Question

The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 41 hours. A single plant operates three galvanizing lines that are assumed to operate independently. Round your answers to four decimal places (e.g. 98.7654).

(a) What is the probability that none of the lines experiences a surface finish problem in 41 hours of operation?
(b) What is the probability that all three lines experience a surface finish problem between 24 and 41 hours of operation?

Answer 1

a) The probability that none of the lines experiences a surface finish problem in 41 hours of operation is 0.0498.

b)The probability that all three lines experience a surface finish problem between 24 and 41 hours of operation is 0.0346.

Explanation:

`Mean = (1)/(\lambda) = 41\\P(X\leq x)= 1-e^(-\lambda x)`

`P(X>x)= e^(-\lambda x)`

a)

`P(x> 41, y>41, Z>41) = (P(X>41))^(3)\\\\P(X>41)=e^{^{-(41)/(41)}}=e^(-1)`

`P(x> 41, y>41, Z>41) = \left (e^(-1) \right )^(3)\\\\P(x> 41, y>41, Z>41) = e^(-3) = 0.0498.`

b)

`\lambda =(24)/(41)\\P(X=1)=e^(-\lambda )\cdot \lambda =\left ( e^(-0.585) \right )\left ( 0.585 \right )\\P(X=1)=0.326`

For 3 where, P(X=1, Y==1, Z=1)

`= (0.326)^(3) \\\\= 0.0346`