A computer disk manufacturer tests disk quality on random basis before approving it. The approval is based on the number of errors in a test area on each disk and follows Poisson distribution with mean = 0.2. What is the percentage of test areas having two or a smaller number?
Answers
Given : The approval is based on the number of errors in a test area on each disk and follows Poisson distribution with mean = 0.2.
To Find : The percentage of test areas having two or a smaller number
Solution:
P(x) = λˣ e^(-λ) / x!
λ = Mean = 0.2
percentage of test areas having two or a smaller number
= P(0) + P(1) + P(2)
= λ⁰ e^(-λ) / 0! + λ¹ e^(-λ) / 1! + λ² e^(-λ) / 2!
= (0.2)⁰ e^(-0.2) / 0! + (0.2)¹ e^(-0.2) / 1! + (0.2)² e^(-0.2) / 2!
= e^(-0.2) + (0.2) e^(-0.2) + 0.04e^(-0.2) /2
= e^(-0.2) (1 + 0.2 + 0.02)
= e^(-0.2) (1.22)
= 0.99885
= 99.885 %
the percentage of test areas having two or a smaller number = 99.885 %
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