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]???.
Hi! I'm krista.
I create online courses to help you rock your math class. Read more.
A probability density function ???f(x)??? must meet these conditions:
???f(x)\ge0??? for all values of ???x???
???\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
Take the course
Want to learn more about Calculus 2? I have a step-by-step course for that. :)
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\%???.
