# A process is in statistical control with x = 20 and s = 1.2. Specifications are at LSL = 16 and...

## Question:

A process is in statistical control with x = 20 and s = 1.2. Specifications are at LSL = 16 and USL = 24.

(a) Estimate the process capability with an appropriate process capability ratio.

(b) Items that are produced below the lower specification limit must be scrapped, while items that are above the upper specification limit can be reworked. What proportion of the process output is scrap, and what proportion is rework?

(c) Suppose that scrap is four times as expensive as rework. Does this suggest that moving the processing center could reduce overall costs? What value of the process target would you recommend?

## Process capability

Process capability is a statistical tool which is used to measure the ability to produce the output within specific limits of a product parameter. The value of Process capability(Cp) is calculated by subtracting the LSL from USL and divide by six times the standard deviation.

## Answer and Explanation:

Given information

{eq}\begin{align*} {\rm{mean}}\left( \mu \right) &= 20{\rm{ }}\\ {\rm{standard deviation}}\left( \sigma \right) &= 1.2\\ LSL &= 16\\ USL &= 24 \end{align*}{/eq}

a)

The process capability is calculated as;

{eq}\begin{align*} C\left( p \right) &= \dfrac{{USL - LSL}}{{6\left( \sigma \right)}}\\ &= \dfrac{{24 - 16}}{{6\left( {1.2} \right)}}\\ &= 1.11 \end{align*} {/eq}

b)

Items are produced below the lower specification limit must be scrapped so, proportion of the process output is scrapping is calculated as;

{eq}\begin{align*} P\left( {LSL < 16} \right) &= P\left( {\dfrac{{LSL - \mu }}{\sigma } < \dfrac{{16 - 20}}{{1.2}}} \right)\\ &= P\left( {Z < - 3.33} \right)\\ &= 0.00043 \end{align*}{/eq}

Items are produced above the upper specification limit must be scrapped so, proportion of the process output is scrapping is calculated as;

{eq}\begin{align*} P\left( {USL < 24} \right) &= P\left( {\dfrac{{USL - \mu }}{\sigma } < \dfrac{{24 - 20}}{{1.2}}} \right)\\ &= P\left( {Z < 3.33} \right)\\ &= 0.9995 \end{align*}{/eq}

The proportion of the each items are scrapped and reworked are 0.00043 and 0.9995 of the process output.

c)

The value of process mean is at center of LCL and UCL and the value of proportion of the process output is scrapping is less than 5%, therefore it is not suggest that moving the processing center could reduce overall costs. The value of process target should be 20.

#### Learn more about this topic:

from Business 112: Operations Management

Chapter 4 / Lesson 5