Question

6.

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two 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 one non-work-related e-mail between 4 p.m. and 5 p.m. yesterday? (Round your answer to 4 decimal places.)

Probability

b.	What is the probability she received five or more non-work-related e-mails during the same period? (Round your answer to 4 decimal places.)

Probability

c.	What is the probability she did not receive any non-work-related e-mails during the period? (Round your answer to 4 decimal places.)

Probability

7.

A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 p.m. on weekdays there is an average of four customers waiting in line. What is the probability that you visit Safeway today during this period and find:

a.	No customers are waiting? (Round your answer to 4 decimal places.)

Probability

b.	Four customers are waiting? (Round your answer to 4 decimal places.)

Probability

c.	Four or fewer are waiting? (Round your answer to 4 decimal places.)

Probability

d.	Four or more are waiting? (Round your answer to 4 decimal places.)

Probability

Answer 1

6)λ =2

a)

probability =

P(X=1)=

{e-λ*λx/x!}=

0.2707

b)

probability =

P(X>=5)=

1-P(X<=4)=

1-∑x=0x-1   {e-λ*λx/x!}=

0.0527

c)

probability =

P(X=0)=

{e-λ*λx/x!}=

0.1353

7)λ =4

a)

probability =

P(X=0)=

{e-λ*λx/x!}=

0.0183

b)

probability =

P(X=4)=

{e-λ*λx/x!}=

0.1954

c)

probability =

P(X<=4)=

∑x=0x   {e-λ*λx/x!}=

0.6288

d)

probability =

P(X>=4)=

1-P(X<=3)=

1-∑x=0x-1   {e-λ*λx/x!}=

0.5665