Answer: 0.2205
Step-by-step explanation:
Given : Technology Services department at Lahey Electronics revealed company employees receive an average of "2.7" non-work-related e-mails per hour.
i.e. [tex]\lambda = 2.7[/tex]
If the arrival of these e-mails is approximated by the Poisson distribution.
Then , the required probability is given by :-
[tex]P(X=x)=\dfrac{\lambda^xe^{-\lambda}}{x!}\\\\P(X=3)=\dfrac{(2.7)^3e^{-2.7}}{3!}\\\\=0.22046768454\approx0.2205[/tex]
Hence, the probability Linda Lahey, company president, received exactly 3 non-work-related e-mails between 4 P.M. and 5 P.M. yesterday =0.2205
