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A process that polishes a mirrored surface leaves an

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 12SE Chapter 4

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 12SE

A process that polishes a mirrored surface leaves an average of 2 small flaws per 5 \(m^{2}\) of surface. The number of flaws on an area of surface follows a Poisson distribution.

a. What is the probability that a surface with area 3m ✕ 5m will contain more than 5 flaws?

b. What is the probability that a surface with area 2m ✕ 3m will contain no flaws?

c. What is the probability that 50 surfaces, each with dimensions 3m ✕ 6 m, will contain more than 350 flaws in total?

Equation transcription:

Text transcription:

m^{2}

Step-by-Step Solution:
Step 1 of 4

Answer:

Step1 of 4:

             Given, a process that polishes a mirrored surface leaves an average of 2 small flaws per of surface. The number of flaws on an area of surface follows a poisson distribution.


 

Step2 of 4: 

a). We need to find the probability that a surface with area  will contain more than 5 flaws.

          Let denote the number of flaws on the surface with the area

         The mean concentration of flaws = so  

        Then the mean number of flaws in a  area is 15(0.4) = 6

     

         Therefore, the mean number of flaws for this surface is 6.    

        Now, find

        That is,

               

                                = 1- [ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)]

                               

                         

                                 

                         

                         

                          =

                          = 1- 0.4457

                                   = 0.5543

      Therefore, the probability that a surface with area  will contain more than 5 flaws is 0.5543.


Step 3 of 4: 

b). To find the probability that a surface with area  will contain no flaws.

       Let denote the number of flaws on the surface with the area

        The mean concentration of flaws is 2.4, so

        Then the mean number of flaws in a  area is 6(2.4) = 14.4.

       Now, find

      That is,

            

                       =

                              = 0.0907

  Therefore, the probability that a surface with area  will contain no flaws is 0.0907.


Step 4 of 4: 

c). To find the probability that fifty surfaces, each with dimensions will contain more than 350 flaws in total?

          Let denote the number of flaws on the surface with the area            

           The mean concentration of flaws is 360, so

      Now, to find  

          Since, is very large. We use the normal approximation of , it follows from the Central Limit Theorem.

    That is, . We use the z-score for 350.

  Therefore,

                     

                    = -0.53

 =  P(> -0.53)

                        = 1- P(Z< -0.53)

                        = 1- 0.2981

                        = 0.7019.

 Therefore, the probability that more than 350 flaws is 0.7019.


 

Step 2 of 4

Chapter 4, Problem 12SE is Solved
Step 3 of 4

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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