Respuesta :
Answer:
a) P(8) = 0.1395
b) P(at most three) = 0.0423684
c) P(X > 8) = 0.41
Step-by-step explanation:
Data provided in the question:
Mean, μ = 8
Now,
Probability that the number of oil tankers on any given day is
a) exactly eight
using Poisson distribution
we have
P(x) = [tex]\frac{\mu^xe^{-\mu}}{x!}[/tex]
for x = 8
P(8) = [tex]\frac{8^8e^{-8}}{8!}[/tex]
or
P(8) = [tex]\frac{16777216\times e^{-8}}{40320}[/tex]
or
P(8) = 0.1395
b) at most three
i.e P(0) + P(1) + P(2) + P(3)
thus,
P(0) = [tex]\frac{8^0e^{-8}}{0!}[/tex] = 0.0003354
P(1) = [tex]\frac{8^1e^{-8}}{1!}[/tex] = 0.002683
P(2) = [tex]\frac{8^2e^{-8}}{2!}[/tex] = 0.01073
P(3) = [tex]\frac{8^3e^{-8}}{3!}[/tex] = 0.02862
⇒ P(at most three) = 0.0003354 + 0.002683 + 0.01073 + 0.02862
= 0.0423684
c) more than eight.
P(X > 8) = 1 - P(X ≤ 8)
Now,
P(0) = [tex]\frac{8^0e^{-8}}{0!}[/tex] = 0.0003354
P(1) = [tex]\frac{8^1e^{-8}}{1!}[/tex] = 0.002683
P(2) = [tex]\frac{8^2e^{-8}}{2!}[/tex] = 0.01073
P(3) = [tex]\frac{8^3e^{-8}}{3!}[/tex] = 0.02862
P(4) = [tex]\frac{8^4e^{-8}}{4!}[/tex] = 0.05725
P(5) = [tex]\frac{8^5e^{-8}}{5!}[/tex] = 0.091603
P(6) = [tex]\frac{8^6e^{-8}}{6!}[/tex] = 0.1221
P(7) = [tex]\frac{8^7e^{-8}}{7!}[/tex] = 0.13958
P(8) = [tex]\frac{8^8e^{-8}}{8!}[/tex] = 0.13758
Thus,
P(X > 8) = 1 - [ 0.0003354 + 0.002683 + 0.01073 + 0.02862 + 0.05725 + 0.091603 + 0.1221 + 0.13958 + 0.13758 ]
P(X > 8) = 1 - 0.5904814
or
P(X > 8) = 0.41
