Question

Based on past experience, it is assumed that the number of flaws of per metre in rolls of wrapping paper follows a Poisson distribution with a mean of 2 flaws per 4 metres of paper. The probability (correct to 2 decimal places) that more than 2 flaws will be observed in 5 metres of wrapping paper produced is:

Answer 1

$\lambda=2(\dfrac{5}{4})=2.5$

$$P(X>2)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dfrac{e^{-2.5}(2.5)^0}{0!}$$

$$-\dfrac{e^{-2.5}(2.5)^1}{1!}-\dfrac{e^{-2.5}(2.5)^2}{2!}\approx0.46$$