ISS (Statistical Services) Statistics Paper IV: Questions 28  30 of 92
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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  280  0.14 
8  306  0.153 
9  337  0.1685 
10  305  0.1525 
11  356  0.178 
12  402  0.201 
13  216  0.108 
14  264  0.132 
15  126  0.063 
16  409  0.2045 
… (243 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 
7  4  0.00267 
Total  41  0.02733 
Control limits for charts
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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 z
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