1. $f(x)\ge0$ for all values of $x$

2. $\int^\infty_{-\infty}f(x)\ dx=1$

The equation for probability density is

$P(a\le{X}\le{b})=\int^b_af(x)\ dx$

where $P(a\le{X}\le{b})$ is the probability that $X$ exists in $[a,b]$.

How to identify, and then solve a probability density function

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Showing that f(x) is a probability density function, then finding the probability that X lies in an interval

Example

Let $f(x)=\left(\frac{x^3}{5,000}\right)(10-x)$ for $0\le{x}\le{10}$ and $f(x)=0$ for all other values of $x$. Show that $f(x)$ is a probability density function and find $P(1\le{X}\le{4})$.

The first thing we need to do is show that $f(x)$ is a probability density function. We can see that the interval $0\le{x}\le{10}$ is positive. For all other possibilities we know that $f(x)=0$. This means we’ve satisfied the first criteria for a probability density equation. Now we need to verify that

$\int^\infty_{-\infty}f(x)\ dx=1$

We can set the interval to $[0,10]$ since it’s only in this interval that the equation doesn’t equal $0$.

$\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\left(\frac{x^3}{5,000}\right)(10-x)\ dx$

$\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\frac{x^3}{500}-\frac{x^4}{5,000}\ dx$

$\int^\infty_{-\infty}f(x)\ dx=\int^{10}_0\frac{x^3}{500}\ dx+\int^{10}_0-\frac{x^4}{5,000}\ dx$

$\int^\infty_{-\infty}f(x)\ dx=\frac{x^4}{2,000}-\frac{x^5}{25,000}\bigg|^{10}_0$

$\int^\infty_{-\infty}f(x)\ dx=\left[\frac{(10)^4}{2,000}-\frac{(10)^5}{25,000}\right]-\left[\frac{(0)^4}{2,000}-\frac{(0)^5}{25,000}\right]$

$\int^\infty_{-\infty}f(x)\ dx=1$

The equation has met both of the criteria, so we’ve verified that it’s a probability density function.

using the probability density equations will tell us the likelihood of an x existing in the interval [a,b].

In order to solve for $P(1\le{X}\le{4})$, we’ll identify the interval $[1,4]$ and plug it into the probability density equation.

$P(a\le{X}\le{b})=\int^b_af(x)\ dx$

$P(1\le{X}\le{4})=\int^4_1\left(\frac{x^3}{5,000}\right)(10-x)\ dx$

$P(1\le{X}\le{4})=\int^4_1\frac{x^3}{500}-\frac{x^4}{5,000}\ dx$

$P(1\le{X}\le{4})=\int^4_1\frac{x^3}{500}\ dx+\int^{10}_0-\frac{x^4}{5,000}\ dx$

$P(1\le{X}\le{4})=\frac{x^4}{2,000}-\frac{x^5}{25,000}\bigg|^4_1$

$P(1\le{X}\le{4})=\left[\frac{(4)^4}{2,000}-\frac{(4)^5}{25,000}\right]-\left[\frac{(1)^4}{2,000}-\frac{(1)^5}{25,000}\right]$

$P(1\le{X}\le{4})=0.0866$

The answer tell us that the probability of $X$ existing between $1$ and $4$ is about $8.66\%$.

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