A. Lahey Electronics: Email Delivery Times & Probabilities

[trastic]

"An internal study at Lahey Electronics,a large software development company,revealed the mean time for an internal email message to arrive at its destination was two seconds. Further, the distribution of the arrival times followed a Poisson distribution.
a.What is the probability a message takes exactly one second to arrive at its destination?
b.What is the probability it takes more than four seconds to arrive at its destination?
c.What is the probability it takes virtually no time, i.e., “zero” seconds?"

 

[MarkFL]

I would begin with:

[box=green]

The Poisson Probability Formula​

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\(\displaystyle P(x)=e^{-\lambda}\frac{\lambda^x}{x!}\tag{1}\)[/box]

In this problem, we are given \(\displaystyle \lambda=2\text{ s}\).

Can you now use this formula to answer the given questions?

 

[MarkFL]

Using (1) and $\lambda=2$, we find:

a.What is the probability a message takes exactly one second to arrive at its destination?

\(\displaystyle P(1)=e^{-2}\frac{2^1}{1!}=\frac{2}{e^2}\approx0.270670566473225\)

b.What is the probability it takes more than four seconds to arrive at its destination?

\(\displaystyle P(>4)=1-(P(0)+P(1)+P(2)+P(3)+P(4))=1-\frac{1}{e^2}\left(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}\right)=1-\frac{1}{e^2}\left(1+2+2+\frac{4}{3}\right)=1-\frac{19}{3e^2}\approx0.142876539501453\)

c.What is the probability it takes virtually no time, i.e., “zero” seconds?"

\(\displaystyle P(0)=e^{-2}\frac{2^0}{0!}=\frac{1}{e^2}\approx0.135335283236613\)