Question 1105923: An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 3.2 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.
What is the probability Linda Lahey, company president, received exactly 2 non-work-related e-mails between 4 P.M. and 5 P.M. yesterday? (Round your probability to 4 decimal places.)
Probability
What is the probability she received 6 or more non-work-related e-mails during the same period? (Round your probability to 4 decimal places.)
Probability
What is the probability she received two or less non-work-related e-mails during the period? (Round your probability to 4 decimal places.)
Probability
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! P(2)=e^(-3.2)*(3.2)^2/2! from the Poisson formula,
P(x)= e^(-lambda)*lambda^x/x!
=0.2087
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For 6, e^(-3.2)*3.2^6/6!=0.0608
For 7, it is 0.0278
For 8, it is 0.0111
For 9, it is 0.0040
For 10, it is 0.0013
For 11, it is 0.0004
For 12, it is 0.0001
0.1055 is the probability.
It would have been a little faster to do 5, 4,3,1,and 0 and have subtracted that sum from 1.
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0 is e^-3.2 or 0.0408
1 is 0.1304
2 is 0.2087 from above
Two or fewer emails would have probability of 0.3799.
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