Answer to Question #244724 in Statistics and Probability for Naresh

Question

No

Answer 1

Assume the question as follows:

A certain kind of sheet metal has, on average, 3 defects per 19 square feet.

Assuming a Poisson distribution, find the probability that a 33 square foot metal sheet has at least 4 defects. Round your answer to four decimals.

Solution:

There are on average $\lambda=3\times\frac{33}{19}=5.21$ defects per 33 square feet.

$P(X\ge4)=1-P(X<4)=1-e^{-\lambda}\sum_{i=0}^{3}\frac{\lambda^i}{i!}=$

$=1-e^{-5.21}\sum_{i=0}^{3}\frac{5.21^i}{i!}=1-0.4044=0.5956$