Answer :
The probability that there will be 4 failure in particular week is 0.00622.
e(x) = 0.75 = v(x)
Given,
If electricity power failures occur according to a Poisson distribution with an average of 3 failures over 4 weeks, what is the probability that there will be 4 failures during a particular week?
To know e(x) are v(x);
From the given info,
So 3/4 failure per 4/4 weeks.
That means on average 0.75 failures per week.
The probability that there will be 4 failure in particular week,
[tex]\frac{e^{-0.75}0.75^{4}}{4!}=\frac{0.4723*0.3164}{24}=0.00622[/tex]
e(x) = Average = 0.75
Poisson distribution have mean and variance same.
So v(x) = 0.75
Hence, the probability that there will be 4 failure in particular week is 0.00622.
e(x) = 0.75 = v(x)
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