 # IGNOU MST-014 Solved Assignment 2023 | MSCAST

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MST-014 Assignment Question Paper 2023

1. (a) State whether the following statements are True or False. Give reason in support of your answer:
(i) The $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$ – chart is suitable when subgroup size is greater than 10 .
(ii) In single sampling plan, if we increase acceptance number then the OC curve will be steeper.
(iii) If the effect of summer and winter is not constant on the sale of AC then we use the additive model of the time series.
(iv) If a researcher wants to find the relationship between today’s unemployment and that of 5 years ago without considering what happens in between then the partial autocorrelation is the better way in comparison to autocorrelation.
(v) A system has four components connected in parallel configuration with reliability $0.2,0.4,0.5,0.8$$0.2,0.4,0.5,0.8$0.2,0.4,0.5,0.80.2,0.4,0.5,0.8$0.2,0.4,0.5,0.8$. To improve the reliability of the system most, we have to replace the component which reliability is 0.2 .
(b) Differentiate between the autoregressive and moving average models of time series.
2(a) 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$$c=2$c=2c=2$c=2$ as the acceptance number. Suppose 3% nonconforming zippers are acceptable to the manufacturer and $8\mathrm{%}$$8\mathrm{%}$8%8 \%$8\mathrm{%}$ nonconforming zippers are not acceptable. Find
(i) Probability of accepting a lot of incoming quality 0.04 .
(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).
(b) 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 nonconforming, 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\mathrm{%}$$4\mathrm{%}$4%4 \%$4\mathrm{%}$.
(i) What is the probability of accepting the lot at the first sample?
(ii) What is the probability of accepting the lot at the second sample?
3(a) A system has seven independent components and reliability block diagram of it shown as follows:
Find reliability of the system.
(b) The failure data for 40 electronic components is shown below:
 $\begin{array}{l}\text{Operating Time (in}\\ \text{hours)}\end{array}${:[” Operating Time (in “],[” hours) “]:}\begin{array}{l}\text { Operating Time (in } \\ \text { hours) }\end{array} $0-5$$0-5$0-50-5$0-5$ $5-10$$5-10$5-105-10$5-10$ $10-15$$10-15$10-1510-15$10-15$ $15-20$$15-20$15-2015-20$15-20$ $20-25$$20-25$20-2520-25$20-25$ $25-30$$25-30$25-3025-30$25-30$ Number of Failures 5 7 6 4 5 4 $\begin{array}{l}\text{Operating Time (in}\\ \text{hours)}\end{array}${:[” Operating Time (in “],[” hours) “]:}\begin{array}{l}\text { Operating Time (in } \\ \text { hours) }\end{array} $30-35$$30-35$30-3530-35$30-35$ $35-40$$35-40$35-4035-40$35-40$ $40-45$$40-45$40-4540-45$40-45$ $45-50$$45-50$45-5045-50$45-50$ $\ge 50$$\ge 50$>= 50\geq 50$\ge 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 | $\begin{array}{l}\text { Operating Time (in } \\ \text { hours) }\end{array}$ | $0-5$ | $5-10$ | $10-15$ | $15-20$ | $20-25$ | $25-30$ | | :— | :—: | :—: | :—: | :—: | :—: | :—: | | Number of Failures | 5 | 7 | 6 | 4 | 5 | 4 | | $\begin{array}{l}\text { Operating Time (in } \\ \text { hours) }\end{array}$ | $30-35$ | $35-40$ | $40-45$ | $45-50$ | $\geq 50$ | | | 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 $\mathbf{1}$$\mathbf{1}$1\mathbf{1}$\mathbf{1}$ $\mathbf{2}$$\mathbf{2}$2\mathbf{2}$\mathbf{2}$ $\mathbf{3}$$\mathbf{3}$3\mathbf{3}$\mathbf{3}$ $\mathbf{4}$$\mathbf{4}$4\mathbf{4}$\mathbf{4}$ $\mathbf{5}$$\mathbf{5}$5\mathbf{5}$\mathbf{5}$ 9 a.m 8 9 15 4 11 10 7 10 7 6 8 11 11 12 10 9 10 12 12 8 6 9 12 1 p.m. 11 10 6 14 11 2 7 7 10 4 11 3 10 7 4 10 10 4 8 11 11 11 7
Time Sample Number 1 2 3 4 5 9 a.m 8 9 15 4 11 10 7 10 7 6 8 11 11 12 10 9 10 12 12 8 6 9 12 1 p.m. 11 10 6 14 11 2 7 7 10 4 11 3 10 7 4 10 10 4 8 11 11 11 7| Time | Sample Number | | | | | | :—: | :—: | :—: | :—: | :—: | :—: | | | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | | 9 a.m | 8 | 9 | 15 | 4 | 11 | | 10 | 7 | 10 | 7 | 6 | 8 | | 11 | 11 | 12 | 10 | 9 | 10 | | 12 | 12 | 8 | 6 | 9 | 12 | | 1 p.m. | 11 | 10 | 6 | 14 | 11 | | 2 | 7 | 7 | 10 | 4 | 11 | | 3 | 10 | 7 | 4 | 10 | 10 | | 4 | 8 | 11 | 11 | 11 | 7 |
(i) Use the data to construct control charts for mean and comments about the process. If process is out of control, then calculate the revised control limits.
(ii) Construct the CUSUM chart when the process is under control and draw the conclusion about the process.
(iii) If the specification limits as the $8±2$$8±2$8+-28 \pm 2$8±2$, then calculate the process capability index ${\mathrm{C}}_{\mathrm{p}\mathrm{k}}$${\mathrm{C}}_{\mathrm{p}\mathrm{k}}$C_(pk)\mathrm{C}_{\mathrm{pk}}${\mathrm{C}}_{\mathrm{p}\mathrm{k}}$ and impetrate the result.
(iv) Also find the percentage of calls lie outside the specification limits assuming that calls follow the normal distribution.
5(a) Consider the time series model
${\mathrm{y}}_{\mathrm{t}}=10+0.5{\mathrm{y}}_{\mathrm{t}-1}-0.8{\mathrm{y}}_{\mathrm{t}-2}+{\epsilon }_{\mathrm{t}}$${\mathrm{y}}_{\mathrm{t}}=10+0.5{\mathrm{y}}_{\mathrm{t}-1}-0.8{\mathrm{y}}_{\mathrm{t}-2}+{\epsilon }_{\mathrm{t}}$y_(t)=10+0.5y_(t-1)-0.8y_(t-2)+epsi_(t)\mathrm{y}_{\mathrm{t}}=10+0.5 \mathrm{y}_{\mathrm{t}-1}-0.8 \mathrm{y}_{\mathrm{t}-2}+\varepsilon_{\mathrm{t}}${\mathrm{y}}_{\mathrm{t}}=10+0.5{\mathrm{y}}_{\mathrm{t}-1}-0.8{\mathrm{y}}_{\mathrm{t}-2}+{\epsilon }_{\mathrm{t}}$
where ${\epsilon }_{\mathrm{t}}\sim \mathrm{N}\left[0,1\right]$${\epsilon }_{\mathrm{t}}\sim \mathrm{N}\left[0,1\right]$epsi_(t)∼N[0,1]\varepsilon_{\mathrm{t}} \sim \mathrm{N}[0,1]${\epsilon }_{\mathrm{t}}\sim \mathrm{N}\left[0,1\right]$
(i) Is this a stationary time series?
(ii) What are the mean and variance of the time series?
(iii) Calculate the autocorrelation function.
(iv) Plot the correlogram.
(b) The marketing manager of a company recorded the number of mobiles sold quarterly for which are given in the following table:
 Quarter ${\mathbf{Q}}_{1}$${\mathbf{Q}}_{1}$Q_(1)\mathbf{Q}_1${\mathbf{Q}}_{1}$ ${\mathbf{Q}}_{\mathbf{2}}$${\mathbf{Q}}_{\mathbf{2}}$Q_(2)\mathbf{Q}_{\mathbf{2}}${\mathbf{Q}}_{\mathbf{2}}$ ${\mathbf{Q}}_{\mathbf{3}}$${\mathbf{Q}}_{\mathbf{3}}$Q_(3)\mathbf{Q}_{\mathbf{3}}${\mathbf{Q}}_{\mathbf{3}}$ ${\mathbf{Q}}_{\mathbf{4}}$${\mathbf{Q}}_{\mathbf{4}}$Q_(4)\mathbf{Q}_{\mathbf{4}}${\mathbf{Q}}_{\mathbf{4}}$ $\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{8}$$\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{8}$2018\mathbf{2 0 1 8}$\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{8}$ 48 41 60 65 $\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{9}$$\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{9}$2019\mathbf{2 0 1 9}$\mathbf{2}\mathbf{0}\mathbf{1}\mathbf{9}$ 58 52 68 74 $\mathbf{2}\mathbf{0}\mathbf{2}\mathbf{0}$$\mathbf{2}\mathbf{0}\mathbf{2}\mathbf{0}$2020\mathbf{2 0 2 0}$\mathbf{2}\mathbf{0}\mathbf{2}\mathbf{0}$ 60 56 75 78
Quarter Q_(1) Q_(2) Q_(3) Q_(4) 2018 48 41 60 65 2019 58 52 68 74 2020 60 56 75 78| Quarter | $\mathbf{Q}_1$ | $\mathbf{Q}_{\mathbf{2}}$ | $\mathbf{Q}_{\mathbf{3}}$ | $\mathbf{Q}_{\mathbf{4}}$ | | :—: | :—: | :—: | :—: | :—: | | $\mathbf{2 0 1 8}$ | 48 | 41 | 60 | 65 | | $\mathbf{2 0 1 9}$ | 58 | 52 | 68 | 74 | | $\mathbf{2 0 2 0}$ | 60 | 56 | 75 | 78 |
(i) Find the quarterly seasonal indexes for the mobile sold using the ratio to trend method.
(ii) Do seasonal forces significantly influence the sale of mobile? Comment.
(iii) Also find the deseasonalised values.
$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$

## MST-014 Sample Solution 2023

### Question:-01

1. (a) State whether the following statements are True or False. Give reason in support of your answer:
(i) The $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$ – chart is suitable when subgroup size is greater than 10 .
The statement “The $R$$R$RR$R$ – chart is suitable when subgroup size is greater than 10″ is generally considered to be False.

### Justification:

The $R$$R$RR$R$ – chart, or range chart, is used in statistical process control to help evaluate the stability of a process in terms of the variation among the values within a sample (subgroup). However, there are some considerations regarding subgroup size:
1. Small Subgroup Sizes: The $R$$R$RR$R$ – chart is most effective and commonly used when the subgroup size ($n$$n$nn$n$) is small, typically between 2 and 10. This is because the range is a simpler and more efficient calculation when dealing with smaller subgroup sizes.
2. Large Subgroup Sizes: When subgroup sizes are larger than 10, the $R$$R$RR$R$ – chart may not be the most appropriate choice for evaluating within-group variability due to the following reasons:
• Sensitivity: The range is less sensitive to the spread of data as the subgroup size increases. It only considers the largest and smallest values and ignores the distribution of all the values in between.
• Normality Assumption: The distribution of the range is highly dependent on the underlying distribution of the process data, and this dependency increases with subgroup size. For larger subgroups, if the data is not normally distributed, the control limits calculated for the $R$$R$RR$R$ – chart may not be accurate or reliable.
• Alternative Charts: The $S$$S$SS$S$ – chart (standard deviation chart) is often recommended for larger subgroup sizes because it considers all data points in the subgroup, providing a more accurate and reliable measure of dispersion, especially when $n>10$$n>10$n > 10n > 10$n>10$.

### Example:

Consider a subgroup of size 15 with the following values:
$5,6,7,7,8,8,9,9,10,10,11,11,12,12,13$$5,6,7,7,8,8,9,9,10,10,11,11,12,12,13$5,6,7,7,8,8,9,9,10,10,11,11,12,12,135, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13$5,6,7,7,8,8,9,9,10,10,11,11,12,12,13$
• The range ($R$$R$RR$R$) would be calculated as the difference between the largest and smallest values: $R=13-5=8$$R=13-5=8$R=13-5=8R = 13 – 5 = 8$R=13-5=8$.
• However, this doesn’t give any insight into the variability of all the other values in the subgroup.
• If there were outliers or shifts within the subgroup, the $R$$R$RR$R$ – chart might not detect them effectively due to its insensitivity to the distribution of internal values.

### Conclusion:

While the $R$$R$RR$R$ – chart is a valuable tool in statistical process control, its suitability is generally constrained to scenarios with smaller subgroup sizes. For larger subgroup sizes, alternative charts, such as the $S$$S$SS$S$ – chart, are typically recommended to provide a more accurate and reliable analysis of process variability.
$$\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}$$

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$$sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta$$

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