Question

Problem 1 (25%) due 05-16-2024 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 50 meters of wire. One day an inspector has to take an emergency phone call and is missing from his post for ten minutes. If the machine turns out 7 meters of wire per minute, what is the probability that the inspector will miss more than one fracture? Hint:Poisson Process with length (you need to replace the time with the length.)

          Problem 1 (25%) due 05-16-2024
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 50 meters of wire. One day an inspector
has to take an emergency phone call and is missing from his post for ten minutes. If the
machine turns out 7 meters of wire per minute, what is the probability that the inspector will
miss more than one fracture?
Hint:Poisson Process with length (you need to replace the time with the length.)

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$Problem 1 (25%) due 05-16-2024 A wire manufacturing company has inspectors to examine the wire for fractures as it comes out of a machine. The number of fractures is distributed in accordance with a Poisson process, having one fracture on the average for every 50 meters of wire. One day an inspector has to take an emergency phone call and is missing from his post for ten minutes. If the machine turns out 7 meters of wire per minute, what is the probability that the inspector will miss more than one fracture? Hint:Poisson Process with length (you need to replace the time with the length.)$

Added by Claudia W.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Suman K

06/28/2022

Step 1

Step 1: The inspector is missing for 10 minutes, and the machine produces 7 meters of wire per minute, so the total length of wire produced during the inspector's absence is 10 minutes * 7 meters/minute = 70 meters. Show more…

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Problem 1 (25%) due 05-16-2024 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 50 meters of wire. One day an inspector has to take an emergency phone call and is missing from his post for ten minutes. If the machine turns out 7 meters of wire per minute, what is the probability that the inspector will miss more than one fracture? Hint:Poisson Process with length (you need to replace the time with the length.)

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Transcript

00:01 In the question we are given that the inspector finds one fracture, one fracture every 60 meters of wire.

00:12 Every 60 meters of wire.

00:14 So let's start the first question.

00:16 So first question is given that the machine turns 7 meter of wire in one minute.

00:26 And we have to find, we have to find probability that, probability that inspector will miss will miss more than one fracture more than one fracture okay so first of all we will calculate expected number of fractures in 10 minutes in 10 minutes this would be equal to 10 times 7 divided by 60 that is 7 by 6 7 by 6 okay so let's move on.

01:05 Now, probability of x greater than 1 will be equal to 1 minus probability of x less than equal to 1, which will be equal to 1 minus probability of x equals 0 minus probability of x equals 1.

01:19 This will be equal to 1 minus e raised par minus 7 divided by 6 times 7 divided by 6 raced by 6 raised 1 divided by 1 factorial so when we calculate this this will come out into this will come out as probability of more than 1 fracture will be come out as 0 .3 2 .53 so we have calculated the probability for the first part now let's move on to second part now in second part we have to calculate probability probability that the next fracture next fracture occurs occurs in the next 10 minutes in the next 10 minutes okay so this probability will be equal to 1 minus probability of no fracture no fracture in 10 minutes right so when we put the formula this will result to do but when we put the values this will be equal to 1 minus e rest per minus 7 divided by 6 times 7 divide by 6 0 divided by 0 factorial so we can write that probably this probability equals 0 0 .6686...

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