The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceedin

The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.256 ohms and 5% having a resistance smaller than 9.671 ohms. What are the mean value and standard deviation of the resistance distribution? slader

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  1. Explanation:

    Formula for the given probability is as follows.

               [tex]P(z > \frac{x – \mu}{\sigma})[/tex] = 0.10

           [tex]P(z > \frac{10.256 – \mu}{\sigma})[/tex] = 0.10

    According to the normal table area we have,

               P(z < 1.28) = 0.10

    Hence,  [tex]\frac{10.256 – \mu}{\sigma} = 1.28[/tex]

                 [tex]\mu = 10.256 – 1.28 \times \sigma[/tex] ………. (1)

    Also, the given probability is as follows.

               [tex]P(z < \frac{x – \mu}{\sigma})[/tex] = 0.05

              [tex]P(z < \frac{9.671 – \mu}{\sigma})[/tex] = 0.05

    Hence,     [tex]\frac{9.671 – \mu}{\sigma} = -1.645[/tex]

                   [tex]\mu = 9.671 + 1.645 \times \sigma[/tex] ……. (2)

    Now, substitute the value of [tex]\mu[/tex] from equation (1) into equation (2) as follows.

          [tex]10.256 – 1.28 \times \sigma = 9.671 + 1.645 \times \sigma[/tex]    

          [tex]-2.925 \sigma = -0.585[/tex]

                 [tex]\sigma[/tex] = 0.2

    Putting the value of [tex]\sigma[/tex] into equation (2) we will find the value of [tex]\mu[/tex] as follows.

                 [tex]\mu = 9.671 + 1.645 \times \sigma[/tex]

                           = [tex]9.671 + 1.645 \times 0.2[/tex]

                           = 10

    Thus, we can conclude that the value of  [tex]\sigma[/tex] is 0.2 and the value of [tex]\mu[/tex] is 10.      

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