Solution:
Given in the question lambda = 2
We need to find P(X=0)
Probability of Poission distribution can be calculated as
P(X=0) = e^(-lambda) * lambda^x / X! = e^-2 * (2^0)/0! =
0.1353
So its answer is B.i.e. 0.1353
Solution(b)
P(X<2) = P(X=0) + P(X=1) = 0.1353 + e^(-1) * (2^1)/1! = 0.1353 +
(0.3678*2) = 0.1353 + 0.7356 = 0.8709
Use Poisson Distribution to solve problems 6-7 6. Suppose that the average number of accidents occurring...
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