An internal study by the Technology Services department at Lahey Electronics rev
ID: 3040383 • Letter: A
Question
An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 3.7 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 non-work-related e-mail between 4 P.M. and 5 P.M. yesterday? (Round your probability to 4 decimal places.) Probability b. What is the probability she received 5 or more non-work-related e-mails during the same period? (Round your probability to 4 decimal places.) Probability c. What is the probability she received five or less non-work-related e-mails during the period? (Round your probability to 4 decimal places.)Explanation / Answer
A) P(1) = e(-3.7)*(3.7)^1/1!
=0.091477
B) P(5) = e(-3.7)*(3.7)^5/5!
= 0.142869
P(6) = e(-3.7)*(3.7)^6/6!
= 0.088103
P(7) = e(-3.7)*(3.7)^7/7!
= 0.046568
P(8) = e(-3.7)*(3.7)^8/8!
= 0.021538
P(9) = e(-3.7)*(3.7)^9/9!
= 0.008854
P(10) = e(-3.7)*(3.7)^10/10!
= 0.003276
P(11) = e(-3.7)*(3.7)^11/11!
= 0.001102
P(12) = e(-3.7)*(3.7)^12/12!
= 0.0003398
Therefore P(5 or more) = 0.3126498
C) P(5) = 0.142869 (Solved Above)
P(4) = e(-3.7)*(3.7)^4/4!
= 0.1930661
P(3) = e(-3.7)*(3.7)^3/3!
= 0.2087201
P(2) = e(-3.7)*(3.7)^2/2!
= 0.1692325
P(1) = e(-3.7)*(3.7)^1/1!
= 0.091477
P(0) = e(-3.7)*(3.7)^0/0!
= 0.0247235
Therefore P(5 or less) = 0.8300882
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