ISS (Statistical Services) Statistics Paper IV: Questions 28  30 of 79
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 79 questions. Access all new questions we will add tracking exampattern and syllabus changes. View Sample Explanation or View Features.
Rs. 300.00 or
Question number: 28
» Statistical Quality Control » Control Charts » Attributes
Describe in Detail
The following are the figures of defectives in 22 lots each containing 2, 000 bulbs. Ind control limit for pChart.
425

430

216

341

225

322

280

306

337

305

356

402

216

264

126

409

193

326

280

389

451

420 
Explanation
We have n = 2000
S. No.

d


1

425

0.2125

2

430

0.215

3

216

0.108

4

341

0.1705

5

225

0.1125

6

322

0.161

7… (105 more words) … 
Question number: 29
» Statistical Quality Control » Control Charts » Attributes
Describe in Detail
Each day a sample of 1500 items from a production process was examined. The number of defectives found in each sample was as follows. With a suitable control chart, check out for control 4, 6, 6, 2, 15, 4, 4
Explanation
We have n = 30
S. No.

d


1

4

0.00267

2

6

0.00400

3

6

0.00400

4

2

0.00133

5

15

0.01000

6

4

0.00267… (61 more words) … 
Question number: 30
» Statistical Quality Control » Control Charts » Variable
Describe in Detail
a) The number of customer complaints are recorded at a television manufacturing company. Complaints have been recorded over the past 20 weeks. Construct 3σ control limits for the cchart.
Week

No. of complaints

Week

No. of complaints

1

0

11

4

2

3

12

3

3

4

13

1

4

1

14

1

5

0

15

1

6

0

16

0

7

3

17

2

8

1

18

1

9

1

19

2

10

0

20

2

b) A machine is manufacturing mica discs. samples of size 5 are drawn every hour and their thickness in units are recorded as follows. From this data construct 3σ control limits for chart.
Sample No.

Observstions
 
1

8

9

15

4

11

2

7

10

7

6

8

3

11

12

10

9

10

4

12

8

6

9

12

5

11

10

6

14

11

6

7

7

10

4

11

7

10

7

4

10

10

8

8

11

11

7

7

9

8

11

8

14

12

10

12

9

12

17

11

11

7

7

9

17

13

12

9

9

4

4

11

13

10

12

12

12

12

14

8

11

9

6

8

15

10

13

9

4

9

16

9

11

8

5

11 
Explanation
a) Average number of defects per sample is
=1.5+ =1.5 + 3 (1.22) =5.174
= →0
Since number of defects cannot be negative, can be considered as zero. (rounded up to zero)
b) Second question: