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(x)=3x−4, for x>1, f(x)=0, elsewhere.

(a) Verify that this is a valid density function.

(b) Evaluate F(x).

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

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dfbustos dfbustos · Answer 1 · 2020-02-26T11:21:19+01:00

Answer:

a) For this case we need to veriffy that:

[tex] \int_{1}^{\infty} f(x) dx =1[/tex]

If we replace the density function given we have this:

[tex]\int_{1}^{\infty} 3x^{-4} dx = -\frac{1}{x^3} \Big|_1^{\inty} = 0 -(-1) =1[/tex]

So then we satisfy that the total area below the probability density function is 1 and we satisfy also that [tex] f(X_i) \geq 0 , \forall X_i[/tex]

So then we can conclude that f(x) is a density function

b) [tex] F(x) = 3 \int_{1}^{x} u^{-4} du = - \frac{1}{u^3} \Big|_1^x = -\frac{1}{x^3} +1 = 1-x^{-3}[/tex]

c) [tex] P(X >4)[/tex]

And we can use the complement rule and we got:

[tex]P(X >4) = 1-P(X \leq 4)[/tex]

And using the result from part b we got:

[tex]P(X >4) = 1-P(X \leq 4)=1- [1- 4^{-3}]=1-0.984375=0.015625[/tex]

Step-by-step explanation:

Part a

For this case we need to veriffy that:

[tex] \int_{1}^{\infty} f(x) dx =1[/tex]

If we replace the density function given we have this:

[tex]\int_{1}^{\infty} 3x^{-4} dx = -\frac{1}{x^3} \Big|_1^{\inty} = 0 -(-1) =1[/tex]

So then we satisfy that the total area below the probability density function is 1 and we satisfy also that [tex] f(X_i) \geq 0 , \forall X_i[/tex]

So then we can conclude that f(x) is a density function

Part b

For this case we can calculate the cumulative distribution function with the following integral:

[tex] F(x) = 3 \int_{1}^{x} u^{-4} du = - \frac{1}{u^3} \Big|_1^x = -\frac{1}{x^3} +1 = 1-x^{-3}[/tex]

Part c

For this case we want this probability:

[tex] P(X >4)[/tex]

And we can use the complement rule and we got:

[tex]P(X >4) = 1-P(X \leq 4)[/tex]

And using the result from part b we got:

[tex]P(X >4) = 1-P(X \leq 4)=1- [1- 4^{-3}]=1-0.984375=0.015625[/tex]