MATH SOLVE

3 months ago

Q:
# 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 3xβ4, 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?

Accepted Solution

A:

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 functionb) [tex] F(x) = 3 \int_{1}^{x} u^{-4} du = - \frac{1}{u^3} \Big|_1^x Β = -\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 aFor 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 functionPart bFor 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 Β = -\frac{1}{x^3} +1 = 1-x^{-3}[/tex]Part cFor 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]