Let X, the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with parameter μ = 5. Use the cumulative Poisson probabilities from the Appendix Tables to compute the following probabilities. (Round your answers to three decimal places.)

(a) P(X ≤ 8)
(b) P(X = 8)
(c) P(9 ≤ X)
(d) P(5 ≤ X ≤ 8)
(e) P(5 < X < 8)

Respuesta :

Answer:

[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]

e. P(5<X<8)=0.251

Step-by-step explanation:

Given that boiler follows a Poisson distribution,$\sim$

[tex]X $\sim$Poi(\mu=5)[/tex]

Poisson Distribution formula is given by the expression:-

[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]

Our probabilities will be calculated as below:

a.

[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]

b.

[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]

c.

[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]

d.

[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]

e.P(5<X<8)

[tex]P(X=6)+P(X=7)\\=0.251[/tex]