Poisson Distribution

In a factory, the number of times that a certain machine breaks down per year has a Poisson distribution with mean 2.2.

a) If the repairing cost for each breakdown of the machine is \$1000, find the mean and the standard deviation (S.D.) of the total repairing cost in one year.

b) A company provides maintenance service for the machine with the following charge sceme: In each year, the service charges for the first three breakdowns are \$1600, \$1000 and \$800 respectively, and the service for the subsequent breakdowns is free of charge. Find the mean and the S.D. of the total service charge in one year.

(Give the answers correct to the nearest dollar.)

1 個解答

• wy
Lv 7
9 年前
最愛解答

(a) Mean = 2.2

so mean repairing cost = 2.2 x \$1000 = \$2200.

Variance = 2.2, s.d. = sqrt 2.2 = 1.4832

so s.d. of repairing cost = 1.4832 x \$1000 = \$1483 (nearest dollar).

(b)

P(1 breakdown) = P(X = 1) = e^(-2.2) x 2.2 = 0.24377

P(2 breakdowns) = P(X = 2) = e^(-2.2) x 2.2^2/2! = 0.26814

P(3 breakdowns) = P(X = 3) = e^(-2.2) x 2.2^3/3! = 0.19664

so Mean service charge, E(X) = 0.24377 x \$1600 + 0.26814 x \$1000 + 0.19664 x \$800 = 815.48 = \$815 ( nearest dollar).

E(X^2) = 0.24377 x 1600^2 + 0.26814 x 1000^2 + 0.19664 x 800^2

= 1018040.8

so Variance = 1018040.8 - 815.48^2 = 353033.17

so S.D. = sqrt 353033.17 = 594.1 = \$594 ( nearest dollar).