Question

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standar deviation will be needed to achieve a process capability index

Answer 1

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is  $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

   The upper specification is  $USL = 1.68 \ mm$

    The lower specification is  $LSL = 1.52 \ mm$

     The sample mean is  $\mu = 1.6 \ mm$

     The standard deviation is  $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

             $Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

Now what min means is that the value of  CPk is the minimum between the value is the bracket

          substituting value given in the question

           $Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=>      $Cpk = min[ 0.88 , 0.88 ]$

So

         $Cpk = 0.88$

Now from the question we are asked to evaluated the value of  standard deviation that will produce a  capability index of 2

Now let assuming that

         $\frac{\mu - LSL }{ 3 * \sigma } = 2$

So

         $\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=>    $0.08 = 6 \sigma$

=>     $\sigma = 0.0133$

So

        $\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=>      $2$

Hence

      $Cpk = min[ 2, 2 ]$

So

    $Cpk = 2$

So    $\sigma = 0.0133$ is  the value of standard deviation required