Question

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

Answer 1

Explanation:

Formula for the given probability is as follows.

           $P(z > \frac{x - \mu}{\sigma})$ = 0.10

       $P(z > \frac{10.256 - \mu}{\sigma})$ = 0.10

According to the normal table area we have,

           P(z < 1.28) = 0.10

Hence,  $\frac{10.256 - \mu}{\sigma} = 1.28$

             $\mu = 10.256 - 1.28 \times \sigma$ .......... (1)

Also, the given probability is as follows.

           $P(z < \frac{x - \mu}{\sigma})$ = 0.05

          $P(z < \frac{9.671 - \mu}{\sigma})$ = 0.05

Hence,     $\frac{9.671 - \mu}{\sigma} = -1.645$

               $\mu = 9.671 + 1.645 \times \sigma$ ....... (2)

Now, substitute the value of $\mu$ from equation (1) into equation (2) as follows.

      $10.256 - 1.28 \times \sigma = 9.671 + 1.645 \times \sigma$    

      $-2.925 \sigma = -0.585$

             $\sigma$ = 0.2

Putting the value of $\sigma$ into equation (2) we will find the value of $\mu$ as follows.

             $\mu = 9.671 + 1.645 \times \sigma$

                       = $9.671 + 1.645 \times 0.2$

                       = 10

Thus, we can conclude that the value of  $\sigma$ is 0.2 and the value of $\mu$ is 10.