An important factor in solid missile fuel is the particle size distrib

Ashley Bell 2021-12-24 Answered
An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by \(\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{-{4}}},{f}{\quad\text{or}\quad}\ {x}{>}{1},{f{{\left({x}\right)}}}={0}\), elsewhere.
a) Verify that this is a valid density function.
b) Evalute F(x).
c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

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Expert Answer

Cheryl King
Answered 2021-12-25 Author has 4943 answers
Step 1
a) In order to verify that f is a valid density function, we need to check if the condition \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={1}\) is satisfied.
\(\displaystyle{\int_{{-\infty}}^{{\infty}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={\int_{{{1}}}^{{\infty}}}{3}{x}^{{-{4}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left(-{\frac{{{1}}}{{{x}^{{{3}}}}}}\right)}_{{{1}}}^{{\infty}}}\)
\(\displaystyle={0}-{\left(-{\frac{{{1}}}{{{1}^{{{3}}}}}}\right)}\)(1)
\(\displaystyle={1}\)
In (1) we used the fact:
\(\displaystyle{\left(-{\frac{{{1}}}{{{x}^{{{3}}}}}}\right)}_{{\infty}}={0}\)
which is true because of:
\(\displaystyle{\left(-{\frac{{{1}}}{{{x}^{{{3}}}}}}\right)}_{{\infty}}=\lim_{{{x}\Rightarrow\infty}}-{\frac{{{1}}}{{{x}^{{{3}}}}}}={0}\).
The last equality is obtained from the fact:
\(\displaystyle\lim_{{{x}\Rightarrow\infty}}{x}^{{{3}}}=\infty\)
Therefore, f truly is a valid density function
Step 2
b) Let X be the random variable which represents the particle size in micrometers.
By the definition of the cumulative distribution function F, for
\(\displaystyle{x}{>}{1}\)
we have: \(\displaystyle{F}{\left({x}\right)}={P}{\left({X}\le{x}\right)}\)
\(\displaystyle={\int_{{\infty}}^{{{x}}}}{f{{\left({t}\right)}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{{1}}}^{{{x}}}}{3}{t}^{{-{4}}}{\left.{d}{t}\right.}\)
\(\displaystyle={{\left(-{\frac{{{1}}}{{{t}^{{{3}}}}}}\right)}_{{{1}}}^{{{x}}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{x}^{{{3}}}}}}-{\left(-{\frac{{{1}}}{{{1}^{{{3}}}}}}\right)}\)
\(\displaystyle={1}-{\frac{{{1}}}{{{x}^{{{3}}}}}}\)
Step 3
c) We need to find the probability \(\displaystyle{P}{\left({X}{>}{4}\right)}\).
\(\displaystyle{P}{\left({X}{>}{4}\right)}={1}-{P}{\left({X}\le{4}\right)}\)
\(\displaystyle={1}-{F}{\left({4}\right)}\)
\(\displaystyle={1}-{\left({1}-{\frac{{{1}}}{{{4}^{{{3}}}}}}\right)}\)
\(\displaystyle={1}-{1}+{\frac{{{1}}}{{{64}}}}\)
\(\displaystyle={\frac{{{1}}}{{{64}}}}\)
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habbocowji
Answered 2021-12-26 Author has 5225 answers

a) To verify whether f(x) is a density function.
We know that, if f(x) is a density function then,
\(\displaystyle\int_{{{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={1}\)
Consider,
\(\displaystyle\int_{{{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={\int_{{{1}}}^{{\infty}}}{3}{x}^{{-{4}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{{1}}}^{{\infty}}}{x}^{{-{4}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{{\left[{\frac{{{x}^{{-{4}+{1}}}}}{{-{4}+{1}}}}\right]}_{{{1}}}^{{\infty}}}\)
\(\displaystyle={\frac{{{3}}}{{-{3}}}}{{\left[{\frac{{{1}}}{{{x}^{{{3}}}}}}\right]}_{{{1}}}^{{\infty}}}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{\infty^{{{3}}}}}}-{\frac{{{1}}}{{{1}^{{{3}}}}}}\right]}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{\infty}}}-{1}\right]}\)
\(\displaystyle=-{1}{\left[{0}-{1}\right]}{\left(\therefore{\frac{{{1}}}{{\infty}}}={0}\right)}\)
\(\displaystyle={1}\)
Therefore, we get.
\(\displaystyle{\int_{{{1}}}^{{\infty}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={1}\)
Hence, given f(x) is a density function.
b) To find F(x), i.e, distribution function.
We know that, distribution is obtained as,
\(\displaystyle{F}{\left({t}\right)}={\int_{{{0}}}^{{{x}}}}{f{{\left({t}\right)}}}{\left.{d}{t}\right.}\)
\(\displaystyle{F}{\left({t}\right)}={\int_{{{1}}}^{{{x}}}}{3}{t}^{{-{4}}}{\left.{d}{t}\right.}\)
\(\displaystyle={3}{{\left[{\frac{{{t}^{{-{4}+{1}}}}}{{-{4}+{1}}}}\right]}_{{{1}}}^{{{x}}}}\)
\(\displaystyle={\frac{{{3}}}{{-{3}}}}{{\left[{\frac{{{1}}}{{{t}^{{{3}}}}}}\right]}_{{{1}}}^{{{x}}}}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{{x}^{{{3}}}}}}-{\frac{{{1}}}{{{1}^{{{3}}}}}}\right]}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{{x}^{{{3}}}}}}-{1}\right]}\)
\(\displaystyle={1}-{\frac{{{1}}}{{{x}^{{{3}}}}}}\)
Hence, \(\displaystyle{F}{\left({x}\right)}={1}-{\frac{{{1}}}{{{x}^{{{3}}}}}};{x}{>}{1}\)
c) To find the probability that a random particle from the manufactured fuel exceeds 4 micrometres
Let X: size of a random particle from the manufactured fuel
We need to find \(\displaystyle{P}{\left({X}{>}{4}\right)}\), which is obtained as,
\(\displaystyle{P}{\left({X}{>}{4}\right)}={\int_{{{4}}}^{{\infty}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{4}}}^{{\infty}}}{3}{x}^{{-{4}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{{4}}}^{{\infty}}}{x}^{{-{4}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{{\left[{\frac{{{x}^{{-{4}+{1}}}}}{{-{4}+{1}}}}\right]}_{{{4}}}^{{\infty}}}\)
\(\displaystyle={\frac{{{3}}}{{-{3}}}}{{\left[{\frac{{{1}}}{{{x}^{{{3}}}}}}\right]}_{{{4}}}^{{\infty}}}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{\infty^{{{3}}}}}}-{\frac{{{1}}}{{{4}^{{{3}}}}}}\right]}\)
\(\displaystyle=-{1}{\left[{\frac{{{1}}}{{\infty}}}-{\frac{{{1}}}{{{64}}}}\right]}\)
\(\displaystyle=-{1}{\left[{0}-{0.015625}\right]}{\left(\because{\frac{{{1}}}{{\infty}}}={0}\right)}\)
\(\displaystyle={0.015625}\)
Therefore, probability that a random particle from the manufactured fuel exceeds 4 micrometres is 0.0156

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karton
Answered 2021-12-30 Author has 8659 answers

a) Solution: To verify it is a valid density function, we take the integral and see that it is one. I.e.
\(\int_{-\infty}^{\infty}f(x)dx=\int_{1}^{\infty}3x^{-4}dx=(-x^{-3})_{1}^{\infty}=-0-(-1)=1\)
b) Solution: To evaluate F(x) we use the definition of CDF of a continuous RV,
\(F(x)=P(X \le x)=\int_{-\infty}^{x}f(t)dt\)
Because f is piecewise, we have
\(F(x)=\begin{cases}1-x^{-3} & x > 1\\0 & elsewhere\end{cases}\)
c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?
Solution:We want to find \(P(X>4)=1-P(X \le 4)\). This can be done quickly with the CDF. That is,
\(P(X>4)=1-P(X \le 4)=1-F(4)=1-(1-4^{-3})=4^{-3}\)

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