1 If a keyboard operator averages 2 errors per page of newsp

1. If a keyboard operator averages 2 errors per page of newsprint and if these errors follow a Poisson process, what is the probability that

a. exactly 4 errors will be found on a given page?

b. at least 2 errors will be found on a given page?

2. Crossville police records show that there has been an average of 4 accidents per week on

Crossville\'s new freeway. If these accidents follow a Poisson process, what\'s the probability that

a. the police must respond to exactly 6 accidents in a week?

b. that the police must respond to fewer than two accidents in a week?

3. Bigrig Trailer Corporation uses large panels of sheet metal in the manufacture of tandem trailers. If there is an average of 3 blemishes per panel and if the blemishes follow a Poisson process, what is the probability that

a. there will be no blemishes on a given panel?
b. there will be exactly 2 blemishes on a given panel?

4. The student health center at Oklahoma State University in Stillwater treats an average of 10 cases of severe alcohol poisoning a semester. Assuming a Poisson distribution, find the probability that

a. that the health center treats twelve cases of severe alcohol poisoning a semester.

b. the health center treats fewer than 5 cases of severe alcohol poisoning a semester.

c. the health center treats more than 15 cases of severe alcohol poisoning a semester.

Solution

I)
Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
a)
P( X = 4 ) = e ^-2 * 2^4 / 4! = 0.0902

b)

P( X < 2) = P(X=1) + P(X=0)   
= e^-2 * 4 ^ 1 / 1! + e^-2 * ^ 0 / 0!   
= 0.406
P( X > = 2 ) = 1 - P (X < 2) = 0.594

II)
a)
P( X = 6 ) = e ^-4 * 4^6 / 6! = 0.1042

b)
P( X < 2) = P(X=1) + P(X=0)   
= e^-4 * 6 ^ 1 / 1! + e^-4 * ^ 0 / 0!
= 0.0916

III)
a)
P( X = 0 ) = e ^-3 * 3^0 / 0! = 0.0498

b)
P( X = 2 ) = e ^-3 * 3^2 / 2! = 0.224