# Specifications for a DVD player state that the part should weight between 24 and 25 ounces. The...

## Question:

Specifications for a DVD player state that the part should weight between 24 and 25 ounces. The process that produces the parts has a mean if 24.5 ounces and a standard deviation of .2 ounce. The distribution of output is normal. A. What percentage of parts will not meet the weight specs? B.Within what values will 95.44 percent of sample means of this process fall if samples of n=16 are taken and the process is in control (random)?

## Process Capability and Control

The process capability shows the degree to which the process output meets the specifications. At the same time, the process control is used to determine the presence of assignable causes of variability in the process.

A. Determining the percentage of parts not meeting the specs

We start with finding the z-value which shows how many standard deviations of the process fall within the specs limits. We find it as

{eq}z=\frac{\text{Upper Specification}-\text{Mean}}{\text{Standard Deviation}} \\ =\frac{25-24.5}{0.2} = 2.5 {/eq}

This z-value means that the output above z and below -z will not meet the specifications. Now we can find the probability that a random part will not meet the specifications as

{eq}q=\text{NORM.S.DIST}(-z)+1-\text{NORM.S.DIST}(z)\\ =2\times \text{NORM.S.DIST}(-z)\\ =2\times \text{NORM.S.DIST}(-\frac{4.55}{2}) = 0.0124 {/eq}

Therefore, {eq}2.29 {/eq} percent of parts will not meet the specs.

B. Determining the Control Limits

We find the z-value for the probability of 0.9544 as

{eq}\text{NORM.S.INV}(0.9544)=1.69 {/eq}

Now we can find the range as

{eq}\text{Mean Value}\pm z \times \frac{\text{St. Dev.}}{\sqrt{n}} \\ =24.5 \pm 1.69 \times \frac{0.2}{\sqrt{16}}\\ =24.5 \pm 0.08 {/eq}

Therefore, the sample means will fall in the range between 24.42 and 24.58