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:
"\\lambda=2(\\dfrac{5}{4})=2.5"
"-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"
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