When a computer disk manufacturer tests a disk, it writes to the disk and then tests it using a certifier. The certifier counts the number of missing pulses or errors. The number of errors on a test area on a disk has a Poisson distribution with ???? = 0.2. (a) What is the expected number of errors per test area? (b) What percentage of test areas have two or fewer errors?

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 = 0.2a) 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 errors100*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%