Question 1

Suppose that $\overline { \mathrm { X } }$
And R charts based on subgroups of size 4 are being used to monitor the tire diameter of a new radial tire.The $\overline { \mathrm { X } }$
And R charts are found to be in statistical control with $\overline { \overline { X } } = 50.25 , \bar { R } = .8$

Inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.If the tire diameter specifications are 50 inches ± 1 inch,is the process capable of meeting the specifications?

Accepted Answer

The answer of Suppose that \(\overline { \mathrm { X...

Question 2

Suppose that $\overline { \mathrm { X } }$
And R charts based on subgroups of size 4 are being used to monitor the tire diameter of a new radial tire.The $\overline { \mathrm { X } }$
And R charts are found to be in statistical control,with $\overline { \overline { X } } = 50.5 , \bar { R } = 1.25$

Inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Compute the natural tolerance limits for this process.

Accepted Answer

The natural tolerance limits are given by:$$\overline { \mathrm { X } } \pm 3 \bar { R }$$$$= 50.5 \pm 3(1.25)$$$$= [48.68, 52.32]$$

Question 3

A foreman wants to use an $\overline { \mathrm { X } }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c } 
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Calculate the mean of the means $\overline { \overline X }$
)

Accepted Answer

The mean of each sample is calculated and then averaged: (2.01 + 2.005 + 1.99 + 2.01 + 1.995 + 2.02) / 6 = 2.005.

Question 4

Use this information about 10 shipments of light bulbs: $\begin{array}{lcccccccccc}
\text { Shipment } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
\text { \# of defective units } & 0 & 1 & 2 & 0 & 4 & 3 & 2 & 1 & 3 & 4
\end{array}$

If 400 bulbs are selected at random from each of 10 shipments and the number of defectives in each shipment is given above,find the appropriate center line.

Accepted Answer

The center line is the average number of defectives per shipment. The average number of defectives per shipment is 2.1. Therefore, the center line is 2.1/400 = 0.005.

Question 5

Among other quality measures,the quality of an electronic component used in manufacturing stereo speakers is monitored with a control chart.In the last 25 days,daily samples of 60 units resulted in the following number of defective units per sample: 8,4,3,7,6,2,5,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,5,6,and 0.Determine the appropriate upper and lower control limits for this process.

Accepted Answer

The control limits for a p-chart are calculated using the formula: p ± 3√(p(1-p)/n), where p is the average proportion defective and n is the sample size. Here, p = total defects/total samples = 81/1500 = 0.054. The control limits are 0.054 ± 3√(0.054(1-0.054)/60), which results in approximately [0, 0.152].

Question 6

Suppose that $\overline { \mathrm { X } }$
And R charts based on subgroups of size 4 are being used to monitor the tire diameter.The $\overline { \mathrm { X } }$
And R charts are found to be in statistical control,with $\overline { \overline { X } } = 50.25 , \bar { R } = .8$

Inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Find the sigma level capability of the process.

Accepted Answer

The sigma level capability of the process is given by:$$\sigma_{level}=\frac{\overline{X}}{R}=\frac{50.25}{0.8}=62.81$$The sigma level capability is 62.81, which is closest to 1.93.

Question 7

A fastener company produces bolts with a nominal (target)length of 2.00 inches.The specifications are 2.00 ± .006 inches.Determine the upper specification limit and the lower specification limit for this process.

Accepted Answer

The answer of A fastener company produces bolts with a...

Question 8

Use this information about 10 shipments of light bulbs: $\begin{array}{lcccccccccc}
\text { Shipment } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
\text { \# of defective units } & 0 & 1 & 2 & 0 & 4 & 3 & 2 & 1 & 3 & 4
\end{array}$

If 200 light bulbs are selected at random from each of 10 shipments and the number of defectives in each shipment is given above,find the LCL and the UCL for the p chart.

Accepted Answer

The LCL (Lower Control Limit) and UCL (Upper Control Limit) for a p-chart are calculated using the formula: $ \bar{p} \pm 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}} $, where $\bar{p}$ is the average proportion of defectives and $n$ is the sample size. Here, $\bar{p} = \frac{20}{2000} = 0.01$ and $n = 200$. Calculating gives LCL = 0 and UCL ≈ 0.0311.

Question 9

Assume that 25 samples of 50 each are taken and the total number of defectives is 34.Find the LCL and UCL for the p chart.

Accepted Answer

The answer of Assume that 25 samples of 50 each...

Question 10

A foreman wants to use an $\overline { \mathrm { X } }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c } 
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Calculate the average range.

Accepted Answer

To find the average range, we first need to calculate the range for each sample. Range is the difference between the largest value and the smallest value in the sample.

Sample 1: 2.02 - 1.99 = 0.03
Sample 2: 2.01 - 2.00 = 0.01
Sample 3: 2.01 - 1.98 = 0.03
Sample 4: 2.02 - 1.99 = 0.03
Sample 5: 2.01 - 1.99 = 0.02
Sample 6: 2.04 - 1.98 = 0.06

Now we can calculate the average range by adding up the ranges and dividing by the number of samples:

(0.03 + 0.01 + 0.03 + 0.03 + 0.02 + 0.06) / 6 = 0.18 / 6 = 0.03

Therefore, the average range is 0.03, which corresponds to answer choice C.

Question 11

A fastener company produces a certain type of bolt for the automobile industry with a nominal (target)length of 2.00 inches.The specifications for the length of the bolt are 2.00 ± .006 inches.An automobile manufacturer will only purchase from this company if the sigma level of capability of the process is at least 4.If the process mean is equal to 2.001,determine the maximum process standard deviation necessary for the fastener manufacturing company in order to qualify as a supplier for the automobile manufacturing company.

Accepted Answer

To calculate the maximum process standard deviation necessary, we need to first calculate the process capability (CP) and then use it to calculate the process standard deviation (sigma).

Process Capability (CP) can be calculated as:
CP = (Upper Specification Limit - Lower Specification Limit) / (6 * Standard Deviation)

Given that:
Nominal Length (Target) = 2.00 inches
Specification Limit = 2.00 ± 0.006 inches
Mean = Process Mean = 2.001 inches
Sigma = Standard Deviation (to be calculated)

Upper Specification Limit = 2.006 inches
Lower Specification Limit = 1.994 inches

CP = (2.006 - 1.994) / (6 * sigma) = 0.012 / sigma

We know that the automobile manufacturer will only purchase from this company if the sigma level of capability of the process is at least 4. Therefore, we need to set CP = 4 and solve for sigma.

4 = 0.012 / sigma

sigma = 0.012 / 4 = 0.003 = 0.00125 (rounded to nearest 0.00001)

Therefore, the maximum process standard deviation necessary for the fastener manufacturing company is 0.00125 inches. The correct choice is C).

Question 12

A foreman wants to use an $\overline { \mathrm { X } }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c } 
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Determine the LCL and the UCL for the $\bar { X }$
Chart.

Accepted Answer

To find the LCL and UCL for the $\bar{X}$ chart, we need to use the following formulas:
$$ 	ext{LCL} = \bar{X} - A_2\frac{\sigma}{\sqrt{n}} $$
$$ 	ext{UCL} = \bar{X} + A_2\frac{\sigma}{\sqrt{n}} $$
where $\bar{X}$ is the sample average, $\sigma$ is the population standard deviation (which we do not know, so we will use the sample standard deviation as an estimate), $n$ is the sample size, and $A_2$ is a constant that depends on the sample size and the desired level of control (we will use 0.577, which corresponds to a 99% level of control).

First, we need to calculate the sample averages:
$$ \overline{X}_1 = \frac{1.99 + 2.01 + 2.02 + 2.02}{4} = 2.01 $$
$$ \overline{X}_2 = \frac{2.00 + 2.00 + 2.01 + 2.01}{4} = 2.00 $$
$$ \overline{X}_3 = \frac{1.98 + 1.99 + 2.01 + 1.98}{4} = 1.99 $$
$$ \overline{X}_4 = \frac{2.01 + 2.02 + 2.02 + 1.99}{4} = 2.01 $$
$$ \overline{X}_5 = \frac{1.99 + 1.99 + 2.01 + 1.99}{4} = 1.99 $$
$$ \overline{X}_6 = \frac{2.03 + 1.98 + 2.03 + 2.04}{4} = 2.02 $$

Next, we need to calculate the sample standard deviation:
$$s = \sqrt{\frac{\sum_{i=1}^{6}\sum_{j=1}^{4}(X_{ij}-\overline{X}_i)^2}{(6)(4-1)}} \approx 0.023$$

Using these values and the formulas above, we can calculate the LCL and UCL:
$$ 	ext{LCL} = 2.01 - 0.577\frac{0.023}{\sqrt{4}} \approx 1.983 $$
$$ 	ext{UCL} = 2.01 + 0.577\frac{0.023}{\sqrt{4}} \approx 2.027 $$

Therefore, the answer is C, [1.983, 2.027].

Question 13

A motorcycle manufacturer produces the parts for its vehicles in different locations and transports them to its plant for assembly.In order to keep the assembly operations running efficiently,it is vital that all parts be within specification limits.One important part used in the assembly is the engine camshaft,and one important quality characteristic is the case hardness depth.Specifications state that the hardness depth must be between 3.0 mm and 6.0 mm.To investigate the process,the quality control engineer selected 25 daily subgroups of n = 5 and measured the hardness depth.The process yielded a mean of the means $\overline { \overline{ x } }$

= 4)50 and an average range = 1.01.Calculate the center line for the R chart.

Accepted Answer

The center line for the R chart is calculated using the formula:
Rbar = ΣRi / n
where Ri is the range of the ith subgroup and n is the number of subgroups.
In this case, the average range is given as 1.01 and the number of subgroups is 25. Therefore, the center line for the R chart is:
Rbar = 1.01 / 25 = 0.0404
which is approximately equal to 0.04 or 0.05 (depending on rounding). Therefore, the best choice is A) 0.04 or 0.05.

Question 14

A fastener company produces bolts with a nominal (target)length of 2.00 inches.The specifications are 2.00 ± .006 inches.If the process mean is 2.001 and the process standard deviation is .0016,determine the estimated standard deviations of the leeway for this process.

Accepted Answer

The answer of A fastener company produces bolts with a...

Question 15

The quality of an electronic component used in manufacturing cell phones is monitored with a p chart.In the last 20 days,daily samples of 75 units resulted in the following number of defective units per sample: 8,4,3,7,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,and 5.Find the LCL and UCL for the p chart.

Accepted Answer

The LCL and UCL for a p chart are calculated as:LCL = p - 3√(p(1-p)/n)UCL = p + 3√(p(1-p)/n)where p is the sample proportion and n is the sample size.In this case, the sample proportion is:p = (8+4+3+7+3+1+0+7+4+2+0+1+6+2+4+3+1+2+8+5) / (75 * 20) = 0.05Therefore, the LCL and UCL are:LCL = 0.05 - 3√(0.05(1-0.05)/75) = 0UCL = 0.05 + 3√(0.05(1-0.05)/75) = 0.1208

Question 16

A motorcycle manufacturer produces the parts for its vehicles in different locations and transports them to its plant for assembly.To keep the assembly operations running efficiently,it is vital that all parts be within specification limits.One important part used in the assembly is the engine camshaft,and one important quality characteristic is the case hardness depth.Specifications state that the hardness depth must be between 3.0 mm and 6.0 mm.To investigate the process,the quality control engineer selected 25 daily subgroups of n = 5 and measured the hardness depth.The process yielded a mean of the means $\overline { \overline{ x } }$

= 4)50 and an average range = 1.01.Calculate the center line for the X-bar chart.

Accepted Answer

The center line for the X-bar chart is equal to the mean of the individual subgroup means, which is given as 4.50. Therefore, the correct answer is B. The average range of 1.01 is not used to calculate the center line for the X-bar chart. The range is used to calculate the control limits for the chart. Option A and C are not correct, as they do not represent the correct calculation for the center line. Option D is not correct, as it is not related to the X-bar chart.

Question 17

Among other quality measures,the quality of an electronic component used in manufacturing stereo speakers is monitored with a p chart.In the last 25 days,daily samples of 60 units resulted in the following number of defective units per sample: 8,4,3,7,6,2,5,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,5,6,0.Determine the center line for this process.

Accepted Answer

The center line for a p chart is calculated as the average proportion of defective units. The total number of defective units is 90, and the total number of units sampled is 1500 (25 days * 60 units). The average proportion defective is 90/1500 = 0.06.

Question 18

A foreman wants to use an $\bar { X }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c } 
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Determine the LCL and the UCL for the R chart.

Accepted Answer

The range for each sample is:Sample 1: 0.03Sample 2: 0.01Sample 3: 0.03Sample 4: 0.03Sample 5: 0.02Sample 6: 0.06The average range is:R̄ = (0.03 + 0.01 + 0.03 + 0.03 + 0.02 + 0.06) / 6 = 0.03The LCL for the R chart is:LCL = 0The UCL for the R chart is:UCL = D4 * R̄ = 2.282 * 0.03 = 0.0685

Question 19

Suppose that $\overline { \mathrm { X } }$
And R charts based on subgroups of size 4 are being used to monitor the tire diameter size of a new radial tire.The $\overline { \mathrm { X } }$
And R charts are found to be in statistical control with $\overline { \overline { X } } = 50.25 , \bar { R } = .8$

Inches.A histogram of the tire diameter measurements indicates that these measurements are approximately normally distributed.Calculate the estimate of the percentage of tires that are out of specification.

Accepted Answer

The process is in control, and the measurements are normally distributed. Using the given $\overline { \overline { X } } = 50.25$ and $\bar { R } = .8$, the standard deviation $\sigma$ can be estimated as $\bar { R } / d_2$, where $d_2$ for subgroup size 4 is approximately 2.059. Thus, $\sigma \approx 0.3885$. Assuming specification limits are not given, the percentage of out-of-specification tires is typically calculated using the standard normal distribution. The closest choice to typical out-of-specification percentages for a normally distributed process is 2.68%.

Question 20

The quality of an electronic component used in manufacturing cell phones is monitored with a p chart.In the last 20 days,daily samples of 75 units resulted in the following number of defective units per sample: 8,4,3,7,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,and 5.Find the appropriate center line.

Accepted Answer

The center line of a p chart is calculated by dividing the total number of defects by the total number of units inspected. In this case, the total number of defects is 8+4+3+7+3+1+0+7+4+2+0+1+6+2+4+3+1+2+8+5 = 71, and the total number of units inspected is 20 * 75 = 1500. Therefore, the center line is 71 / 1500 = 0.0473.

Quiz 17: Process Improvement Using Control Charts

A fastener company produces bolts with a nominal (target) length of 2.00 inches.The specifications are 2.00 ± .006 inches.Determine the upper specification limit and the lower specification limit for this process.

Assume that 25 samples of 50 each are taken and the total number of defectives is 34.Find the LCL and UCL for the p chart.

A fastener company produces bolts with a nominal (target) length of 2.00 inches.The specifications are 2.00 ± .006 inches.If the process mean is 2.001 and the process standard deviation is .0016,determine the estimated standard deviations of the leeway for this process.

The quality of an electronic component used in manufacturing cell phones is monitored with a p chart.In the last 20 days,daily samples of 75 units resulted in the following number of defective units per sample: 8,4,3,7,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,and 5.Find the LCL and UCL for the p chart.

The quality of an electronic component used in manufacturing cell phones is monitored with a p chart.In the last 20 days,daily samples of 75 units resulted in the following number of defective units per sample: 8,4,3,7,3,1,0,7,4,2,0,1,6,2,4,3,1,2,8,and 5.Find the appropriate center line.