Respuesta :
Answer:
a) [tex]P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288[/tex]
[tex]P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335[/tex]
b) [tex]P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452[/tex]
c) [tex]P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[0.0183+0.0733+ 0.1465+0.1954+0.1954+0.1563+ 0.1042+0.0595]=0.0511[/tex]
d) [tex]P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559[/tex]
Step-by-step explanation:
Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that [tex]X \sim Poisson(\lambda=4)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda=4[/tex] , [tex]Var(X)=\lambda=2[/tex], [tex]Sd(X)=2[/tex]
a. Compute both P(X≤4) and P(X<4).
[tex]P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)[/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183[/tex]
[tex]P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733[/tex]
[tex]P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465[/tex]
[tex]P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954[/tex]
[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]
[tex]P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646[/tex]
[tex]P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)[/tex]
[tex]P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692[/tex]
b. Compute P(4≤X≤ 8).
[tex]P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)[/tex]
[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]
[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]
[tex]P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042[/tex]
[tex]P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595[/tex]
[tex]P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298[/tex]
[tex]P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452[/tex]
c. Compute P(8≤ X).
[tex]P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)+P(X=5)+ P(X=6)+P(X=7)][/tex]
[tex]P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[0.0183+0.0733+ 0.1465+0.1954+0.1954+0.1563+ 0.1042+0.0595]=0.0511[/tex]
d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?
The mean is 4 and the deviation is 2, so we want this probability
[tex]P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)[/tex]
[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]
[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]
[tex]P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042[/tex]
[tex]P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559[/tex]
The probabilities P(X ≤ 4); P(X < 4); P(4 ≤ X ≤ 8) and P(8 ≤ X) are respectively; 0.6288, 0.4335,
How to calculate poisson's distribution?
We are given that;
µ = 4.
In poisson distribution this is equivalent to λ. Thus, λ = 4
Formula for poisson distribution is;
P(X = x) = [(e^(-λ)) * (λˣ)]/x!
A) We want to find the probabilities P(X ≤ 4) and P(X < 4)
P(X ≤ 4) = P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)
From online binomial probability calculator, we have;
P(X = 4) = 0.1954
P(X = 3) = 0.1954
P(X = 2) = 0.1465
P(X = 1) = 0.0733
P(X = 0) = 0.01832
P(X ≤ 4) = 0.1954 + 0.1954 + 0.1465 + 0.0733 + 0.01832
P(X ≤ 4) = 0.6288
Also; P(X < 4) = P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)
P(X < 4) = 0.4335
B) We want to find the probability P(4 ≤ X ≤ 8);
P(4 ≤ X ≤ 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
From online binomial probability calculator, we have;
P(4 ≤ X ≤ 8) = 0.5452
C) We want to find P(8 ≤ X);
P(8 ≤ X) = 1 - P(X < 8)
P(8 ≤ X) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
P(8 ≤ X) = 0.05113
Read more about Poisson's Distribution at; https://brainly.com/question/14802212
