Question

It is estimated that the average number of surface defects in 20 m2 of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40 m² of paper through random selection?

Answer 1

The probability of finding no more than 2 defects in 40 m² of paper through random selection is approximately 0.6767.

Step-by-step explanation:

To determine the probability, we can use the Poisson distribution, which is suitable for modeling the number of events in a fixed interval of time or space. In this case, the average number of defects (\(\lambda\)) is given as 3 per 20 m². To find the average number of defects in 40 m², we can scale up the average by doubling the area, resulting in \(\lambda = 6\).

Now, we can use the Poisson probability formula
`\(P(X=k) = (e^(-\lambda) \cdot \lambda^k)/(k!)\)`, where (k) is the number of defects (0, 1, or 2).

For 0 defects:

`\[ P(X=0) = (e^(-6) \cdot 6^0)/(0!) \approx 0.00247875 \]`

For 1 defect:

`\[ P(X=1) = (e^(-6) \cdot 6^1)/(1!) \approx 0.0148725 \]`

For 2 defects:

`\[ P(X=2) = (e^(-6) \cdot 6^2)/(2!) \approx 0.0446175 \]`

Finally, to find the probability of no more than 2 defects, we sum these probabilities:

`\[ P(X \leq 2) \approx 0.00247875 + 0.0148725 + 0.0446175 \approx 0.6767 \]`