Possibilities Density Functions are a set of data measures that can be used to anticipate that a discontinuous value will turn out as the following calculation:
Density function calculated value:
Given function= [tex]\frac{1}{5}[/tex]
interval= [4,9]
Assuming that the given function that is [tex]fx) = 0[/tex] .
For this, the x outside the interval is [4,9].
Equation:
[tex]E[X^k] = \int^{9}_{4} x^k\ f(x) \ dx\\\\[/tex]
[tex]Var(X) = E(X)^2 - (E[X])^2[/tex]
The values are:
Standard deviation [tex]= \sqrt{Var(X)}[/tex]
Mean [tex]= E[X][/tex]
Solving the equation then:
[tex]E[X] =\int^{9}_{4} \frac{1}{5}\ dx[/tex]
[tex]= \frac{9^2-4^2}{2\cdot 5} \\\\ = \frac{81-16}{10} \\\\ = \frac{65}{10} \\\\=\frac{13}{2} \\\\[/tex]
[tex]E[X^2] =\int^{9}_{4} \frac{x^2}{5}\ dx[/tex]
[tex]= \frac{9^3-4^3}{3\cdot 5} \\\\= \frac{729-64}{15} \\\\ = \frac{665}{15}\\ \\=\frac{133}{3} \\\\[/tex]
[tex]\to Var(x) = \frac{133}{3} - (\frac{13}{2})^2 = \frac{25}{12}\\\\[/tex]
Therefore the standard deviation value is [tex]\frac{5}{2\sqrt{3}}[/tex]
Find out more about the probability here:
brainly.com/question/11234923