ISS (Statistical Services) Statistics Paper IV: Questions 18  21 of 92
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Question number: 18
» Operations Research and Reliability » Finding Solutions in 22, 2xm and Mxn Games
Appeared in Year: 2015
Describe in Detail
Prove that the number of basic variables is a balanced transportation problem is at most , where m is the number of origins and n is the number of destinations.
Explanation
First note that there are in all m + n constraints. We shall show that one of these is redundant so that there are in effect equation in mn variables and hence at most basic variable are there. We note
Summing the m cons
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Question number: 19
» Statistical Quality Control » Control Charts » Variable
Describe in Detail
What are control charts for variables? Find 3 σ control limits on the chart. If we are given that μ=15.95 σ=0.14 and n = 4 construct control limits for ?
Explanation
Control chart for variables are used to monitor characteristics that can be measured, e. g. length, weight, diameter, time etc. When a record is made of an actual measured quality characteristic like dimension expressed as hundredths of a millimeter, the quality is expressed by variables. For quality control of such data, two types of control chart
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Question number: 20
» Statistical Quality Control » Control Charts » Attributes
Describe in Detail
The following are figures of defective bulbs in 10 lots each containing 30 bulbs. Construct 3σ control limits for the pchart.
Sample  No. of defective (d) 
1  1 
2  3 
3  3 
4  1 
5  0 
6  5 
7  1 
8  1 
9  1 
10  1 
Explanation
We have n = 30
Sample  No. of defective (d) 

1  1  0.033 
2  3  0.100 
3  3  0.100 
4  1  0.033 
5  0  0.000 
6  5  0.167 
7  1  0.033 
8  1  0.033 
9  1  0.033 
10  1  0.033 
Total  0.567 
Control limits for charts
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Question number: 21
» Statistical Quality Control » Control Charts » Variable
Describe in Detail
Obtain the control limits for for the following data of 25 samples with 4 observations on the basis of volume of the bottles filled.
Sample No.  Observations  Sample No.  Observations  
1  15.85  16.02  15.83  15.93  13  15.85  15.76  15.74  15.98  
2  16.12  16.00  15.85  16.01  14  15.73  15.84  15.96  16.10  
3  16.00  15.91  15.94  15.83  15  16.20  16.01  16.10  15.89  
4  16.20  15.85  15.74  15.93  16  16.12  16.08  15.83  15.94  
5  15.74  15.86  16.21  16.10  17  16.01  15.93  15.81  15.68  
6  15.94  16.01  16.14  16.03  18  15.78  16.04  16.11  16.12  
7  15.75  16.21  16.01  15.86  19  15.84  15.92  16.05  16.12  
8  15.82  15.94  16.02  15.94  20  15.92  16.09  16.12  15.93  
9  16.04  15.98  15.83  15.98  21  16.11  16.02  16.00  15.88  
10  15.64  15.86  15.94  15.89  22  15.98  15.82  15.89  15.89  
11  16.11  16.00  16.01  15.82  23  16.05  15.73  15.73  15.93  
12  15.72  15.85  16.12  16.15  24  16.01  16.01  15.89  15.86  
25  16.08  15.78  15.92  15.98 
Explanation
From the above data we get
From the tables, for n = 4, we have
Control limits for :
And
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