the expected number of defects in a particular size is 5.
What is the probability of getting one or less defects in the sample
Five defects out of how many? What size sample?
To find the probability of getting one or less defects in a sample, we can use the Poisson distribution formula:
P(x ≤ k) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!)
Where:
- P(x ≤ k) represents the probability of getting k or fewer defects
- e is the base of the natural logarithm (approximately 2.71828)
- λ is the expected number of defects in the sample
In this case, the expected number of defects is 5. So the probability of getting one or fewer defects would be:
P(x ≤ 1) = e^(-5) * (5^0/0!) + e^(-5) * (5^1/1!)
= e^(-5) * (1/1) + e^(-5) * (5/1)
= e^(-5) * (1 + 5)
= e^(-5) * 6
≈ 0.0337
Therefore, the probability of getting one or fewer defects in the sample is approximately 0.0337 or 3.37%.
To calculate the probability of getting one or less defects in the sample, we need to use the Poisson distribution. The Poisson distribution models the probability of a certain number of events occurring within a fixed interval of time or space when the events are independent and occur at a constant average rate.
In this case, the average number of defects in a particular size is given as 5. The Poisson distribution is defined by the following formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- P(X = k) is the probability of getting k defects in the sample
- e is the base of the natural logarithm (approximately 2.71828)
- λ is the average number of defects
- k is the number of defects we are interested in calculating the probability for
- k! represents k factorial, which is the product of all positive integers less than or equal to k
To find the probability of getting one or less defects in the sample, we need to sum the probabilities of getting exactly zero defects and exactly one defect.
P(X ≤ 1) = P(X = 0) + P(X = 1)
Let's calculate it step by step:
1. For k = 0:
P(X = 0) = (e^(-5) * 5^0) / 0! = e^(-5) ≈ 0.00674
2. For k = 1:
P(X = 1) = (e^(-5) * 5^1) / 1! = 5 * e^(-5) ≈ 0.0337
3. Summing the probabilities:
P(X ≤ 1) = P(X = 0) + P(X = 1) ≈ 0.00674 + 0.0337 ≈ 0.04044
Therefore, the probability of getting one or less defects in the sample is approximately 0.04044, or 4.044%.