Question

2. It has been established that the number of defective stereos produced daily at a certain plant is Poisson distributed with mean 4. Over a 2 days span, what is the probability that the number of defective stereos does not exceed 3?

Answer 1

2/3 if answer is correct but I think I can do it

Answer 2

Answer:

0.43347

Step-by-step explanation:

Let X be the random variable representing number of defective stereos.

The number of defective stereos produced daily at a certain plant follows Poisson  distribution with mean = 4

\lambda = 4

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

We need to find, P(X ≤ 3) = P(X=0) + P(X=1) +P(X=2)+P(X=3)

                                         = $\frac{e^{-4}(4)^{0}}{0!}$ + $\frac{e^{-4}(4)^{1}}{1!}$ + $\frac{e^{-4}(4)^{2}}{2!}$ + $\frac{e^{-4}(4)^{3}}{3!}$

                                        = e^{-4}[1+4+8+$\frac{32}{3}$]

                                       = e^{-4}$\frac{71}{3}$

                                       = 0.43347