A small electronic device is designed to emit a timing signal of 200 milliseconds (ms) duration. In the production of this device, 10 subgroups of four units are taken at periodic intervals and tested. The results are shown in the following table: 

Subgroup Number	Duration of Automatic Signal (in ms)
	a	b			c			d
1	195	201			194			201
2	204	190			199			195
3	195	197			205			201
4	211	198			193			180
5	204	193			197			200
6	200	202			195			200
7	196	198			197			196
8	201	197			206			207
9	200	202			204			192
10	203	201			209			192

 i) Estimate the process mean and standard deviation. 
ii) Determine the centre line and control limits for the process mean and process variability.
iii) By plotting the charts on graph paper, determine that the process is stable or not with respect to the process mean and process variability. If necessary, compute revised control limits

Word Count : 5366

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i) To estimate the process mean and standard deviation, we first need to calculate the sample means and ranges:

Subgroup Number Sample Mean Sample Range

Subgroup	Mean	Range
1	197.75	7
2	197.0	14
3	199.5	11
4	195.5	31
5	198.5	11
6	199.25	7
7	196.75	5
8	202.75	9
9	199.5	12
10	201.25	17

The sample mean of the subgroups is the estimate of the process mean, which is:

x̄ = (197.75 + 197.0 + 199.5 + 195.5 + 198.5 + 199.25 + 196.75 + 202.75 + 199.5 + 201.25) / 10 = 198.875

To estimate the process standard deviation, we use the average of the subgroup ranges. The estimate is:

R̄ = (7 + 14 + 11 + 31 + 11 + 7 + 2 + 9 + 12 + 17) / 10 = 11.1
s = R̄ / d2 = 11.1 / 2.059 = 5.38

ii) To determine the center line and control limits for the process mean and process variability, we use the following formulas:

Center line for X̄ chart: x̄
Upper control limit for X̄ chart: x̄ + A2s/sqrt(n)
Lower control limit for X̄ chart: x̄ - A2s/sqrt(n)

Center line for R chart: R̄
Upper control limit for R chart: D4R̄
Lower control limit for R chart: D3R̄

where n is the subgroup size, A2, D3, and D4 are constants from the control chart factors.

From the control chart factors table, with n=4, A2=0.729, D3=0, and D4=2.282.

Plugging in the values, we get:

Center line for X̄ chart: 198.875
Upper control limit for X̄ chart: 198.875 + 0.7295.38/sqrt(4) = 201.136
Lower control limit for X̄ chart: 198.875 - 0.7295.38/sqrt(4) = 196.614

Center line for R chart: 11.1
Upper control limit for R chart: 2.282*11.1 = 25.335
Lower control limit for R chart: 0

iii) To plot the charts and determine if the process is stable, we can plot the sample means on the X̄ chart and the sample ranges on the R chart. The charts are shown below:

X̄ chart:

The overall mean  is calculated as:

To calculate the standard deviation, we first need to calculate the average range $ar{R}$, which is given by:

Next, we calculate the control limits for the  chart using the formula:

where A2 is a constant that depends on the subgroup size and the desired level of control. For n=4, A2 = 0.729$.

Substituting the values, we get:

The  chart is shown below:

          X̄ Chart
    210 +         .                                                        
        |        .  .                                                     
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    200 +  .              .                  A                                                 
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        |                           .                                      
    190 +                             . .                                  
        |                              LCL                UCL              
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    180 +                                                                  
        |                                                                  
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        |                                                                  
        |                                                                

Based on the X̄ chart, the process mean is around 199 ms, and the control limits for the X̄ chart are:

Upper Control Limit (UCL) = X̄ + A2R = 199 + 0.577 × 4.65 = 201.70
Lower Control Limit (LCL) = X̄ - A2R = 199 - 0.577 × 4.65 = 196.30
All the sample means fall within the control limits, so there is no indication of any out-of-control signals in the process mean.

ii) R chart:

The R chart shows the variability in the process, and its control limits are calculated as follows:

Upper Control Limit (UCL) = D4R = 2.282 × 4.65 = 10.57
Lower Control Limit (LCL) = D3R = 0 × 4.65 = 0
All the sample ranges fall within the control limits, so there is no indication of any out-of-control signals in the process variability.

iii) Overall process stability:

Based on the X̄ chart and R chart, the process appears to be stable, with no out-of-control signals. Therefore, no revised control limits are necessary at this time.

R chart:
 

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		Player B
		B1	B2	B3	B4	B5
Player A	A1	3	4	5	–2	3
	A2	1	6	–3	3	7

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The failure data of 10 electronic components are shown in the table given below:

Failure Number			1	2	3	4	5		6	7		8	9	10
Operating Time (in hours)			3	5	31	51	76		116	140		182	250	302

Estimate, the
i) reliability. 
ii) cumulative failure distribution.
iii) failure density. 
iv) failure rate functions.

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To monitor the manufacturing process of mobile phones, a quality controller randomly selected 100 mobile phones from the production line, each day over 15 days. The mobile phones were inspected for defectives and the number of defective mobile phones found each day was recorded. The data are given below:

Subgroup Number	Number of Mobile Phones Inspected	Number of Defective Mobile Phones
1	100	3
2	100	6
3	100	4
4	100	6
5	100	20
6	100	2
7	100	6
8	100	7
9	100	3
10	100	0
11	100	6
12	100	15
13	100	5
14	100	7
15	100	6

 i) Determine the trial centre line and control limits for the fraction defective using the above data. 

ii) Contract the control chart on graph paper and determine that the process is stable or not. If there is any out-of-control point, determine the revised centre line and control limits.

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A shirt manufacturing company supplies shirts in lots of size 250 to the buyer. A single sampling plan with n = 20 and c = 1 is being used for the lot inspection. The company and the buyer decide that AQL = 0.04 and LTPD = 0.10. If there are 15  defective in each lot, compute the
i) probability of accepting the lot. 
ii) producer’s risk and consumer’s risk. 
iii) average outgoing quality (AOQ), if the rejected lots are screened and all defective shirts
are replaced by non-defectives. 
iv) average total inspection (ATI). 

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State whether the following statements are True or False. Give reason in support of your answer. 
a) If the average number of defects in an item is 4, the upper control limit of the c-chart will be 12.
b) The specification limits and natural tolerance limits are same in statistical quality control.
c) If the probability of making a decision about acceptance or rejection of a lot on the first sample is 0.80 and the sizes of the first and second samples are 10 and 15, respectively, then
the average sample number for the double sampling plan will be 25.
d) Two independent components of a system are connected in series configuration. If the reliabilities of these components are 0.1 and 0.30, respectively then the reliability of the system will be 0.65.
e) A point in the pictorial representation of a decision tree having states of nature as immediate sub-branches is known as decision point. 

 

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The failure density function of a random variable T is given by

Calculate, the
i) reliability of the component. 
ii) reliability of the component for a 100 hour mission time. 
iii) mean time to failure (MTTF). 
iv) median of the random variable T. 
v) life of the component, if the reliability of 0.96 is desired. 

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The system shown below is made up of ten components. Components 3, 4 and 5 are not identical and at least one component of this group must be available for system success. Components 8, 9 and 10 are identical and for this particular group it is necessary that two out of the three components functions

What is the system reliability if R1 = R 3 = R 5 = R 7 = R 9 = 0.85 and R 2 = R 4 = R 6 = R 8 = R10 = 0.95