Answer:
The answer is 0.002341 and 0.000000
Explanation
From the question stated we recall the following:
Let us find the lower and upper and control limits for their chart control if their mean sample mirrors their historical process average?
Now,
The number of sample size n =5000
The number of sample k =10
The total number of observations = n x k = 5000 x 10 = 50000
The proportion defective displays p = 0.1% which is =0.001
The standard deviation, Sp = √p (1-p)/n = √0.001 x (1-0.001)/5000 =0.000447
The Upper control limit is UCL = p+3 x Sp =0.001+3 x 0.000447 =0.002341
The Lower control limit is UCL = p -3 x Sp = 0.001 - 3 x 0.000447 = -0.000341 which is 0
Therefore the LCL is 0 which is seen as negative