Question

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Answer 1

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week, $\mu = \frac{3}{20} = 0.15$

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607$

$P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898$

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.