Question

The maintenance department in a factory claims that the number of breakdowns of a particular machine follows a Poisson distribution with a mean of 3 breakdowns every 588 hours. Let x denote the time (in hours) between successive breakdowns. (a) Find and ux. (Write the fraction in reduced form.) = My = (b) Write the formula for the exponential probability curve of x. f(x) = 1 e-x/ for x 2 (d) Assuming that the maintenance department's claim is true, find the probability that the time between successive breakdowns is at most 4 hours. (Round your answer to 4 decimal places.) P(x < 4) (e) Assuming that the maintenance department's claim is true, find the probability that the time between successive breakdowns is between 161 and 320 hours. (Round your answer to 4 decimal places.) P(161 < x < 320) (f) Suppose that the machine breaks down 4 hours after its most recent breakdown. Based on your answer to part d, do you believe the maintenance department's claim? (Round your answer to 4 decimal places.) (Click to select) A ; the probability of this result is quite (Click to select) ( ) if the claim is true.

Answer 1

a)

$\mu$x =588/3 =196

$\lambda$ =1/196

b)

f(X) =(1/196)e^-x/196 for x>=0

d)

probability =

P(X<4)=

1-exp(-4/196)=

0.0202

e)

probability =

P(161<X<320)=

(1-exp(-320/196)-(1-exp(-161/196))=

0.2444

f)

No, the probability of this claim is quite small (0.0202) if the claim is true