Question

Specifications for a part for a 3-D printer state that the part should weigh between 24.5 and 25.5 ounces. The process that produces the parts has a mean of 25.0 ounces and a standard deviation of .25 ounce. The distribution of output is normal. Use Table-A.

a.What percentage of parts will not meet the weight specs? (Round your "z" value and final answer to 2 decimal places.)

b.Within what values will 99.74 percent of the sample means of this process fall if samples of n = 8 are taken and the process is in control (random)? (Round your answers to 2 decimal places.)

lower value: upper value:

c. Using the control limits from part b, would the following sample means be in control? 24.52, 24.53, 24.44, 24.51, 24.41, 24.39

• This process is in control.

• This process is not in control.

Answer 1

(A).

Given,

Mean \((X)=25\)

Lower Specification limit (LSL) \(=24.5\)

Upper Specification limit (USL) \(=25.5\)

Standard deviation \((S)=0.25\)

\(\mathbf{Z 1}=(\mathbf{U S L}-\mathbf{X}) / \mathbf{S} ; \mathbf{Z} 2=(\mathbf{X}-\mathbf{L S L}) / \mathbf{Z}\)

\(Z 1=(25.5-25) / 0.25 ; Z 2=(25-24.5) / 0.25\)

\(Z 1=2 ; Z 2=2\)

*Use Z - table to find value of probability at \(Z=2\)

Probability \(=0.9772\)

Percentage of parts that will not meet the weight specifications \(=2 \times(1-\) Probability) \(\times 100 \%\)

Percentage of parts that will not meet the weight specifications \(=2 \times(1-0.9772) \times 100 \%\)

Percentage of parts that will not meet the weight specifications \(=4.56 \%\)

(B).

For \(99.74\) percent, the value of \(Z\) is 3 .

Sample size \((n)=8\)

Uppervalue \(=X+Z \cdot \frac{S}{\sqrt{n}}\)

Uppervalue \(=25+3 * \frac{0.25}{\sqrt{8}}\)

Upper Value \(=25.27\)

Lower Value \(=X-Z \cdot \frac{S}{\sqrt{n}}\)

Lower Value \(=25-3 * \frac{0.25}{\sqrt{8}}\)

Lower Value \(=24.73\)

(C).

Given Samples \(=24.52,24.53,24.44,24.51,24.41,24.39\)

Average of samples \(=(24.52+24.53+24.44+24.51+24.41+24.39) / 6\)

Average of samples \(=24.47\)

Since the average of samples doesn't lie in between the calculated lower value and upper value, therefore,

THE PROCESS IS NOT IN CONTROL.