Question

Can you help me for this question ?

A wire manufacturing company has inspectors to examine the wire for fractures as it comes out of a machine. The number of fractures is distributed in accordance with a Poisson process, having one fracture on the average for every 60 meters of wire. One day an inspector has to take an emergency phone call and is missing from his work for ten minutes. a. If the machine turns out 7 meters of wire per minute, what is the probability that the inspector will miss more than one fracture? b. When the inspector returns to his work, what is the probability that the next fracture occurs in the next 10 minutes? c. What are the expected time and standard deviation of the time of the first fracture?

Answer 1

a.

in 10 minutes :

7*10 meteres of wire turned = 70 meter

poisson formula :

rate of fractures = 1 in 60 meters = 1/60

P(k fractures in x meters) = e^(-(1/60)*x) * ((1/60)*x)^k / (k!)

we have x = 70 meter

P(more than 1) = 1 - P(0) - P(1)

= 1 - [e^(-(1/60)*70) * ((1/60)*70)^0 / (0!)] - [e^(-(1/60)*70) * ((1/60)*70)^1/(1!)]

= 1 - 0.3114 - 0.3633

= 0.3253

b.

in 10 minutes wire turned = 70 meter

P(next fracture in 10 minutes) = 1 - P(0 fracture in 10 minutes)

= 1 - P(0 fracture in 70 meter)

= 1 - [e^(-(1/60)*70) * ((1/60)*70)^0 / (0!)]

= 0.6886

c.

in poisson

expected length = 1 / rate

variance of expected length = 1 / (rate)^2

expected time = expected length / rate of turninng

= (1/(rate of fractures))/7

= (1 / (1/60))/7

= 60/7

= 8.5714 min

SD of length = var^0.5 = [ 1 / (1/60)^2 ] ^0.5 = 60

SD of expected time = SD of length / rate of turning wire

= 60 / 7

= 8.57 min

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