In: Statistics and Probability
| 
 Day  | 
 1  | 
 2  | 
 3  | 
 4  | 
 5  | 
 6  | 
 7  | 
 8  | 
 9  | 
 10  | 
 11  | 
 12  | 
 13  | 
 14  | 
 15  | 
| 
 Number of Aides Absent  | 
 5  | 
 8  | 
 11  | 
 15  | 
 4  | 
 2  | 
 7  | 
 1  | 
 4  | 
 6  | 
 14  | 
 19  | 
 3  | 
 5  | 
 8  | 
Solution:
Pbar can be calculated as
Pbar = Total no. of absenteeism/Total no. of sample =
(5+8+11+15+15+4+2+7+1+4+6+14+19+3+5+8)/1500 = 112/1500 =
0.0747
UCL ca be calculated as
Pbar + 3*sqrt(Pbar*(1-pbar)/n) = 0.0747 +
3*sqrt(0.0747*(1-0.0747)/100) = 0.1536
So its answer is B. i.e. 0.1536
LCL can be calculated as
Pbar - 3*sqrt(Pbar*(1-pbar)/n) = 0.0747 -
3*sqrt(0.0747*(1-0.0747)/100) = 0
So its answer is A. i.e. o to 0.0030
| 
 Day  | 
 No. Of Absent  | 
 Proportion Of Absentism (No. Of Absent/100)  | 
| 
 1  | 
 5  | 
 0.05  | 
| 
 2  | 
 8  | 
 0.08  | 
| 
 3  | 
 11  | 
 0.11  | 
| 
 4  | 
 15  | 
 0.15  | 
| 
 5  | 
 4  | 
 0.04  | 
| 
 6  | 
 2  | 
 0.02  | 
| 
 7  | 
 7  | 
 0.07  | 
| 
 8  | 
 1  | 
 0.01  | 
| 
 9  | 
 4  | 
 0.04  | 
| 
 10  | 
 6  | 
 0.06  | 
| 
 11  | 
 14  | 
 0.14  | 
| 
 12  | 
 19  | 
 0.19  | 
| 
 13  | 
 3  | 
 0.03  | 
| 
 14  | 
 5  | 
 0.05  | 
| 
 15  | 
 8  | 
 0.08  | 
From the Proportion we can say that all absent proportion are
between 0 to 0.1536 except day12.
so we can say that absenteeism is our of statistical control
because you found one day out of the control limits.
So its answer is B.