A telephone company\'s goal is to have no more than 3 monthly line failures on a

Question

A telephone company's goal is to have no more than 3 monthly line failures on any 100 miles of line. The company currently experiences an average of 5 monthly line failures per 50 miles of line. Let xdenote the number of monthly line failures per 100 miles of line. Assuming x has a Poisson distribution:

(a) Find the probability that the company will meet its goal on a particular 100 miles of line. (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Probability:

(b) Find the probability that the company will not meet its goal on a particular 100 miles of line. (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Probability:   

(c) Find the probability that the company will have no more than 3 monthly failures on a particular 200 miles of line. (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Probability:

Explanation / Answer

The company currently experiences an average of 5 monthly line failures per 50 miles of line.

mean = 5/50 = 0.1

therfore there are 10 monthly line failures per 100 miles of line

mean = 10 / 100 = 0.1 _______________eq1

Implies  there are 20 monthly line failures per 200 miles of line

mean = 20 / 100 = 0.1 ______________eq2

x denote the number of monthly line failures per 100 miles of line and x has poisson distribution with mean 0.1

a) Using eq 1

P(The company will meet its goal on a particular 100 miles of line) = P(x < 3)

= exp(-0.1)* (0.1)3 /3!

= 0.0002

P(The company will meet its goal on a particular 100 miles of line) = 0.0002

b) P( The company will not meet its goal on a particular 100 miles of line) = 1 - P(x < 3)

= 1 - exp(-0.1)* (0.1)3 /3!

= 1 - 0.9048 * 0.001 / 6

= 1 - 0.0002

= 0.9998

P( The company will not meet its goal on a particular 100 miles of line) = 0.9998

c) Using eq 2

P(The company will have no more than 3 monthly failures on a particular 200 miles of line) = P(x < 3)

= exp(-0.1)* (0.1)3 /3!

= 0.0002

P(The company will have no more than 3 monthly failures on a particular 200 miles of line) = 0.0002

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