A metal sheet has ,on the average , 5 defects per 1.0 m^2. Assuming a poisson distribution , calculate the probability that a 1.5 m^2 piece of the metal sheet will have at least 4 defects.
To calculate the probability that a 1.5 m^2 piece of the metal sheet will have at least 4 defects, we can use the Poisson distribution formula. The Poisson distribution is commonly used to model events that occur randomly over a fixed interval of time or space. In this case, the defects on the metal sheet can be modeled using the Poisson distribution.
The Poisson distribution formula is:
P(x; λ) = (e^-λ * λ^x) / x!
Where:
P(x; λ) represents the probability of having exactly x events occur;
e is the base of the natural logarithm (approximately 2.71828);
λ is the average rate at which the events occur within the given interval.
In this case, the average number of defects per 1.0 m^2 is given as 5. Thus, λ (the average rate) is equal to 5.
To calculate the probability of having at least 4 defects on a 1.5 m^2 piece of the metal sheet, we need to sum up the individual probabilities of having 4, 5, 6, and so on, defects.
P(at least 4 defects) = P(4 defects) + P(5 defects) + P(6 defects) + ...
Since the question asks for "at least 4 defects", we need to calculate the probabilities of having 4, 5, 6, and so on, defects, and then sum them up.
Let's calculate the probability for each value of x and sum them up until a reasonable number of terms (e.g., until the probabilities become negligible).
P(at least 4 defects) = P(4 defects) + P(5 defects) + P(6 defects) + ...
To calculate each individual probability, we can use the Poisson distribution formula:
P(x; λ) = (e^-λ * λ^x) / x!
For x = 4:
P(4 defects) = (e^-5 * 5^4) / 4!
For x = 5:
P(5 defects) = (e^-5 * 5^5) / 5!
Let's calculate the probabilities for each value of x.
For x = 4:
P(4 defects) = (e^-5 * 5^4) / 4!
≈ (2.71828^-5 * 5^4) / (4 * 3 * 2 * 1)
≈ 0.17547
For x = 5:
P(5 defects) = (e^-5 * 5^5) / 5!
≈ (2.71828^-5 * 5^5) / (5 * 4 * 3 * 2 * 1)
≈ 0.17547
Now, we have the probabilities for 4 defects and 5 defects. We need to sum them up to get the probability of having at least 4 defects.
P(at least 4 defects) = P(4 defects) + P(5 defects)
≈ 0.17547 + 0.17547
≈ 0.35094
Therefore, the probability that a 1.5 m^2 piece of the metal sheet will have at least 4 defects is approximately 0.35094, or 35.09%.