Part a:
Given that the mean distance between consecutive flaws on a roll of sheet aluminum is 3 meters, thus the rate parameter (the mean number of flaws per meter) is given by 1 divided by the mean distance = 1 / 3 = 0.333
Part b:
[tex]P(x>k)=e^{-\lambda k}[/tex]
Thus,
[tex]P(x>6)=e^{- \frac{1}{3} (6)} \\ \\ = e^{-2}= 0.1353[/tex]
Therefore, the probability that we inspect the next six meters before finding a flaw is 0.1353.
Part c:
In a 5 metre length of aluminium, the mean distance between consecutive flaws is given by 5 x 1/3 = 5/3.
The probability that the five meter length of aluminum contains exactly two flaws is given by the poisson distribution with a mean of 5/3.
[tex]P(x=2)=\frac{\left( \frac{5}{3} \right)^2e^{-\frac{5}{3} }}{2!} \\ \\ = \frac{2.7778\times0.1889}{2} = \frac{0.5247}{2} \\ \\ =0.2623[/tex]
