The number of power outages follows a Poisson Distribution. P(x,mu), where x = number of outages and mu equals mean number of outages = e^-mu * mu^x/x!.
mu = 4. For part a), x = 3. For part b) x = 3 and 4, then add probabilities together.
Luqman M.
asked 06/12/20In Islamabad city the mean number of power outages is 4 per month. Find the probability that in a given month, a. There are exactly 3 outages b. There are more than 2 outages.
The number of power outages follows a Poisson Distribution. P(x,mu), where x = number of outages and mu equals mean number of outages = e^-mu * mu^x/x!.
mu = 4. For part a), x = 3. For part b) x = 3 and 4, then add probabilities together.
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