Answer :
Answer:
(a) f(x) is a valid density function.
(b) [tex]F(x)=\left \{ {1-x^{-3}}\qquad {\:x \:> \:1} \atop {0}\quad\qquad{elsewhere} \right[/tex]
(c) The probability that a random particle from the manufactured fuel exceeds 4 micrometers is [tex]P(X>4)=0.0156[/tex].
Step-by-step explanation:
We know the particle size (in micrometers) distribution is characterized by
[tex]f(x)=\left \{ {{3x^{-4}}\quad {if \:x \:> \:1} \atop {0}\quad{\:elsewhere} \right.[/tex]
(a) For f(x) to be a legitimate probability density function, it must satisfy the following two conditions:
- [tex]f(x)\geq 0[/tex] for all x
- [tex]\int\limits^{\infty}_{-\infty} {f(x)} \, dx =[/tex] area under the entire graph of f(x) = 1
It holds that [tex]f(x)\geq 0[/tex] for all [tex]x \in \mathbb{R}[/tex] and
[tex]\int\limits^{\infty}_{-\infty} {f(x)} \, dx = \int\limits^{\infty}_{-\infty} {3x^{-4}} \, dx=\int\limits^{\infty}_{1} {3x^{-4}} \, dx \\\\\int\limits^{\infty}_{1} {3x^{-4}} \, dx = [-x^{-3}]_{\infty}^{1}} \right = 0-(-1)=1[/tex]
Therefore f(x) is a valid density function.
(b) The cumulative distribution function (CDF) F(x) for a continuous rv X is defined for every number x by
[tex]F(x)= P(X\leq x)=\int\limits^x_{-\infty} {f(y)} \, dy[/tex]
Applying the CDF definition we get:
For [tex]x\leq 1[/tex] F(x) = 0, while for [tex]x>1[/tex]
[tex]\int\limits^x_{-\infty} {f(y)} \, dy=\int\limits^x_{-\infty} {3y^{-4}} \, dy=\int\limits^x_{1} {3y^{-4}} \, dy\\\\\int\limits^x_{1} {3y^{-4}} \, dy=[-y^{-3}]_{1}^{x}} \right=1-x^{-3}[/tex]
Because f(x) is a piece-wise function, we have
[tex]F(x)=\left \{ {1-x^{-3}}\qquad {\:x \:> \:1} \atop {0}\quad\qquad{elsewhere} \right[/tex]
(c) To find the probability that a random particle from the manufactured fuel exceeds 4 micrometers, you need to use the CDF,
[tex]P(X>4)=1-P(X\leq 4)=1-F(4)=1-(1-4^{-3})=4^{-3}=0.0156[/tex]