Answer:
c) 2
d) 0.96
Step-by-step explanation:
We are given the following in the question:
[tex]f(x) = 2x^{-3}, x > 1\\~~~~~~~= 0, x \leq 1[/tex]
a) probability density function.
[tex]\displaystyle\int^{\infty}_{\infty}f(x) dx = 1\\\\\displaystyle\int^{\infty}_{-\infty}2x^{-3}dx = 1\\\\\displaystyle\int^{\infty}_{1}2x^{-3}dx\\\\\Rightarrow \big[-x^{-2}\big]^{\infty}_1\\\\\Rightarrow -(0-1) = 1[/tex]
Thus, it is a probability density function.
b) cumulative distribution function.
[tex]P(X<x) = \displaystyle\int^{x}_{1}2x^{-3}dx\\\\P(X<x)=\big[-(x^{-2})\big]^{x}_{1}\\\\P(X<x)=-\bigg(\dfrac{1}{x^2}-1\bigg) = 1 - \dfrac{1}{x^2}[/tex]
c) mean of the distribution
[tex]\mu = \displaystyle\int^{\infty}_{-\infty}xf(x) dx\\\\\mu = \int^{\infty}_12x^{-2}dx\\\\\mu = (-\frac{2}{x})^{\infty}_{1}\\\\\mu = 2[/tex]
d) probability that the size of random particle will be less than 5 micrometers
[tex]P(X<5)= 1 - \dfrac{1}{(5)^2} = 0.96[/tex]
