Question

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 4.3 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 7 or more non-work-related e-mails during the same period? (Round your probability to 4 decimal places.) What is the probability she received three or less non-work-related e-mails during the period? (Round your probability to 4 decimal places.)

Answer 1

Answer:

P(4 non-work-related e-mails)=0.1933

P(7 or more non-work-related e-mails)=0.1442

P(3 or less non-work-related e-mails)=0.3771

Step-by-step explanation:

A Poisson distribution is useful to measure the probability of a number of events in a given period, in this problem, receiving a non-work-related e-mail would be the event and an hour would be the period of time.

Its probability mass function is $f(k; \lambda) = P(x=k) = \frac{\lambda^k e^{-\lambda}}{k!}$, being k the number of events and  λ how many sucesses (events) are per period of time (4.3 in this case)

First, we calculate what is the probability of receiving exactly 4 e-mails:

$f(k; \lambda) =P(k=4) =\frac{4.3^4 e^{-\4.3}}{4!}=0.1933$

As k is a discrete variable, to know the joint probability of different number of events we add the probability of each value.

The probability of receiving 7 or more is the sum of P(7), P(8), ..., P(∞):

$P(k\geq 7 ) =\sum_{i=7}^{\infty}\frac{4.3^i e^{-\4.3}}{i!}=0.1442$

Last, the probability of receiving 3 or less is the sum of P(0), P(1), P(2) and P(3):

$P(k\leq3 ) =\sum_{i=0}^{3}\frac{4.3^i e^{-\4.3}}{i!}=0.3771$