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## Karnataka 2nd PUC Statistics Question Bank Chapter 7 Statistical Quality Control

Section – A

### 2nd PUC Statistics Statistical Quality Control One Mark Questions and Answers

Question 1.

What is meant by Statistical Quality Control?

Answer:

Statistical Quality Control is the method of controlling the quality of the product using statistical techniques.

Question 2.

Mention the three stages of production process.

Answer:

- specification
- production
- inspection

Question 3.

Mention two uses of statistical quality control.

Answer:

- To improve and maintain the quality standards by decreasing the proportion of defectives.
- To provide greater quality assurance at lower inspection cost.

Question 4.

Mention two types of causes for variation in a manufacturing process?

Answer:

- Chance cause of variation.
- Assignable cause of variation.

Question 5.

What are chance and assignable causes?

Answer:

A-small amount of variation for which no specific cause can be identified and is beyond the control of human hand is named as chance variation.

Assignable causes : Any one cause can result in a large amount of variation and the presence of these variation can be detected and eliminated.

Question 6.

Statistical quality control helps in detecting which type of variation?

Answer:

Assignable causes of variation.

Question 7.

What do you mean by product control and process control?

Answer:

Process control: controlling the quality of the goods, during the manufacturing process itself is named as process control.

Product control: The process of inspection of manufactured lot for acceptability is named as product control or acceptance sampling.

Question 8.

Name the control chart used in case of defective in statistical quality control.

Answer:

‘np’ chart or ‘d’ chart is used in case of defective.

Question 9.

What is a control chart?

Answer:

Control chart is a graphical device which is used to study and verify whether the production process is in statistical control or not.

Question 10.

What are upper and lower control limits? What is the conclusion if an observation falls outside the control limits?

Answer:

The maximum acceptable variation from the mean for a process, is called The upper control limit

1. Lower control limit (LCL): Which indicates the minimum acceptable variation from the average quality level.

2. Upper control limit (UCL): Is the maximum acceptable variation from the mean.

If the sample point falls outside the control limits the process is said to be out of statistical control.

Question 11.

Mention different types of control charts?

Answer:

- Control charts for variables, x̄ chart, R-chart
- Control charts for attributes np chart, c-chart

Question 12.

What are defect and defectives?

Answer:

Defect: It is a quality characteristic which does not conform to specifications

Defective: An item having one or more defects is a defective item

Question 13.

Give an example for defect in a product?

Answer:

- A blade without sharpness.
- A garment with a tear.

Question 14.

What are single and double sampling plans?

Answer:

SSP : In a SSP decision about accepting or rejecting a lot manufactured lot based on one sample.

DSP: Decisionaboutacceptingorrejectinga manufactured lot based on two samples.

Question 15.

Mention two differences between single sampling plan and double sampling plan?

Answer:

SSP |
DSP |

1. Decision of accepting or rejecting the manufactured lot on single sample | 1. Decision of accepting or rejecting the manufacturing lot based on two samples. |

2. It is less economical | 2. It is more economical |

Question 16.

Define acceptance sampling

Answer:

The process of inspection of manufactured lot for acceptability is called product control or acceptance sampling

Question 17.

Briefly explain single sampling plan and double sampling plan.

Answer:

In single sampling, a random sample is drawn from a lot. Each item in the sample is examined. Depending on the number of defectives found, the conclusion will be drawn.

In double sampling plan the conclusion will be based on two samples if it is not possible to come to a conclusion on the basis of first sample, then the second sample will be selected from the remaining lot.

Question 18.

Mention two advantages of acceptance sampling1.

Answer:

- It is used when the items are of destructive in nature.
- It is less expensive as 100% inspection is more expensive

Question 19.

Mention two disadvantages of acceptance sampling plan

Answer:

- There is a risk of accepting a bad lot and reject¬ing a good lot, since verification is done only on the basis of samples.
- Timely identification of the production of defectives cannot be achieved.

Question 20.

Write down the control limits for X̄ and R charts when standards known and not known.

Answer:

Control limits for X̿ chart

Control limits for R chart

Question 21.

Write down the control limits for np chart (d chart) and c charts when standards are given and not given control limits for np chart:

Answer:

Control limits for np chart

Where p̄ = \(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}\)

Control limits for c chart

Question 22.

Write the upper control limit and lower control limit for d chart when standard P’ is given

Answer:

LCL – nP’ – \(3 \sqrt{\mathrm{np}^{\prime} \mathrm{Q}^{\prime}}\)

UCL – nP’ + \(3 \sqrt{\mathrm{np}^{\prime} \mathrm{Q}^{\prime}}\)

Question 23.

Measurements of metal holders for different samples of size 5 each gave the results X̿ = 80.3 R̄ = 5.2. Find the values of control limits for drawing X̄-chart. Given A = 0.577.

Answer:

Control limits for x̄ chart:

Given X̿ = 80 3 R̄ = 5 2 A = 0.577 n = 5

CL = X̿ = 80.3

LCL = X̿ – A_{2}R̄ = 80.3 – 0.577(5.2)

80.3 – 3.0004 = 77.3.

UCL = X̿ + A_{2}R̄ ⇒ 80.3 + 0.577(5.2)

⇒ 80.3 + 3.0004 ⇒ 83.3

Question 24.

Calculate control limits for d-chart given p̄ = 0.05 and sample size n = 5.

Answer:

Given p̄ = 0.05 n = 5

∴ q̄ = 1 – p̄ = 1 – 0.05 = 0.95

CL = n . p̄ = (0.05)5 = 0.25

LCL = np̄ – 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\) = 5(0.05) – 3\(\sqrt{5(0.05)(0.95)}\)

⇒ 0.25 – 3 (0.487)

⇒ 0.25 – 1.462 = -1.212 = 0

UCL = np + 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\)

⇒ 5(0.05) + 3\(\sqrt{5(0.05)(0.95)}\)

⇒ 0.25 + 1.462 = 1.712

Question 25.

In a light bulb manufacturing plant, a products on manager has inspected a sample of 30 bulbs at regular intervals. If the average fraction defective is p̄ = 0.057 write down the control limits for number of defectives.

Answer:

Given p̄ = 0.057 q̄ = 1 – p̄ = 1 – 0.057 ⇒ 0.943, n = 30

Control limits for d chart (no. of defectives)

CL = np̄ = 30(0.057) = 1.71

LCL = np̄ – 3\(\sqrt{\mathrm{n} \overline{\mathrm{pq}}}\) ⇒ 30(0.0571) – 3\(\sqrt{(30)(0.0571)(0.943)}\)

⇒ 1.71 – 3(1.271)

⇒ 1.71 – 3.813 ⇒ – 2.103 = 0

UCL = np̄ + 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\) ⇒ 1.71 + 3.813 = 5.523

Question 26.

In a mobile phone manufacturing plant, a production manager has inspected a sample of 10 phone at regular intervals. If the average fraction defectives is P’ = 0.027 write down the control limits for number of defectives.

Answer:

Here the standard ‘P’ is given tives?

∴ The control limits are

Given n = 10 P’ = 0.027 Q’= 1 – P’

⇒ 1 – 0.027 ⇒ 0.973

CL = n.P’ = (10)0.027 ⇒ 0.27

LCL = nP’ – 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\)

⇒ 10(0.027) – 3\(\sqrt{10(0.027)(0.973)}\)

⇒ 0. 27 – 1.5377 = -1.267 = 0

UCL = nP’ + 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\)

⇒ 10(0.027) + 3\(\sqrt{10(0.027)(0.973)}\)

⇒ 0.27 + 3\(\sqrt{0.26271}\) = 0.27 + (0.51255)3 ⇒ 1.8077

Question 27.

In a fish net manufacturing process, the proportion of defectives is 0.03, if process control is based on samples of size 150 each. Write down control limit for np-chart.

Answer:

Given n = 150, P’ = 0.03, Q’ = 1 – P’ = 1 – 0.03 → 0.97 control limits for np-chart.

CL = n.P’ ⇒ (150)0.03 ⇒ 4.5

LCL = nP’ – 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\)

⇒ (150)(0.03) – \(\sqrt{(150)(0.03)(0.97)}\)

⇒ 4.5 – 6.2677 ⇒ 1.767 = 0

UCL = nP’ + 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\)

⇒ (150)(0.03) + 3\(\sqrt{150(0.03)(0.93)}\)

⇒ 4.5 + 6.2677 ⇒ 10.7677

Question 28.

One meter of cloth was inspected for weaving defects. Total number of defects for 10 such samples was 24. Find the control limits for defects.

Answer:

Given k = 10 ΣC = 24

c̄ = \(\frac { 24 }{ 10 }\) = 2.4;

CL = c̄ =2.4

LCL = c̄ – 3\(\sqrt{\bar{c}}\) ⇒ 2.4 – 3\(\sqrt{2.4}\)

⇒ 2.4 – 6475 = -2.247 = 0

UCL ⇒ c̄ + 3\(\sqrt{\bar{c}}\)

⇒ 2.4 + 3\(\sqrt{2.4}\) = 2.4 + 4.6475 = 7.0475

Question 29.

In a palm oil packaging company the palm oil is filled in tins. Samples of 4 tins are selected at regular intervals. The following data reveals the mean and rearrange of weight of tins.

Construct (i) x̄-chart (ii) R-chart and comment given A_{2} = 0.729 D_{4} = 2.282 and D_{3} = 0

Answer:

Given n = 4; k = 6

1. Here the standards are not given

∴ The control limits for X̄ chart are

CL = X̄ = \(\frac{\Sigma \bar{X}}{K}=\frac{81.5}{6}\) = 13.583;

LCL = X̿ – A_{2}R̄

LCL = 13.583 (0.729) (7.167) = 8.35

Where R̄ = \(\frac{\Sigma R}{K}=\frac{43}{6}\) = 7.167

UCL = X̿ + A_{2}R̄ = 13.583 +5.224 = 18.807

(i)

Conclusion: One of the sample points lie outside the control limits.

∴ The production process is out of statistical control

(ii) R chart:

Given K = 6, n = 5

Here standards are not given

∴ The control limits, are

CL = R̄ = \(\frac{\Sigma R}{K}=\frac{43}{6}\) = 7.16

LCL = D_{3}R̄ = 0

DCL = D_{4}R̄ = 2.282 × 7.169

= 16.355

In R-chart all the sample points lie inside the control limits. Therefore the production process is said to be in statistical control.

Question 30.

From the following data calculate consumer price index number for 2008 and 2009

Draw the control charts for mean and range.

Answer:

Given = 5, k = 9, X̿ = \(\frac{\sum \bar{x}}{k}\) = 24.289

Here the standards are not given

∴ The control limits for X̄ chart (mean) are

(i) CL = X̄ where X̄ = \(\frac{\Sigma \bar{X}}{K}=\frac{218.6}{9}\) = 24.28

(ii) LCL = X̿ – A_{2}R̄ where R̄ = \(\frac{\Sigma R}{K}=\frac{34}{9}\) = 3.78

LCL = 24.289 – (0.577 × 3.78) = 24.289 – 2.18106 = 22.1076

(Sample size is 5 ∴ A_{2} = 0.577 from chart)

(iii) UCL = X̿ + A_{2}R̄ = 24.289 + 2.18106 = 26.47

Conclusion: Two of the sample points lie out side the control limits.

∴ The production process is out of statistical control.

(ii) R chart

Here standards are not given .

∴ Control limits are

(a) CL = R̄ = \(\frac{\Sigma R}{K}=\frac{34}{9}\) = 3.78

(b) LCL = D_{3}R̄ = 0

(from table for n = 5, D_{3} = 0, D_{4} = 2.115)

(c) UCL = D_{4}R̄ = 2.115 × 3.78 = 7.99

Conclusion: All of the sample points lie inside the control limits. Therefore production process is under statistical control.

Question 31.

In a production process 8 samples of size 4 are collected. Their means and ranges are given below. Construct x – R charts.

Answer:

Given n = 4, K = 8,

Here standards are not given

CL for X̄ chart

(i) CL = X̿ where X̿ = \(\frac{\Sigma \bar{X}}{K}=\frac{101}{8}\) = 12.625

where R̄ = \(\frac{\Sigma R}{K}=\frac{25}{8}\) = 3.125 K 8

(from table A_{2} = .729 for n = 4)

(ii) LCL = X̿ – A_{2}R̄ (From Chart A, = .729)

= 12.625 – (0.729 × 3.125)

= 12.625 – 2.278 = 10.347

(iii) UCL = X̿ + A_{2}R̄

= 12.625 + 2.278 = 14.903

Conclusion: One of the sample point falls out side the control limits. Therefore production process is out is statistical control

R – chart

The control limits are

(i) CL = R̄ =3.125

(ii) LCL = D_{3}R̄ = 0 × 3.125 = 0

(From table for n = 4, D_{3} = 0, D_{4} = 2.282)

(iii) UCL = D_{4}R̄ = 2.282 × 3.125 = 7.1312

Conclusion: All of the sample points lie within the control limits.

∴ The production process is in statistical control.

Question 32.

The following table gives the number of defectives found during inspection of 8 samples of size 100 each. Find the suitable control limits.

Answer:

Given n = 100, k = 8

Here standards are not given

∴ The control limits are

(i) CL = np̄ where p̄ = \(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}=\frac{12}{100 \times 8}\) = 0.015

∴ q̄ = 1 – p̄ = 1 – 0.015 = 0.985

∴ LCL = (100) (0:015) = 1.5

LCL = np̄ – 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\)

= 1.5 – 3\(\sqrt{(1.5)(0.985)}\)

= 1.5 – 3.6465

= -2.1465 = 0

UCL = np̄ + 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\)

= 1.5 + 3\(\sqrt{(1.5)(0.985)}\)

= 5.1465

Question 33.

10 samples each of size 5 were inspected from the number of defectives and the following data is obtained. Given P1 = 0.32 draw a suitable control chart and comment.

Answer:

Given n = 5, K = 10, P’ – 0.32, Q’ = 1 – P’ = 1- 0.32 = 0.68

Here standards are given

∴ The control limits are

(i) CL = nP’ = 5 × 0.32 =1.6

(ii) LCL = nP’ – 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\)= 5 × 0.32 – 3\(\sqrt{5 \times 0.32 \times 0.68}\)

= 1.6 – 3.129 = -1.529 = 0

(iii) UCL = nP’+ 3\(\sqrt{\mathrm{nP}^{\prime} \mathrm{Q}^{\prime}}\) = 5 × 0.32 + 3\(\sqrt{5 \times 0.32 \times 0.68}\)

= 1.6 + 3.129 = 4.729

Conclusion: All of the sample points lie on or within the control limits.

∴ The production process is in statistical control.

Question 34.

In a textile mill cloth is inspected at regular intervals for weaving defects. The results are recorded as below.

(a) Of on an average 0.7 defects are expected per square meters, draw c-chart and interpret.

(b) In the absence of given standard, analyze using c-chart.

Answer:

(i) Given λ’ = 0.7 (Here standard is given)

∴ The control limits are

(i) CL = λ’ = 0.7

CL = λ’ – 3\(\sqrt{\lambda^{\prime}}\) = 0.7 – 3\(\sqrt{0.7}\) = 0.7 – 3(0.8367)

(ii) ⇒ 0.7 – 2.5099 = -1.809 = 0

(iii) UCL = λ’ + 3\(\sqrt{\lambda^{\prime}}\) = 0.7 – 3\(\sqrt{0.7}\)

= 0.7 + 2.5099 = 3.2099

Conclusion: Three of the sample points lie outside the control limits.

∴ The production process is out of statistical control.

(ii) In the absence of given standards

The control limits are

(i) CL = c̄ = \(\frac{\Sigma c}{k}=\frac{15}{10}\) = 1.5

LCL = c̄ – 3\(\sqrt{\bar{c}}\) = 1.5 – 3\(\sqrt{1.5}\)

(ii) ⇒ 1.5 – 3(1.2247) ⇒ 1.5 – 3.67 = -2.1741 = 0

(iii) UCL = c̄ + 3\(\sqrt{\bar{c}}\) = 1.5 + 3\(\sqrt{1.5}\) ⇒ 15 + 3.67 ⇒ 5.178

Conclusion: All of the sample points lie inside the control limits. Therefore production process is under statistical control.

Question 35.

From a razor blade production process, the following data regarding number of defective blades in a sample containing 100 razor blades each is obtained. Draw np-chart. Is the process under control?

Answer:

Given n = 100, k = 10

Here standard is not known

CL for np -chart

(0 CL = np̄ = where p̄ =\(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}=\frac{150}{100 \times 10}\) = 0.15

∴ q̄ = 1 – p̄ = 1 – 0.15 = 0.85

∴CL = 100 × 0.15 = 15

LCL= np̄ – 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\)

(ii) = 100 × 0.15 – 3\(\sqrt{100 \times 0.15 \times 0.85}\)

⇒ 15 – 3\(\sqrt{12.75}\) ⇒ 15 – 10.71 = 4.288

(iii) UCL = np̄ + 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\) = 15 + 10.712 = 25.712

Conclusion: One of the sample point lie outside the control limit

∴ The production process is out of statistical control.

Question 36.

Glass tables are manufactured by a firm. A crack or a bubble or an improper spread of tint is considered as defect. The number of defects in randomly selected glass tables is noted below. Draw control chart for defects and analyze the data.

Answer:

Given, K = 12

Here standard is not known

∴ The control limits for c chart are

CL for c-chart

(i) CL = pc̄ = where c̄ = \(\frac{\Sigma c}{K}=\frac{20}{12}\) =1.667

LCL = c̄ – 3√c̄ = 1.667 – 3\(\sqrt{1.667}\)

(ii) ⇒ 1.667 – 3(1.291) ⇒ 1.667 – 3.8734

= – 2.206 = 0

(iii) UCL = c̄ + 3√c̄ = 1.667 + 3\(\sqrt{1.667}\)

= 1.667 + 3.8734 = 5.54

Conclusion: All of the sample points lie on or within the control limits.

∴ The production process is in statistical control.

Question 37.

Following are the defects found in the assembly of ten IC’s. (In an electrician board).

2, 3, 0,1, 2, 5, 0, 2,1, 3

(i) Draw a C-chart and state whether the process is in control

(ii) If on an average 1 defect is expected, draw c-chart and conclude.

Answer:

(i) Given K = 10, Here standard is not known

∴ The CL’s for c chart are

(a) CL = c̄ = where c̄ = \(\frac{\Sigma c}{k}=\frac{19}{10}\) = 1.9

LCL = c̄ – 3√c̄ = 1.9 – 3\(\sqrt{1.9}\)

⇒ 1.9 – 3(1.378) ⇒ 1.9 – 4.1352

(b) = -2.2352 = 0

(iii) UCL = c̄ + 3√c̄ = 1.9 + 3\(\sqrt{1.9}\) ⇒ 1.9 + 4.1352 = 6.03

Conclusion: All of the sample points lie on or within the control limits.

∴ The production process is in statistical control.

(i) Given λ’ = 1 (Here standard is Known)

∴ The control limit are CL = λ’ = 1

(ii) LCL = λ’ – 3\(\sqrt{\lambda^{\prime}}\)

= 1 – 3√1

LCL = -2 = 0

UCL = λ’ + 3\(\sqrt{\lambda^{\prime}}\) = 1 + 3√1 UCL = 4

Conclusion: One of the sample point lie outside the control limit.

∴ The production process is out of statistical control.

### 2nd PUC Statistics Statistical Quality Control : I (X̄ and R – Charts) Exercise Problems

Question 1.

If X̄^{1} = 4.5, σ^{1} = 1.5 for sample size of n = 4 write down 3 σ – control limits for drawing x̄ – chart.

Answer:

Given : X̄^{1} = 4.5, σ^{1} = 1.5, n = 4- standards are known.

The 3 σ – control limits are ; c.L = X̄^{1} = 4.5

L.C.L = X̄^{1} – Aσ^{1} = 4.5 – 1.5(1.5) = 2.25

U.C.L = X̄^{1} + Aσ^{1} =4.5 + 1.5(1.5) = 6.75

Question 2.

If X̿ = 40, R̄ = 2.5 for sample size of n = 5 find 3σ – control limits for drawing x̄ – chart.

Answer:

Given: X̿ = 40, R̄ = 2.5, n = 5, standards are not known

The 3σ – control limits for x̄ – chart are : c.L = X̿ = 40

L.C.L = X̿ – A_{2} R̄ = 40 – 0.577(2.5) = 38.5575

U.C.L = X̿ + A_{2}R̄ = 40 + 0.577(2.5) = 41.4425

Question 3.

If the average range for the twenty five samples of 5 each is 0.29 ml (computed) and the average mean of the observation is 15.95 ml, develop three – sigma control limits for the packaging operation.

Answer:

Given: k = 25; n = 5, R̄ = 0.29, X̿ = 15.95,

Standards not known:

The 3 σ – control limits for x̄ – chart are; C.L = X̿ = 15.95

L.C.L = X̿ – A_{2}R̄ = 15.95 – 0.577(0.29) = 15.7827

U.C.L = X + A_{2}R̄ = 15.95 + 0.577(0.29) = 16.1173

Question 4.

A Quality control inspector has taken 4 samples with 5 observations each, measuring the volume of Potato chips per bag. If the average of ranges for the 4 samples is 0.2 gms. And the average of means of the observations is 12.5 gms. Develop three sigma control limits for the x – chart for packing operation.

Answer:

Given: n = 4, k = 5; R̄ = 0.2, X̿ = 12.5

Here standards are not known, the control limits for x̄ – Chart are:

C.L = X̿ = 12.5,

L.C.L = X̿ – A_{2}R̄ = 12.5 – (0.729 × 0.2) = 12.3542

U.C.L = X̿ + A_{2}R̄ = 12.5 + 0.729 × 0.2 = 12.6458

Question 5.

If the standards are known to be X̄^{1} = 20 and σ^{1} = 6, construct x̄ – chart for the following data (sample size n = 5 and A = 1.342).

Answer:

Given: X̄^{1} =20, σ^{1} = 6, A = 1.342, standards are known,

the control limits for x̄ – chart are: C.L = X̄^{1} =20,

L.C.L = X̄^{1} – Aσ^{1} = 20 – (1.342 × 6) = 11.948 = 11.95

U.C.L = X̄^{1} + Aσ^{1} = 20 + (1.342 × 6) = 28.052 = 28

Conclusion: All the sample means lies with in control limits, so, the production process is under control.

Question 6.

Construct x̄ and R – chart for the following data with sample size 4 and comment.

Answer:

Given : n = 4, standards not known ; we need,

X̿ = \(\frac{\Sigma \bar{x}}{k}=\frac{400}{8}\) = 50 and R̄ = \(\frac{\Sigma R}{K}=\frac{80}{8}\) = 10

The control limits for x̄ – charts are : CL = X̿ = 50

L.C.L = X̿ – A_{2}R̄ = 50 – 0.729 × 10 = 42.71 = 42.7

U.C.L = X̿ + A_{2}R̄ = 50 + 0.729 × 10 = 57.29 = 57.3

Comment : From x̄ – chart all sample points lies within the control limits and so, the production process is under control.

For R – chart: The control limits are ;

CL = R̄ = 10 L.C.L = D_{3}R̄ = 0 × 10 = 0

U.C.L = D_{4}R̄ = 2.282 × 10 = 22.82 = 22.8

R- CHART

Comment: From R – chart all sample points lies within the control limits, so the production process under control.

Question 7.

The following table gives means (x̄) and Ranges (R) of 6 samples of size 5 each.

Find the control limits for drawing x chart and R-chart.

Answer:

Given : n = 5, k = 8, standards not known, we need

The control limits for x̄ – chart are

C.L = x̄ = 6.565 = 6.6

L.C.L = X̿ – A_{2}R̄ = 6.565 – (0.577 × 3.875)

= 4.329 = 4.3

U.C.L = X̿ + A_{2}R̄ = 6.565 + (0.577 × 3.875) = 8.8

The control limits for R̄ – chart are

C.L = R̄ = 3.875

L.C.L = D_{3}R̄ = 0 × 3.875 = 0

U.C.L = D_{4}R̄ = 2.115 × 3.875 = 8.1950

Question 8.

Given n = 4 and R_{1} : 14, 8, 11, 9, 12 and 8. Find the control limits for R – chart.

Answer:

Given: n = 4, standard not known, we need :

R̄ = \(\frac{\Sigma R_{1}}{K}=\frac{62}{6}\) = 10.33,

The control limits for R – chart are

C.L = R̄ = 10.33 ,

L.C.L = D_{3}R̄ = 0 × 10.33 = 0

U.C.L = D_{4}R̄ = 2.282 × 10.33 = 25.573.

Question 9.

Draw R – chart for the following data and give your conclusion.

R : 6, 3, 8, 4,1, 2, 5, 7 and n = 5.

Answer:

For R – chart standard is not known, we need :

R̄ = \(\frac{\Sigma R_{1}}{k}=\frac{36}{8}\) = 4.5

The control limits are : C.L = R̄ = 4.5

L.C.L = D_{3}R̄ = 0 × 4.5 = 0

U.C.L = D_{4}R̄ = 2.155 × 4.5 = 9.5175 = 9.5

R- CHART

Comment : From R – chart all sample points lies within the control limits, so the production process is under control

Question 10.

Given D_{3} = 0, D_{4} = 2.115 and R̄ = 4, draw R – chart for the following data.

Answer:

Given: R = 4,D_{3 }= 0, D_{4} =2.115 ; standard is not known

The control limits are :

C.L = R̄ = 4,

L.C.L = D_{3}R̄ = 0 × 4 = 0;

U.C.L = D_{4}R̄ = 2.115 × 4 = 8.46

R- CHART

Comment: All sample points lies within the control limits, so the production process is under control.

Question 11.

A quality control inspector of a Nature’s Shampoo Company has made 4 observations of 4 bottle each to check the volume of shampoo in each bottle. The data collected by the inspector are shown below:

If the standards of the shampoo bottle filling operation is not known, use the information in the table to develop control limits X̄ and R – charts for the filling operation.

Answer:

Given : n = 4, k = 4

From the data sample means and Ranges are obtained as below.

From observation 1:

R_{1} = (Highest value – Lowest value) = (19.9 -19.5)

= 0.4, From the observations 1,2,3 and 4

R_{2} = (20.6 – 18.7) = 1.9 ; R_{3}= 21.6 – 18.9 =2.7 ;

R_{4} = 20.8 – 20 = 0.8.

∴ R̄ = \(\frac{\Sigma R s}{k}=\frac{0.4+1.9+2.7+0.8}{4}=\frac{5.8}{4}\)

∴ R̄ = 1.45

The control limits for x̄ – chart are : C.L = X̿ = 19.99

L.C.L = X̿ – A_{2}R̄ = 19.99 – (0.729 × 1.45) = 18.9329

U.C.L = X̿ + A_{2}R̄ = 19.99 +(0.729 × 1.45) = 21.047

Control limits for R – chart are;

C.L = R̄ = 1.45; L.C.L = D_{3}R̄ = 0 × 1.45 = 0.

U.C.L = D_{4}R̄ = 2.28 × 1.45 = 3.3089

### 2nd PUC Statistics Statistical Quality Control : II (NP And C – charts) Exercise Problems

Question 1.

If P^{1} = 0.02 and n = 100, calculate the control limits for d – chart (np – chart).

Answer:

Standard is known :

Given : p^{1} = 0.02 n – 100

The control limits for np/d – chart are :

C.L = np^{1} = 0.02 × 100 = 2,

L.C.L = np^{1} – 3\(\sqrt{n p^{1} Q^{1}}\) Here Q^{1} = 1 – p^{1} = 1 – 0.02= 0.98

= 100 × 0.02 – 3\(\sqrt{100 \times 0.02 \times 0.98}\)

= 2 – 3(1.4) = -2.2 = 0 Taken as ‘0’

U.C.L = np1 +3\(\sqrt{n p^{1} Q^{1}}\)

= 100×0.02 + 3\(\sqrt{100 \times 0.02 \times 0.98}\)

= 2 + 3(1.4) = 6.2

Question 2.

In a fish net manufacturing process the proportion of defectives p = 0.01 of the process control is based on samples of size 100 each, find the control limits for np – chart.

Ansewr:

p^{1} = 0.01

∴ Q^{1} = 1 – p^{1} = 1 – 0.01 = 0.99 and

n = 100, standard is known,

The control limits for d – chart are:

C.L = nP^{1} = 100 x 0.01 = 1,

L.C.L = nP^{1} + 3\(\sqrt{n p^{1} Q^{1}}\)

= 100 × 0.01 – 3\(\sqrt{100 \times 0.01 \times 0.99}\) = 3.9845

Question 3.

Calculate the control limits for d – chart given p̄ = 0.05 and sample size of 50

Answer:

p̄ = 0.05, n = 50; q̄ = 1 – p̄ = 1 – 0.05 = 0.95

Standard is not known. The control limits for np/d – chart are

C.L = np/ = 50 × 0.05=2.5

L.C.L = n̄p̄/ – 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\) = 50 × 0.05 – 3\(\sqrt{50 \times 0.05 \times 0.95}\)

= 2.5 – 32 (1.5411) = -2.1233 = 0

U.C.L = np /+ 3\(\sqrt{\mathrm{n} \overline{\mathrm{p}} \overline{\mathrm{q}}}\) = 50 x 0.05 + 3\(\sqrt{50 \times 0.05 \times 0.95}\) = 7.1233

Question 4.

A company manufactures flooring tiles. Samples of 100 tiles each are drawn at regular intervals. The number of defective tiles are given below.

Obtain the control limits for the above data.

Answer:

n = 100, k = 10 Standard is not known. We need 28

p̄ = \(\frac{\Sigma d}{n k}=\frac{28}{100 \times 10}\) = 0.028;

q̄ = 1 – p̄ = 1 – 0.028 = 0.972

The control limits for d- chart are

C.L = np̄ = 100 × 0.028 = 2.8.

L.C.L = np̄ – 3\(\sqrt{n \overline{p q}}\) =100 × 0.025 – 3\(\sqrt{100 \times 0.028 \times 0.972}\)

= 2.8 – 3 (1.6497) = -2.1491 = 0

U.C.L = np̄ + 3\(\sqrt{n \overline{p q}}\) =100 × 0.028 – 3\(\sqrt{100 \times 0.028 \times 0.972}\) = 7.7491

Question 5.

Ten samples of size 50 each were inspected and the number of defectives in each of them were as follows :

Obtain the control limits for the above data.

Answer:

k = 10, n = 50, Standard is not known, we need

p̄ = \(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}=\frac{15}{50 \times 100}\) = 0.03

∴ q̄ = 1 – p̄ = 1 – 0.03 = 0.97 ;

The control limits for d – chart are :

C.L = np̄ = 50 × 0.03 = 15

L.C.L = np̄ – 3\(\sqrt{n \overline{p q}}\) = 50 × 0.03 – 3\(\sqrt{50 \times 0.03 \times 0.97}\)

= 1.5 – 3 (1.2062) = -2.1186 = 0

U.C.L = np̄ + 3\(\sqrt{n \overline{p q}}\)

= 1.5 + 3 (1.2062) = 5.1186

Question 6.

Ten samples of 100 P.V.C. pipes manufactured by a firm are inspected for the number of defectives. The number of pipes having defectives are noted as: 2,1, 3, 0, 2, 2,4, 4, 5, 6. Calculate the control limits for np – chart.

Answer:

No. of defective pipes; d/np – chart. Standard is not known,

Given : n = 100, k = 10 we need

p̄ = \(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}=\frac{28}{100 \times 10}\) = 0.028;

∴ q̄ = 1 – p̄ = 1 – 0.029 = 0.971

The control limits for d- chart are:

C.L = np̄ = 100 × 0.029 = 2.9

L.C.L = np̄ – 3\(\sqrt{n \overline{p q}}\)

= 100 × 0.029 – 3\(\sqrt{100 \times .029 \times 0.971}\)

= 2.9 – 3 (1.678) = -2.134

U.C.L = np̄ + 3\(\sqrt{n \overline{p q}}\)

= 100 × 0.029 + 3\(\sqrt{100 \times .029 \times 0.971}\) = 7.9342

Question 7.

A production manager at a tire manufacturing plant has inspected the number of defective tires in ten random samples with twenty observations each. Following are the number of defective tires found in each sample.

Draw a control chart for defectives, when 0.042 fraction defectives are expected.

Answer:

k = 9, n = 120, Expected fraction defectives;

d- chart, standard is known : p^{1} = 0.042 and

Q^{1} = 1 – p^{1} = 1 – 0.042 = 0.958.

The control limits for d- chart are:

C.L = nP^{1} = 120 × 0.042 = 5.04. = 5

L.C.L = nP^{1} – 3\(\sqrt{\mathrm{np}^{1} \mathrm{Q}^{1}}\)

=120 × 0.042 – 3\(\sqrt{120 \times 0.042 \times 0.958}\)

=5.04 – 3 (2.1973) = -1.5519 = 0

U.C.L = nP^{1} + 3\(\sqrt{\mathrm{np}^{1} \mathrm{Q}^{1}}\)

= 120 × 0.042 + 3\(\sqrt{120 \times 0.042 \times 0.958}\)

= 11.632 = 11.6

Comment : Some of the sample point lies outside the control limits, so the production process is not under control.

Question 8.

In a printing industry at regular intervals, cloth is inspected for defects in printing. If on an average 0.5 defects are expected per square meters, obtain suitable control limits

Answer:

λ^{1} = 0.5

Standard is known:

The control limits for defects i.e., C – chart are : C.L = λ^{1} = 0.5

L.C.L = λ^{1} – 3\(\sqrt{\lambda^{1}}\) = 0.5 – 3\(\sqrt{0.5}\)

= 0.5 – 3 (0.7071) = -1.6213 = 0 (Taken as‘O’)

U.C.L = λ^{1} + 3\(\sqrt{\lambda^{1}}\) = 0.5 – 3\(\sqrt{0.5}\) = 2.6213

Question 9.

During an examination of equal length of cloth, the following are the number of defects observed as : 2,3, 4, 0,5, 6, 7,4, 3, 2. Calculate control limits for C – chart.

Answer:

Defects are given so use C – chart, standard is not known,

we need c̄ = \(\frac{\Sigma c}{k}\) = 3.6 The control limits are: k

C.L = C̄ = 3.6,

L.C.L = C̄ – 3√C̄ = 3.6 – 3\(\sqrt{3.6}\) = -2.0921;

U.C.L = C̄ + 3√C̄ =3.6 – 3\(\sqrt{3.6}\) = 9.292

Question 10.

Twenty pieces of cloth out of different rolls contained respectively 1, 4, 3, 2, 5, 4, 6,7, 2, 5, 7, 6, 4, 5, 1, 2, 3, 8 imperfections. Calculate control limits for suitable control chart.

Answer:

Here Imperfections – means defects, use c- chart. Standard is not known;

We need C̄ = \(\frac{\Sigma c}{k}=\frac{80}{20}\) = 4.

The control limits are : C.L = C̄ = 4

L.C.L = C̄ – 3√C̄ = 4.3 – \(\sqrt{4}\) = -2 = 0 (Taken as

U.C.L = C̄ + 3√C̄ = 4 + 3\(\sqrt{4}\) = 10

Question 11.

The number of defects are monitored at a large motor car painting company. The painting defects on the painted chassis have been recorded as below. Using a c- chart develop three – sigma control limits:

Answer:

Here Defects are given so use c- chart. Standard is not known,

we need,

c̄ = \(\frac{\Sigma c}{k}=\frac{24}{10}\) = 2.4

The control limits for c- chart are:

C.L = c̄ = 2.4; L.C.L = c̄ = 3

√C̄ = 2.4 – 3√2.4 = -2.2475 = 0 (Taken as ‘0’)

U.C.L = c̄ + 3√C̄ = 2.4 + 3√2.4 = 7.0476