Question

g Let Y denote the number of oil tankers arriving each day at a certain port and it is a Poisson random variable. Record shows that the average of tankers arriving per day is 10. What is the probability that more than 7 tankers arrives in a certain day.

Answer 1

Answer:

0.77978

Explanation:

This is a Poisson distribution problem

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = mean = 10 tankers per day

x = variable whose probability is required

The probability that more than 7 tankers arrives in a certain day = 1 - (Probability that 7 or less tankers arrive in a certain day)

P(X > 7) = 1 - P(X ≤ 7)

P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)

P(X ≤ 7) = 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)

Computing this,

P(X≤7) = 0.22022

P(X > 7 ) = 1 - P(X≤7) = 1 - 0.22022 = 0.77978