Answer to Question #323662 in Statistics and Probability for Bene

Question

A Biotech company produces enzymes at a rate that can be described by the Poisson process.

Their lipase enzyme from a genetically modified bacteria produces glycerol at an average rate of

3 mg/L per minute.

4.1 Find the probability that the enzyme will produce exactly 5 mg/L in a given minute. (2)

4.2 What is the likelihood that the enzyme will produce four or more mg/L in a given minute? (7)

4.3 What chance is there that no product will be produced in a given minute?

Answer 1

$P(k)=\frac{\lambda^ke^{-\lambda}}{k!}$

$\lambda =3$

1. k=5 $P(5)=\frac{3^5e^{-3}}{5!}=0.1$

2. $P(X \ge 4)=1-P(3)-P(2)-P(1)-P(0)$

$P(3)=\frac{3^3e^{-3}}{3!}=0.224$

$P(2)=\frac{3^2e^{-3}}{2!}=0.224$

$P(1)=\frac{3^1e^{-3}}{1!}=0.149$

$P(0)=\frac{3^0e^{-3}}{0!}=0.05$

$P(X \ge 4)=1-0.224-0.224-0.149-0.05=0.353$

3.$P(0)=0.05$