Question

A company produces optical-fiber cable with a mean of 0.6 flaws per 100 feet. What is the probability that there will be exactly 4 flaws in 1000 feet of cable

Answer 1

Answer:

$P(X = 4) =0.133$

Step-by-step explanation:

From the question we are told that:

Mean $\x=0.6/100$

Flaws $f=4$

Distance  $d_2=1000ft$

Generally the equation for Poisson mean lambda over 1000 is mathematically given by

$\lambda=0.6*1000/100$  

$\lambda = 6$

Therefore

$P(X = 4) = {e^-\lambda * \lambda^{\=x}} {{\=x}!}$

$P(X = 4) =\frac{ e^{-6} * 6^4}{ 4!}$

$P(X = 4) =0.133$