Oil tankers the mean number of oil tankers at a port city is eight per day. find the probability that the number of oil tankers on any given day is (a) exactly eight, (b) at most three, and (c) more than eight.

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) =

for x = 8

P(8) =

or

P(8) =

or

P(8) = 0.1395

b)  at most three

i.e P(0) + P(1) + P(2) + P(3)

thus,

P(0) = = 0.0003354

P(1) = = 0.002683

P(2) = = 0.01073

P(3) = = 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) = = 0.0003354

P(1) = = 0.002683

P(2) = = 0.01073

P(3) = = 0.02862

P(4) = = 0.05725

P(5) = = 0.091603

P(6) = = 0.1221

P(7) = = 0.13958

P(8) = = 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

first one because it can't have a fraction in it

step-by-step explanation:

6(h+1)/2=a

a=area

step-by-step explanation:

its the same thing as a square or rectangle but after multiplying two sides you need to divide it by two.