Ashley Bell

Answered

2021-12-24

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 $f\left(x\right)=3{x}^{-4},f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x1,f\left(x\right)=0$ , 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?

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?

Answer & Explanation

Cheryl King

Expert

2021-12-25Added 36 answers

Step 1

a) In order to verify that f is a valid density function, we need to check if the condition${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(x\right)dx=1$ is satisfied.

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(x\right)dx={\int}_{1}^{\mathrm{\infty}}3{x}^{-4}dx$

$={(-\frac{1}{{x}^{3}})}_{1}^{\mathrm{\infty}}$

$=0-(-\frac{1}{{1}^{3}})$ (1)

$=1$

In (1) we used the fact:

${(-\frac{1}{{x}^{3}})}_{\mathrm{\infty}}=0$

which is true because of:

${(-\frac{1}{{x}^{3}})}_{\mathrm{\infty}}=\underset{x\Rightarrow \mathrm{\infty}}{lim}-\frac{1}{{x}^{3}}=0$ .

The last equality is obtained from the fact:

$\underset{x\Rightarrow \mathrm{\infty}}{lim}{x}^{3}=\mathrm{\infty}$

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

$x>1$

we have:$F\left(x\right)=P(X\le x)$

$={\int}_{\mathrm{\infty}}^{x}f\left(t\right)dt$

$={\int}_{1}^{x}3{t}^{-4}dt$

$={(-\frac{1}{{t}^{3}})}_{1}^{x}$

$=-\frac{1}{{x}^{3}}-(-\frac{1}{{1}^{3}})$

$=1-\frac{1}{{x}^{3}}$

Step 3

c) We need to find the probability$P\left(X>4\right)$ .

$P\left(X>4\right)=1-P(X\le 4)$

$=1-F\left(4\right)$

$=1-(1-\frac{1}{{4}^{3}})$

$=1-1+\frac{1}{64}$

$=\frac{1}{64}$

a) In order to verify that f is a valid density function, we need to check if the condition

In (1) we used the fact:

which is true because of:

The last equality is obtained from the fact:

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

we have:

Step 3

c) We need to find the probability

habbocowji

Expert

2021-12-26Added 22 answers

a) To verify whether f(x) is a density function.

We know that, if f(x) is a density function then,

Consider,

Therefore, we get.

Hence, given f(x) is a density function.

b) To find F(x), i.e, distribution function.

We know that, distribution is obtained as,

Hence,

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

0

karton

Expert

2021-12-30Added 439 answers

a) Solution: To verify it is a valid density function, we take the integral and see that it is one. I.e.

b) Solution: To evaluate F(x) we use the definition of CDF of a continuous RV,

Because f is piecewise, we have

c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

Solution:We want to find

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