Question

A certain kind of sheet metal has, on average, 7 defects per 11 square feet. Assuming a Poisson distribution, find the probability that a 21 square foot metal sheet has at least 11 defects. Round your answer to four decimals.

Answer 1

Answer:

$p(x\geq 11)=0.7782$

Step-by-step explanation:

If a variable x follows a poisson distribution, the probability that occurs x events in a region of size t is:

$p(x)=\frac{e^{-mt}*(mt)^{x}}{x!}$

Where m is the mean per unit. So, replacing m by 7/11 because there are 7 defects per 11 square feet and t by 21 square feet, the probability that the metal sheet has x defects is:

$p(x)=\frac{e^{-(7/11)21}*((7/11)(21))^{x}}{x!}=\frac{e^{-147/11}*(147/11)^{x}}{x!}$

Then, the probability that a 21 square foot metal sheet has at least 11 defects is calcualted as:

$p(x\geq 11)=1-p(x\leq 10)$

Where $p(x\leq 10)=p(0)+p(1)+p(2)+...+p(8)+p(9)+p(10)$

Now, p(0) is equal to:

$p(0)=\frac{e^{-147/11}*(147/11)^{0}}{0!}=0.00000157$

At the same way we can calculated the other probabilities, so:

$p(x\leq 10)=0.2218\\p(x\geq 11)=1-0.2218=0.7782$