Answer:
Step-by-step explanation:
Let X be the random variable representing the number of missing pulses or errors. The number of errors on a test area has Poisson with parameter = 0.2
a) Expected no of errors per test area = [tex]0.2[/tex]
(Since in a Poisson distribution mean = parameter value)
b)percentage of test areas have two or fewer errors
100*P(X≤2)
[tex]P(X\leq 2) = P(0)+P(1)+P(2)\\= \frac{e^{-0.2} 0.2^0}{o!} +\frac{e^{-0.2} 0.2^1}{1!}+\frac{e^{-0.2} 0.2^2}{2!}[/tex]
[tex]=P(X≤2)=0.99885[/tex]
Hence percentage = 99.885%