Question

Can you please assist me with this problem please!

A certain kind of sheet metal has, on average, 4 defects per 18 square feet.

Assuming a Poisson distribution, find the probability that a 33 square foot metal sheet has at least 5 defects. Round your answer to three decimal places.

Answer 1

$Mean \; λ= \frac{4 \times 33}{18}=7.333 \\ P(X=k) = \frac{λ^k \times e^{-λ}}{k!} \\ P(X≥5) = 1 -P(X<5) \\ = 1 -(P(X=0) +P(X=1) + P(X=2) +P(X=3) +P(X=4))\\ = 1 -[ \frac{7.333^0 \times e^{-7.333}}{0!} + \frac{7.333^1 \times e^{-7.333}}{1!} + \frac{7.333^2 \times e^{-7.333}}{2!} + \frac{7.333^3 \times e^{-7.333}}{3!} + \frac{7.333^4 \times e^{-7.333}}{4!}]\\ = 1 -e^{-7.333}[1 + 7.333 + 26.886 + 65.719 + 120.480] \\ = 1 - 0.00065361 \times 221.418 \\ = 1- 0.1447 \\ = 0.8553$

Answer: 0.855