IGNOU MSTE-001 Solved Assignment | 1st January 2025 to 31st December, 2025 | Industrial Statistics-I | PGDAST

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IGNOU MSTE-001 Solved Assignment | 1st January 2025 to 31st December, 2025 | Industrial Statistics-I | PGDAST Sample Solution
  1. State whether the following statements are True or False. Give reason in support of your answer:
(a) Statistical quality control (SQC) is a technique of process control only.
(b) Twenty pieces of different length of cloth contained 2 , 4 , 1 , 3 , 5 , 4 , 2 , 7 , 3 , 5 , 2 , 2 , 4 , 5 2 , 4 , 1 , 3 , 5 , 4 , 2 , 7 , 3 , 5 , 2 , 2 , 4 , 5 2,4,1,3,5,4,2,7,3,5,2,2,4,52,4,1,3,5,4,2,7,3,5,2,2,4,52,4,1,3,5,4,2,7,3,5,2,2,4,5, 6 , 4 , 2 , 1 , 2 , 4 6 , 4 , 2 , 1 , 2 , 4 6,4,2,1,2,46,4,2,1,2,46,4,2,1,2,4 defects respectively. To check the process is under control with respect to the number of defects, we should use np-chart.
(c) If the probabilities are not associated with the occurrence of different states of nature, then the situation is known as decision making under risk.
(d) In single sampling plan, if we increase acceptance number then the OC curve will be steeper.
(e) A system has four components connected in parallel configuration with reliability 0.2 , 0.4 , 0.5 , 0.8 0.4 , 0.5 , 0.8 0.4,0.5,0.80.4,0.5,0.80.4,0.5,0.8. To improve the reliability of the system most, we have to replace the component which reliability is 0.2 .
  1. Twenty samples each of size 10 were inspected. The number of defectives detected in each of them is given below: 0 , 1 , 0 , 3 , 9 , 2 , 0 , 7 , 0 , 1 , 1 , 0 , 0 , 3 , 1 , 0 , 0 , 2 , 1 , 0 0 , 1 , 0 , 3 , 9 , 2 , 0 , 7 , 0 , 1 , 1 , 0 , 0 , 3 , 1 , 0 , 0 , 2 , 1 , 0 0,1,0,3,9,2,0,7,0,1,1,0,0,3,1,0,0,2,1,00,1,0,3,9,2,0,7,0,1,1,0,0,3,1,0,0,2,1,00,1,0,3,9,2,0,7,0,1,1,0,0,3,1,0,0,2,1,0 Find the control limits for the number of defectives and establish quality standards for the future. Plot the graph and interpret.
  2. A manufacturer of men’s jeans purchases zippers in lots of 500 . The jeans manufacturer uses single-sample acceptance sampling with a sample size of 10 to determine whether to accept the lot. The manufacturer uses c = 2 c = 2 c=2\mathrm{c}=2c=2 as the acceptance number. Suppose 3% nonconforming zippers are acceptable to the manufacturer and 8 % 8 % 8%8 \%8% nonconforming zippers are not acceptable. Let incoming quality be 4 % 4 % 4%4 \%4%.
    i) Construct an OC curve.
    ii) Average outing quality (AOQ), if the rejected lots are screened and all defective zippers are replaced by non-defectives.
    iii) Average total inspection (ATI).
  3. An office supply company ordered a lot of 400 printers. When the lot arrives the company inspector will randomly inspect 12 printers. If more than three printers in the sample are non-conforming, the lot will be rejected. If fewer than two printers are non-conforming, the lot will be accepted. Otherwise, a second sample of size 8 will be taken. Suppose the inspector finds two non-conforming printers in the first sample and two in the second sample. Also AQL and LTPD are 0.05 and 0.10 respectively. Let incoming quality be 4 % 4 % 4%4 \%4%.
    i) What is the probability of accepting the lot at the first sample?
    ii) What is the probability of accepting the lot at the second sample?
    iii) Find AQL and ATI
  4. A two-person zero-sum game having the following payoff matrix for player A
I II III IV V
I 2 4 3 8 4
II 5 6 3 7 8
III 6 7 9 8 7
IV 4 2 8 4 3
I II III IV V I 2 4 3 8 4 II 5 6 3 7 8 III 6 7 9 8 7 IV 4 2 8 4 3| | I | II | III | IV | V | | :—: | :—: | :—: | :—: | :—: | :—: | | I | 2 | 4 | 3 | 8 | 4 | | II | 5 | 6 | 3 | 7 | 8 | | III | 6 | 7 | 9 | 8 | 7 | | IV | 4 | 2 | 8 | 4 | 3 |
(i) Check whether saddle point exit or not.
(ii) If saddle point does not exit then determine optimal strategies for both the manufacturers and value of the game.
  1. A system has seven independent components and reliability block diagram of it shown blow:
original image
Find reliability of the system.
  1. The failure data for 40 electronic components is shown below:
Operating Time (in hours) 0 − 5 0 − 5 0-50-50−5 5 − 10 5 − 10 5-105-105−10 10 − 15 10 − 15 10-1510-1510−15 15 − 20 15 − 20 15-2015-2015−20 20 − 25 20 − 25 20-2520-2520−25 25 − 30 25 − 30 25-3025-3025−30
Number of Failures 5 7 6 4 5 4
Operating Time (in hours) 30 − 35 30 − 35 30-3530-3530−35 35 − 40 35 − 40 35-4035-4035−40 40 − 45 40 − 45 40-4540-4540−45 45 − 50 45 − 50 45-5045-5045−50 ≥ 50 ≥ 50 >= 50\geq 50≥50
Number of Failures 4 0 2 1 2
Operating Time (in hours) 0-5 5-10 10-15 15-20 20-25 25-30 Number of Failures 5 7 6 4 5 4 Operating Time (in hours) 30-35 35-40 40-45 45-50 >= 50 Number of Failures 4 0 2 1 2 | Operating Time (in hours) | $0-5$ | $5-10$ | $10-15$ | $15-20$ | $20-25$ | $25-30$ | | :— | :— | :— | :— | :— | :— | :— | | Number of Failures | 5 | 7 | 6 | 4 | 5 | 4 | | Operating Time (in hours) | $30-35$ | $35-40$ | $40-45$ | $45-50$ | $\geq 50$ | | | Number of Failures | 4 | 0 | 2 | 1 | 2 | |
Estimate the reliability, cumulative failure distribution, failure density and failure rate functions.
  1. At a call centre, callers have to wait till an operator is ready to take their call. To monitor this process, 5 calls were recorded every hour for the 8 -hour working day. The data below shows the waiting time in seconds:
Time Sample Number
1 1 1\mathbf{1}1 2 2 2\mathbf{2}2 3 3 3\mathbf{3}3 4 4 4\mathbf{4}4 5 5 5\mathbf{5}5
9 a.m 8 9 15 4 11
10 7 10 7 6 8
11 11 12 10 9 10
12 12 8 6 9 12
Time Sample Number 1 2 3 4 5 9 a.m 8 9 15 4 11 10 7 10 7 6 8 11 11 12 10 9 10 12 12 8 6 9 12| Time | Sample Number | | | | | | :— | :— | :— | :— | :— | :— | | | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | | 9 a.m | 8 | 9 | 15 | 4 | 11 | | 10 | 7 | 10 | 7 | 6 | 8 | | 11 | 11 | 12 | 10 | 9 | 10 | | 12 | 12 | 8 | 6 | 9 | 12 |
1 p.m. 11 10 6 14 11
2 7 7 10 4 11
3 10 7 4 10 10
4 8 11 11 11 7
1 p.m. 11 10 6 14 11 2 7 7 10 4 11 3 10 7 4 10 10 4 8 11 11 11 7| 1 p.m. | 11 | 10 | 6 | 14 | 11 | | :— | :— | :— | :— | :— | :— | | 2 | 7 | 7 | 10 | 4 | 11 | | 3 | 10 | 7 | 4 | 10 | 10 | | 4 | 8 | 11 | 11 | 11 | 7 |
i) Use the data to construct control charts for mean and variability and comments about the process. If process is out of control, then calculate the revised control limits.
ii) If the specification limits as the 8 ± 2 8 ± 2 8+-28 \pm 28±2, then find the process capability. Does it appear that the process is capable of meeting the specification requirements?

Answer:

Question:-01

State whether the following statements are True or False. Give reason in support of your answer:

(a) Statistical quality control (SQC) is a technique of process control only.

Answer:

Statement: Statistical quality control (SQC) is a technique of process control only.
Answer: False
Justification: Statistical quality control (SQC) is not limited to process control only. While it is widely used to monitor and improve processes (e.g., in manufacturing), SQC also encompasses product quality control through techniques like acceptance sampling, where finished products are inspected to ensure they meet standards. Thus, it applies to both process and product control, not just process control.
(b) Twenty pieces of different length of cloth contained 2 , 4 , 1 , 3 , 5 , 4 , 2 , 7 , 3 , 5 , 2 , 2 , 4 , 5 2 , 4 , 1 , 3 , 5 , 4 , 2 , 7 , 3 , 5 , 2 , 2 , 4 , 5 2,4,1,3,5,4,2,7,3,5,2,2,4,52,4,1,3,5,4,2,7,3,5,2,2,4,52,4,1,3,5,4,2,7,3,5,2,2,4,5, 6 , 4 , 2 , 1 , 2 , 4 6 , 4 , 2 , 1 , 2 , 4 6,4,2,1,2,46,4,2,1,2,46,4,2,1,2,4 defects respectively. To check the process is under control with respect to the number of defects, we should use np-chart.

Answer:

Statement: To check the process is under control with respect to the number of defects, we should use an np-chart.
Answer: False
Justification: An np-chart is used for attribute data to monitor the number of defective items in a fixed sample size, assuming a binomial distribution. In this case, the data provided (defects in 20 pieces of cloth: 2, 4, 1, 3, 5, 4, 2, 7, 3, 5, 2, 2, 4, 5, 6, 4, 2, 1, 2, 4) represents the number of defects per piece, and the sample size (one piece) is not fixed across multiple subgroups, nor are the pieces necessarily of equal length. A c-chart, which tracks the count of defects per unit when sample size varies, would be more appropriate here, not an np-chart.
(c) If the probabilities are not associated with the occurrence of different states of nature, then the situation is known as decision making under risk.

Answer:

Statement: If the probabilities are not associated with the occurrence of different states of nature, then the situation is known as decision making under risk.
Answer: False
Justification: Decision making under risk occurs when probabilities are associated with the occurrence of different states of nature, allowing for expected value calculations. If probabilities are not associated with the states of nature, the situation is known as decision making under uncertainty, not under risk. For example, if a manager must choose an action without knowing the likelihood of outcomes (e.g., weather conditions affecting sales), it’s uncertainty, not risk.
(d) In single sampling plan, if we increase acceptance number then the OC curve will be steeper.

Answer:

Statement: In a single sampling plan, if we increase the acceptance number then the OC curve will be steeper.
Answer: False
Justification: In a single sampling plan, the operating characteristic (OC) curve shows the probability of accepting a lot as a function of its quality (proportion defective). Increasing the acceptance number (c) makes the plan less stringent, shifting the OC curve upward and typically making it less steep. This is because a higher c increases the likelihood of acceptance for a wider range of defect rates, flattening the curve rather than steepening it. For example, if c increases from 1 to 2, more lots with higher defect rates are accepted, reducing the curve’s slope.
(e) A system has four components connected in parallel configuration with reliability 0.2 , 0.4 , 0.5 , 0.8 0.4 , 0.5 , 0.8 0.4,0.5,0.80.4,0.5,0.80.4,0.5,0.8. To improve the reliability of the system most, we have to replace the component which reliability is 0.2.

Answer:

Statement: A system has four components connected in parallel configuration with reliability 0.2, 0.4, 0.5, 0.8. To improve the reliability of the system most, we have to replace the component which reliability is 0.2.
Answer: True
Justification: In a parallel configuration, system reliability is 1 minus the product of the failure probabilities (1 – reliability). Here, failure probabilities are 0.8, 0.6, 0.5, and 0.2. System reliability = 1 − ( 0.8 × 0.6 × 0.5 × 0.2 ) = 1 − 0.048 = 0.952 1 − ( 0.8 × 0.6 × 0.5 × 0.2 ) = 1 − 0.048 = 0.952 1-(0.8 xx0.6 xx0.5 xx0.2)=1-0.048=0.9521 – (0.8 \times 0.6 \times 0.5 \times 0.2) = 1 – 0.048 = 0.9521−(0.8×0.6×0.5×0.2)=1−0.048=0.952. Replacing the component with reliability 0.2 (failure 0.8) has the largest impact because it has the highest failure probability. If replaced with a perfect component (reliability 1), new reliability = 1 − ( 0.6 × 0.5 × 0.2 × 0 ) = 1 1 − ( 0.6 × 0.5 × 0.2 × 0 ) = 1 1-(0.6 xx0.5 xx0.2 xx0)=11 – (0.6 \times 0.5 \times 0.2 \times 0) = 11−(0.6×0.5×0.2×0)=1, a bigger jump than replacing any other component (e.g., 0.4 to 1 yields 1 − ( 0.8 × 0.5 × 0.2 ) = 0.984 1 − ( 0.8 × 0.5 × 0.2 ) = 0.984 1-(0.8 xx0.5 xx0.2)=0.9841 – (0.8 \times 0.5 \times 0.2) = 0.9841−(0.8×0.5×0.2)=0.984). Thus, replacing the 0.2 component maximizes improvement.
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