Question #244724
No
1
Expert's answer
2021-10-01T00:21:17-0400

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 λ=3×3319=5.21\lambda=3\times\frac{33}{19}=5.21 defects per 33 square feet.

P(X4)=1P(X<4)=1eλi=03λii!=P(X\ge4)=1-P(X<4)=1-e^{-\lambda}\sum_{i=0}^{3}\frac{\lambda^i}{i!}=

=1e5.21i=035.21ii!=10.4044=0.5956=1-e^{-5.21}\sum_{i=0}^{3}\frac{5.21^i}{i!}=1-0.4044=0.5956

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS