Question

Question 8 2 pts In a Poisson probability problem, the rate of defects is 1 every 2 hours. Find the probability of 3 defects in 4 hours. use poisspdf or poisscdf function Wk5Hw_Pois 1

Answer 1

The rate of defects is 1 every 2 hours, so total 2 defects in 4 hours.

This is a Poisson experiment in which we know the following:

• μ = 2; since 2 defects every 4 hours, on average.
• x = 3; since we want to find the likelihood of 3 defects in 4 hours
• e = 2.71828; since e is a constant equal to approximately 2.71828.

We plug these values into the Poisson formula as follows:

P(x; μ) = (e^-μ) (μ^x) / x!

P(3; 2) = (2.71828^-2) (2³) / 3!

P(3; 2) = (0.13534) (8) / 6

P(3; 2) = 0.180

Thus, the probability of 3 defects per 4 hours is 0.180 .