Answer :
Answer:
75.8% probability that a 31 square foot metal sheet has at least 4 defects.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that:
3 defects per 18 square feet.
So for 31 square feet, we have to solve a rule of three
3 defects - 18 square feet
x defects - 31 square feet
[tex]18x = 3*31[/tex]
[tex]x = \frac{31*3}{18}[/tex]
[tex]x = 5.17[/tex]
So [tex]\mu = 5.17[/tex]
Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.
Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.
So
[tex]P(X \leq 3) + P(X \geq 4) = 1[/tex]
We want [tex]P(X \geq 4)[/tex]
So
[tex]P(X \geq 4) = 1 - P(X \leq 3)[/tex]
In which
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057[/tex]
[tex]P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294[/tex]
[tex]P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760[/tex]
[tex]P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309[/tex]
So
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242[/tex]
Finally
[tex]P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758[/tex]
75.8% probability that a 31 square foot metal sheet has at least 4 defects.
