Question

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 5.0 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 4 non-work-related e-mails between 4 P.M. and 5 P.M. yesterday? (Round your probability to 4 decimal places.)

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.)

What is the probability she received four or less non-work-related e-mails during the period? (Round your probability to 4 decimal places.)

Answer 1

POSSION DISTRIBUTION
pmf of P.D is = f ( k ) = e-λ λx / x!
where
λ = parameter of the distribution.   
x = is the number of independent trials
I.
mean = λ
= 5
a.
EQUAL
P( X = 4 ) = e ^-5 * 5^4 / 4! = 0.1755
b.
GREATER THAN EQUAL
P( X < 6) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-5 * 4 ^ 5 / 5! + e^-5 * 6 ^ 4 / 4! + e^-5 * 4 ^ 3 / 3! + e^-5 * ^ 2 / 2! + e^-5 * ^ 1 / 1! + e^-5 * ^ 0 / 0!
= 0.616,
P( X > = 6 ) = 1 - P (X < 6) = 0.384
c.
LESS THAN EQUAL
P( X < = 4) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-5 * 5 ^ 4 / 4! + e^-5 * 4 ^ 3 / 3! + e^-5 * 4 ^ 2 / 2! + e^-5 * 4 ^ 1 / 1! + e^-5 * ^ 0 / 0!
= 0.4405