(a) and (b) Poisson Distribution requires a constant failure rate. It is also a discrete distribution, rather than continuous, which suits this type of problem.
Po(X=x)=µˣ(e^-)/x! where µ is the average number of failures per day.
=(172×0+131×1+49×2+13×3)/365=268/365=0.734 approx, which we take to be the constant failure rate.
So, Po(X=0)=0.48 approx, corresponding to 175 days, compared with 172 actual days.
Po(X=1)=0.35 approx, corresponding to 129 days, compared with 131.
Po(X=2)=0.13 approx, corresponding to 47 days, compared to 49.
Po(X=3)=0.032 approx, corresponding to 12 days, compared to 13.
This is an unbiased estimate because we are assuming (and have been given) that the failure rate is constant throughout the year.
(c) and (d) The reciprocal of the average failure rate is the average time between failures=365/268=1.36 days (about 32hr 41min). The Poisson Process requires failure events to be independent of one another, which is the case here, and two events can’t be simultaneous—either there’s a failure or there isn’t. Multiple faults are considered as one failure event. Like the failure rate, the inter-failure time is an unbiased estimate, because failure event are random independent events.
If we take a time interval t = 1 week, we can find out how many failures to expect in a week: 268×7/365=5 approx. This would be the parameter to plug into the Poisson distribution function. If t=4 weeks, the parameter would be about 21, and so on.